3.10 Medical physics (A-level only) — revision question pack

19 specification points · notes, questions, answers and worked methods

Checked against AQA 7408 section 3.10. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

How this checking works

Answer all questions in the spaces provided.

3.10.1.1 · Physics of vision

Explanation

The eye is an optical refracting system. The cornea provides most refraction, while the lens changes power during accommodation so that a real, inverted image forms on the retina.
Ray diagrams should show refraction at the eye and convergence onto the retinal surface.
Cones work best in bright light, give colour vision and provide high spatial resolution because there is little convergence of their nerve pathways.
Rods are more sensitive at low light levels and many feed one nerve pathway, increasing sensitivity but reducing spatial resolution; rods do not distinguish colour.
The eye's sensitivity also depends on wavelength, so its spectral response changes with the active receptor cells and lighting conditions.

Parallel rays bend first at the cornea and again through the separate biconvex lens before converging on the retina.

Worked example

Explain why cone-rich vision gives greater spatial resolution than rod-rich vision.

1.Cone pathways show little convergence, so signals from nearby receptors remain distinguishable.
2.Many rods share a nerve pathway, so separate nearby stimuli may produce the same combined signal.
3.The reduced convergence of cone pathways therefore preserves more positional detail.

Answer: Cones give greater spatial resolution because their low pathway convergence keeps signals from neighbouring retinal positions separate.

Common mistakes

Don't state that the lens provides all of the eye's refraction and omit the cornea.
Don't describe rods as producing detailed colour vision rather than sensitive low-light vision.
Don't claim that convergence of many receptors onto one nerve pathway improves spatial resolution.

Exam tip

A compare question should link receptor wiring to performance: greater convergence increases sensitivity but reduces spatial resolution.

Tier 1 · Easy

State the type of retinal receptor that is mainly responsible for vision at very low light intensity and give one property of the resulting image.
[2 marks]
Total for this question: 2
State the retinal receptor used for detailed colour vision in bright light and whether neighbouring receptors usually share one nerve pathway.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain why two faint points of light that fall on nearby rod cells may be seen as one point even though both points are detected.
[3 marks]
Total for this question: 3
Explain how the eye keeps the image of a printed letter focused on the retina as the letter is moved closer.
[3 marks]
Total for this question: 3
A student's ray diagram for a distant object shows no refraction at the cornea and shows the eye lens forming an upright virtual image on the retina. Explain how the diagram should be corrected.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A coloured grid is viewed first in bright light and then after the illumination is greatly reduced. Explain the changes expected in colour, detail and sensitivity, referring to the retinal receptors and their nerve connections.
[5 marks]
Total for this question: 5
A faint star is detected more readily when the observer looks slightly to one side of the star rather than directly at it, but two close bright stars are resolved more clearly when viewed directly. Explain both observations using retinal receptors and their nerve pathways.
[5 marks]
Total for this question: 5
A retinal disorder leaves all rods and cones responsive but exchanges the positions of output pathways between neighbouring cones. Each pathway is exchanged only with a neighbouring cone of the same type, and rod pathways are unaffected. Deduce the results of tests using a large uniform coloured patch, a fine coloured grid in bright light and a faint grey patch in very low light.
[5 marks]
Total for this question: 5
In retinal region X, each cone has a separate nerve pathway. In region Y, signals from groups of 12 rods converge on one pathway. Each illuminated receptor produces a signal of $0.11\,\text{mV}$ , and a pathway fires only if its input reaches $0.90\,\text{mV}$ . Two faint point images each illuminate one cone in X. In Y, the two images fall within one 12-rod group and together illuminate nine of its rods. Deduce whether the points are detected and resolved in each region.
[6 marks]
Total for this question: 6
In a simplified model, the relative sensitivity at $510\,\text{nm}$ is taken to be 16 times that at $650\,\text{nm}$ . The threshold intensity at $510\,\text{nm}$ is $2.0\times10^{-9}\,\text{W m}^{-2}$ . Calculate the threshold intensity at $650\,\text{nm}$ . Explain why the threshold images have no reliable colour and poor spatial resolution, and why both properties improve when the illumination is greatly increased.
[5 marks]
Total for this question: 5

3.10.1.2 · Defects of vision and their correction using lenses

Explanation

Lens power is $P=1/f$ , with $f$ in metres and power in dioptres; a converging lens has positive power and a diverging lens negative power. Using the real-is-positive convention, $1/f=1/u+1/v$ and $m=v/u$ , so a virtual image distance is negative.
Myopia makes distant objects focus in front of the retina and is corrected by a diverging lens.
Hypermetropia prevents nearby objects focusing on the retina and is corrected by a converging lens.
Astigmatism gives different focusing in different planes and is corrected using a cylindrical component; a prescription states its cylindrical power and an axis from $0^\circ$ to $180^\circ$ .
Ray diagrams must show how the correcting lens changes convergence.

Worked example

A myopic eye has a far point $0.80\,\text{m}$ from the eye. Calculate the power of a correcting lens that makes a distant object appear at the far point.

1.For a distant object, $u=\infty$ , and the required virtual image has $v=-0.80\,\text{m}$ .
2.Use $1/f=1/u+1/v=0-1/0.80=-1.25\,\text{m}^{-1}$ .
3.Since $P=1/f$ , the correcting-lens power is $-1.25\,\text{D}$ .

Answer: A diverging lens of power $-1.25\,\text{D}$ is required.

Common mistakes

Don't use a positive image distance for the virtual image at the myopic eye's far point.
Don't calculate $P=1/f$ with focal length in centimetres rather than metres.
Don't correct hypermetropia with a diverging lens instead of a converging lens.

Exam tip

State the sign and lens type with a calculated prescription; a negative dioptre value must be identified as diverging.

Tier 1 · Easy

State the type of spectacle lens used to correct myopia and state the sign of its power.
[2 marks]
Total for this question: 2
State the lens feature used to correct astigmatism and the purpose of the axis stated on the prescription.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A spectacle lens of power $+6.25\,\text{D}$ corrects a hypermetropic eye. An object is $0.400\,\text{m}$ from the lens. Using the real-is-positive convention, calculate the image distance and the magnitude of the magnification.
[3 marks]
Total for this question: 3
A spectacle lens has power $-3.30\,\text{D}$ . Determine its focal length, including its sign, and state the defect of vision it corrects and the type of image it forms for a distant object.
[3 marks]
Total for this question: 3
A spectacle prescription is written as $-2.25\,\text{D}$ sphere with $+0.75\,\text{D}$ cylinder at axis $90^\circ$ . State which planes are corrected by the spherical and cylindrical components, and state what the axis specifies.
[2 marks]
Total for this question: 2

Tier 3 · Hard

A hypermetropic eye has a near point $0.80\,\text{m}$ from a spectacle lens. Determine the lens power needed so that an object $0.25\,\text{m}$ from the lens forms a virtual image at the near point. Use the real-is-positive convention.
[5 marks]
Total for this question: 5
A myopic eye can focus on objects or virtual images between $0.25\,\text{m}$ and $0.90\,\text{m}$ in front of it. A single spectacle lens close to the eye must allow both a distant road sign and a phone $0.40\,\text{m}$ away to be seen. Determine the lens power and show that the phone can still be focused.
[5 marks]
Total for this question: 5
Patient M is myopic and has a far point $0.65\,\text{m}$ from the eye. Patient H is hypermetropic and has a near point $0.90\,\text{m}$ from the eye. Spectacle lenses are close to their eyes. Determine the lens power required for M to view a distant object and the lens power required for H to view an object $0.30\,\text{m}$ away. Identify the lens type in each case.
[5 marks]
Total for this question: 5
A hypermetropic eye has an unaided near point $1.10\,\text{m}$ from the eye. A spectacle lens of power $+2.60\,\text{D}$ is close to the eye. Calculate the distance of the closest object that can be focused and the magnitude of the lens magnification for that object. State why the lens corrects hypermetropia.
[5 marks]
Total for this question: 5
A myopic eye has an unaided far point $0.460\,\text{m}$ in front of the cornea. Calculate the correcting power of a contact lens at the cornea and of a spectacle lens $16\,\text{mm}$ in front of the cornea. For both corrections, require parallel incident rays to appear to originate at the unaided far point. Explain the difference between the powers.
[5 marks]
Total for this question: 5

3.10.2.1 · Ear as a sound detection system

Explanation

The pinna and ear canal direct pressure variations in air to the tympanic membrane, which vibrates at the sound frequency.
The ossicles transmit this mechanical vibration through the middle ear and act as a lever system.
Because the oval window has a smaller area than the tympanic membrane, the transmitted force produces a larger pressure in the inner-ear fluid.
Pressure waves in the inner-ear fluid stimulate receptors, which produce electrical impulses in the auditory nerve.
A complete explanation follows the energy transfer in order: air pressure variation, membrane vibration, ossicle motion, fluid pressure wave, receptor stimulation and electrical nerve signal.

Worked example

Explain why the ossicles and oval window help transfer sound from air into inner-ear fluid.

1.The ossicles act as levers and transmit force from the tympanic membrane.
2.The oval window has a smaller area than the tympanic membrane.
3.For the transmitted force, the smaller area produces a greater pressure in the inner-ear fluid.

Answer: The lever action and area reduction increase fluid pressure, improving transmission into the inner ear.

Common mistakes

Don't send sound directly from the ear canal into the inner ear and omit the tympanic membrane and ossicles.
Don't say that the larger oval-window area increases pressure even though pressure is force divided by area.
Don't describe the auditory nerve signal as a sound wave rather than an electrical signal.

Exam tip

For a transmission question, give the full ordered chain and name the change from mechanical motion to electrical signalling.

Tier 1 · Easy

Name the membrane that vibrates when sound reaches the end of the ear canal and name the three-bone system moved by it.
[2 marks]
Total for this question: 2
The pressure amplitude at an eardrum of area $4.8\times10^{-5}\,\text{m}^2$ is $0.030\,\text{Pa}$ . Calculate the force on the eardrum.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain two features of the middle ear that increase the pressure delivered to the inner-ear fluid.
[3 marks]
Total for this question: 3
Compare the vibration at the oval window with the sound vibration at the tympanic membrane (eardrum), stating what happens to frequency, displacement amplitude and pressure.
[3 marks]
Total for this question: 3
Explain how a pressure variation arriving at the tympanic membrane produces an electrical signal in the auditory nerve.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A sound produces pressure amplitude $0.012\,\text{Pa}$ on a tympanic membrane of area $5.5\times10^{-5}\,\text{m}^2$ . The ossicles increase the force by a factor of $1.4$ and act on an oval window of area $3.2\times10^{-6}\,\text{m}^2$ . Calculate the pressure amplitude delivered to the inner-ear fluid.
[5 marks]
Total for this question: 5
A sound produces pressure amplitude $0.020\,\text{Pa}$ on a tympanic membrane of area $6.4\times10^{-5}\,\text{m}^2$ . The oval-window area is $3.4\times10^{-6}\,\text{m}^2$ and the required fluid pressure is $0.62\,\text{Pa}$ . Calculate the required ossicle force gain. State which contributes more to the pressure gain, the lever or the area reduction, and justify your answer.
[4 marks]
Total for this question: 4
A patient's tympanic membrane and inner-ear receptors work normally, but the ossicles have been removed and there is no mechanical connection between the tympanic membrane and the oval window. A loud tone in air still makes the tympanic membrane vibrate, while a vibrating tuning fork pressed against the skull still produces a weak auditory response. Explain both observations.
[4 marks]
Total for this question: 4
A middle-ear prosthesis restores the force amplitude applied by the ossicles to its healthy value, but its piston is larger than the normal oval-window contact area. Deduce the effect on the pressure amplitude in the inner-ear fluid and on the sound perceived by the patient. Explain the chain from the piston to the auditory nerve. The piston is in direct contact with the inner-ear fluid.
[4 marks]
Total for this question: 4
A powered middle-ear actuator bypasses the tympanic membrane and applies a sinusoidal force of amplitude $1.9\times10^{-6}\,\text{N}$ directly to an oval window of area $2.7\times10^{-6}\,\text{m}^2$ at $1.3\,\text{kHz}$ . Only $72\%$ of the force is transferred to the inner-ear fluid. Calculate the fluid pressure amplitude, state the frequency of the fluid vibration, and explain how this produces an auditory-nerve signal.
[5 marks]
Total for this question: 5

3.10.2.2 · Sensitivity and frequency response

Explanation

Sound intensity is power transferred per unit area, measured in $\text{W m}^{-2}$ ; an isotropic point source gives $I=P/(4\pi r^2)$ . Intensity level is $L=10\log_{10}(I/I_0)$ in decibels, where $I_0=1.0\times10^{-12}\,\text{W m}^{-2}$ .
The logarithmic scale reflects perception and compresses a very wide intensity range.
Equal-loudness curves show the level required at each frequency for the same perceived loudness; lower curves mean greater sensitivity, normally around a few kilohertz.
A relative level is $\Delta L=10\log_{10}(I_2/I_1)$ .
The dBA scale applies frequency weighting to approximate human response, whereas an unweighted dB value represents the physical intensity ratio alone.

Equal-loudness curves showing greatest hearing sensitivity at a few kilohertz.

Worked example

Calculate the intensity level for $I=1.0\times10^{-6}\,\text{W m}^{-2}$ .

1.Use $L=10\log_{10}(I/I_0)$ with $I_0=1.0\times10^{-12}\,\text{W m}^{-2}$ .
2.The ratio is $I/I_0=10^6$ .
3.Therefore $L=10\log_{10}(10^6)=60\,\text{dB}$ .

Answer: The intensity level is $60\,\text{dB}$ .

Common mistakes

Don't use $20\log_{10}$ for an intensity ratio instead of the required factor $10$ .
Don't read the highest point of an equal-loudness curve as the frequency of greatest sensitivity.
Don't treat dBA frequency weighting as identical to an unweighted dB measurement.

Exam tip

When comparing two sounds, use the intensity ratio directly in $\Delta L=10\log_{10}(I_2/I_1)$ .

Tier 1 · Easy

Calculate the intensity level of a sound with intensity $1.0\times10^{-8}\,\text{W m}^{-2}$ . Use $I_0=1.0\times10^{-12}\,\text{W m}^{-2}$ .
[2 marks]
Total for this question: 2
Calculate the sound intensity $2.0\,\text{m}$ from an isotropic source of acoustic power $0.040\,\text{W}$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

A sound detector reads an intensity level of $83\,\text{dB}$ . Its collecting area is $2.0\times10^{-3}\,\text{m}^2$ . Calculate the sound intensity at the detector and the sound power incident on it. Use $I_0=1.0\times10^{-12}\,\text{W m}^{-2}$ .
[3 marks]
Total for this question: 3
An observer moves from $2.5\,\text{m}$ to $10.0\,\text{m}$ from an isotropic sound source. Determine the change in intensity level.
[3 marks]
Total for this question: 3
Two pure tones each have an unweighted intensity level of $74\,\text{dB}$ . The dBA correction is $-16\,\text{dB}$ at $125\,\text{Hz}$ and $+1\,\text{dB}$ at $2.0\,\text{kHz}$ . Determine both dBA readings and explain what their difference represents.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An isotropic sound source has power $0.80\,\text{W}$ . Determine the distance at which its intensity level is $85\,\text{dB}$ . Then explain why a $12\,\text{kHz}$ tone at this level may sound quieter than a $3\,\text{kHz}$ tone at the same level. Use $I_0=1.0\times10^{-12}\,\text{W m}^{-2}$ .
[5 marks]
Total for this question: 5
One equal-loudness curve passes through $53\,\text{dB}$ at $200\,\text{Hz}$ and $40\,\text{dB}$ at $1.0\,\text{kHz}$ . Determine the ratio of the two sound intensities and hence the ratio of acoustic powers incident on equal areas. Explain what the result shows about the ear's sensitivity at these frequencies.
[4 marks]
Total for this question: 4
Before ear defenders are fitted, tones at $250\,\text{Hz}$ and $4.0\,\text{kHz}$ have levels of $92\,\text{dB}$ and $84\,\text{dB}$ respectively. The defenders transmit $8.0\%$ of the $250\,\text{Hz}$ intensity and $0.50\%$ of the $4.0\,\text{kHz}$ intensity. Over the range of levels involved here, equal loudness requires a $250\,\text{Hz}$ tone to have a level $15\,\text{dB}$ greater than a $4.0\,\text{kHz}$ tone. Determine which protected tone is perceived as louder.
[4 marks]
Total for this question: 4
At a worker's ear, two independent $250\,\text{Hz}$ machines produce unweighted intensity levels of $73\,\text{dB}$ and $79\,\text{dB}$ . Determine their combined intensity level. The dBA correction at this frequency is $-8\,\text{dB}$ ; calculate the combined dBA reading and decide whether it exceeds a stated limit of $75\,\text{dBA}$ .
[5 marks]
Total for this question: 5
An isotropic speaker A emits a $160\,\text{Hz}$ tone with acoustic power $0.12\,\text{W}$ and is $6.0\,\text{m}$ from a listener. Speaker B emits a $2.5\,\text{kHz}$ tone and is $9.0\,\text{m}$ away. At the relevant loudness, the $160\,\text{Hz}$ tone must have an intensity level $14\,\text{dB}$ greater than the $2.5\,\text{kHz}$ tone to sound equally loud. Determine the acoustic power required from B for equal perceived loudness.
[4 marks]
Total for this question: 4

3.10.2.3 · Defects of hearing

Explanation

Hearing loss means that a greater sound intensity level is needed to produce the same perceived loudness.
On an equal-loudness or threshold graph, the affected curve shifts upward over the frequencies where sensitivity has fallen.
The hearing loss at a stated frequency is the vertical difference in decibels between normal and impaired curves, not the horizontal frequency separation.
Deterioration with age commonly produces greater loss at high frequencies, while injury from prolonged excessive noise may produce a pronounced loss over a narrower frequency range, often near $4\,\text{kHz}$ .
The required account concerns the change in sensitivity and its representation on equal-loudness curves; unsupported detail about physiological damage is outside this specification point.

Worked example

At $4.0\,\text{kHz}$ , a normal threshold is $5\,\text{dB}$ and an impaired threshold is $35\,\text{dB}$ . Determine the hearing loss.

1.Read both levels at the same frequency.
2.Find the vertical difference: $35-5=30\,\text{dB}$ .

Answer: The hearing loss at $4.0\,\text{kHz}$ is $30\,\text{dB}$ .

Common mistakes

Don't measure a horizontal frequency difference instead of the vertical dB separation between curves.
Don't shift an impaired threshold curve downward even though more intensity is required for detection.
Don't give detailed ear physiology when the question asks for the change shown on equal-loudness curves.

Exam tip

State both the affected frequency range and the upward change in required intensity level when interpreting hearing loss.

Tier 1 · Easy

State the frequency range in which age-related hearing deterioration is usually greatest.
[1 mark]
Total for this question: 1
State the likely cause of a pronounced hearing-loss maximum close to $4\,\text{kHz}$ .
[1 mark]
Total for this question: 1

Tier 2 · Standard

At $4.0\,\text{kHz}$ , a normal threshold is $8\,\text{dB}$ and an impaired threshold is $38\,\text{dB}$ . Calculate the hearing loss and the factor by which the threshold intensity has increased.
[3 marks]
Total for this question: 3
At one frequency, an impaired hearing threshold is $400$ times the normal threshold intensity. Calculate the hearing loss in decibels and state how the impaired equal-loudness curve compares with the normal curve at this frequency.
[3 marks]
Total for this question: 3
An older listener has a much greater hearing loss at high frequencies than at low frequencies. Explain why increasing every frequency component of recorded speech by the same intensity level does not restore its normal perceived balance.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Listener A has hearing losses of $6\,\text{dB}$ at $0.5\,\text{kHz}$ , $15\,\text{dB}$ at $4\,\text{kHz}$ and $44\,\text{dB}$ at $10\,\text{kHz}$ . Listener B has losses of $9\,\text{dB}$ , $48\,\text{dB}$ and $17\,\text{dB}$ at the same frequencies. Deduce the likely cause for each pattern and explain how each pattern changes an equal-loudness curve and perceived sound.
[5 marks]
Total for this question: 5
At $4.0\,\text{kHz}$ , a listener's threshold intensity is $2.0\times10^{-7}\,\text{W m}^{-2}$ rather than the normal value $2.0\times10^{-12}\,\text{W m}^{-2}$ . Thresholds at neighbouring frequencies are much closer to normal. Determine the hearing loss, describe how the listener's threshold curve differs from the normal curve, and deduce the likely cause.
[4 marks]
Total for this question: 4
At one chosen perceived loudness, readings from a normal listener's equal-loudness curve are $38\,\text{dB}$ at $0.50\,\text{kHz}$ and $31\,\text{dB}$ at $4.0\,\text{kHz}$ . The corresponding readings from a listener with hearing damage are $53\,\text{dB}$ and $73\,\text{dB}$ . For each listener, calculate the ratio $I_{4.0\,\text{kHz}}/I_{0.50\,\text{kHz}}$ required for the two tones to sound equally loud. Explain what the change in this ratio shows.
[4 marks]
Total for this question: 4
For the same perceived loudness, a normal listener requires $46\,\text{dB}$ at $0.75\,\text{kHz}$ and $39\,\text{dB}$ at $3.5\,\text{kHz}$ . An impaired listener requires $58\,\text{dB}$ and $76\,\text{dB}$ respectively. A hearing aid provides gains of $10\,\text{dB}$ and $30\,\text{dB}$ at these frequencies. Determine the external levels the aided listener requires and evaluate how well the aid restores the normal frequency balance.
[5 marks]
Total for this question: 5
A model of hearing damage assumes that independent reductions in sensitivity multiply. At $4.0\,\text{kHz}$ , ageing increases the threshold intensity by a factor of $18$ and previous excessive-noise exposure adds a factor of $7.0$ . The normal threshold there is $3.0\times10^{-13}\,\text{W m}^{-2}$ . Determine the combined threshold intensity and hearing loss in decibels. Describe the expected shape of the impaired threshold curve if the age effect is broad at high frequency but the noise effect is localised near $4.0\,\text{kHz}$ .
[5 marks]
Total for this question: 5

3.10.3.1 · Simple ECG machines and the normal ECG waveform

Explanation

Skin electrodes measure small potential differences produced by electrical activity in the heart. Conductive gel, prepared skin and secure contacts reduce resistance and movement artefacts; a high-gain, low-noise differential amplifier and shielded leads are needed because signals are of order millivolts.
On a normal waveform, the P wave represents atrial depolarisation, the QRS complex ventricular depolarisation and the T wave ventricular repolarisation.
Heart period is measured between equivalent points, usually successive R peaks, and $\text{rate}=60/T$ when $T$ is in seconds.
Measuring across several cycles reduces percentage uncertainty.
An ECG records electrical potential difference, not blood pressure, blood flow or the heart's mechanical force.

Normal ECG waveform showing P, QRS and T features and an R-to-R period.

Worked example

Six R-to-R intervals occupy $4.80\,\text{s}$ on an ECG trace. Calculate the heart rate.

1.Find the mean period: $T=4.80/6=0.800\,\text{s}$ .
2.Use $\text{rate}=60/T=60/0.800$ .

Answer: The heart rate is $75.0\,\text{beats min}^{-1}$ .

Common mistakes

Don't count six R peaks as six complete intervals when the trace contains only five gaps between them.
Don't identify the QRS complex as ventricular repolarisation instead of ventricular depolarisation.
Don't describe the vertical ECG signal as blood pressure rather than potential difference.

Exam tip

Measure several R-to-R intervals and divide by the number of intervals before converting period to beats per minute.

Tier 1 · Easy

Identify the electrical events represented by the QRS complex and the T wave in a normal ECG.
[2 marks]
Total for this question: 2
State the event represented by the P wave and the electrical quantity measured vertically on an ECG trace.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An ECG is recorded on paper moving at $25\,\text{mm s}^{-1}$ . The horizontal separation from the start of the P wave to the start of the QRS complex is $4.8\,\text{mm}$ . Calculate the PR interval and determine whether it lies within the supplied normal range $0.12$ to $0.20\,\text{s}$ .
[3 marks]
Total for this question: 3
A patient's heart rate is $84\,\text{beats min}^{-1}$ . Determine the separation of successive R peaks on paper moving at $25\,\text{mm s}^{-1}$ .
[3 marks]
Total for this question: 3
A treatment delays ventricular repolarisation but does not alter atrial depolarisation, ventricular depolarisation or the heart period. Deduce how the timing of the P wave, QRS complex, T wave and successive R peaks changes on the ECG.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An ECG trace contains slow baseline changes when the patient moves and a regular interference signal from nearby mains equipment. Explain how electrode attachment, the leads and the amplifier should be arranged to obtain a clearer normal waveform.
[5 marks]
Total for this question: 5
An ECG signal of amplitude $0.80\,\text{mV}$ is amplified to $1.6\,\text{V}$ . Determine the voltage gain. One R-R interval is measured as $(0.80\pm0.02)\,\text{s}$ , while five consecutive intervals are measured as $(4.00\pm0.02)\,\text{s}$ . Calculate the percentage uncertainty in each timing method and explain why timing several intervals is preferable.
[5 marks]
Total for this question: 5
At one instant, the heart-signal components at ECG electrodes A and B are $+0.70\,\text{mV}$ and $-0.50\,\text{mV}$ . Both leads also pick up the same $+0.180\,\text{V}$ interference. A differential amplifier produces $800(V_A-V_B)$ . Determine the output due to the heart signal and show why the common interference gives no output. Poor contact then adds an extra $+0.20\,\text{mV}$ interference at B only. Determine the resulting output error and explain how conductive gel helps.
[4 marks]
Total for this question: 4
The P wave, QRS complex and T wave in an ECG have input amplitudes $0.18\,\text{mV}$ , $1.4\,\text{mV}$ and $0.32\,\text{mV}$ respectively. An amplifier of voltage gain $1100$ feeds a recorder that can display only between $-1.0\,\text{V}$ and $+1.0\,\text{V}$ . Determine the attempted output amplitude of each feature, identify any distortion, and calculate the greatest gain that records all three without clipping.
[5 marks]
Total for this question: 5
An ECG recorder calibration pulse of $1.00\,\text{mV}$ produces a vertical displacement of $14.0\,\text{mm}$ . On a trace recorded at $30.0\,\text{mm s}^{-1}$ , six successive R peaks span $105\,\text{mm}$ and a QRS complex is $18.2\,\text{mm}$ high. Calculate the mean heart rate and QRS potential amplitude. Identify the broad wave following QRS and state the electrical event it represents.
[6 marks]
Total for this question: 6

3.10.4.1 · Ultrasound imaging

Explanation

Acoustic impedance is $Z=\rho c$ . At a boundary, the reflected intensity fraction is $I_r/I_i=((Z_2-Z_1)/(Z_2+Z_1))^2$ ; a larger mismatch gives a stronger echo and less transmission.
Coupling gel removes air between transducer and skin, reducing mismatch. A piezoelectric crystal converts a short alternating potential difference into an ultrasound pulse and converts returning deformation into a detected potential difference.
Echo depth is $d=ct/2$ because the pulse travels out and back. A new pulse is sent only after the deepest echo returns, so the maximum pulse-repetition frequency is set by the two-way travel time.
Higher frequency improves resolution but increases attenuation. An A-scan displays echo amplitude against time or depth; a B-scan maps echo brightness and position into a two-dimensional image.
Ultrasound is non-ionising and real-time, but air and bone boundaries limit access and resolution.

At normal incidence, part of the ultrasound pulse is transmitted and the echo retraces the outward path to the transducer (drawn offset for clarity).

Worked example

An echo returns $78\,\mu\text{s}$ after emission. Calculate the boundary depth for $c=1540\,\text{m s}^{-1}$ .

1.Convert the time: $t=78\times10^{-6}\,\text{s}$ .
2.Use $d=ct/2$ because the measured time includes both journeys.
3.Calculate $d=1540(78\times10^{-6})/2=6.01\times10^{-2}\,\text{m}$ .

Answer: The boundary is approximately $6.0\,\text{cm}$ deep.

Common mistakes

Don't use $d=ct$ and forget that the echo time covers the outward and return paths.
Don't say that coupling gel increases the air gap instead of removing the impedance mismatch with air.
Don't call an A-scan a two-dimensional brightness image rather than an amplitude trace.

Exam tip

For an echo calculation, state the factor of two explicitly; for a comparison, balance non-ionising real-time imaging against attenuation and resolution.

Tier 1 · Easy

A clinician needs to monitor fetal movement repeatedly during pregnancy and view the movement as it happens. Identify a suitable imaging modality and explain why it is appropriate in terms of image timing and radiation safety.
[2 marks]
Total for this question: 2
State the two energy conversions performed by the piezoelectric crystal in an ultrasound transducer.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Ultrasound passes normally from tissue A, where $\rho=1000\,\text{kg m}^{-3}$ and $c=1500\,\text{m s}^{-1}$ , into tissue B, where $\rho=1060\,\text{kg m}^{-3}$ and $c=1540\,\text{m s}^{-1}$ . Calculate the percentage of incident intensity reflected at the boundary.
[4 marks]
Total for this question: 4
Echoes from the front and back surfaces of a tissue layer return after $52\,\mu\text{s}$ and $88\,\mu\text{s}$ . Calculate the layer thickness when the ultrasound speed is $1540\,\text{m s}^{-1}$ .
[3 marks]
Total for this question: 3
Explain how repeated pulse-echo measurements made as an ultrasound transducer moves across the skin are converted into a B-scan image.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A $4.0\,\text{MHz}$ ultrasound pulse travels through tissue at $1540\,\text{m s}^{-1}$ and returns to the transducer $130\,\mu\text{s}$ after transmission. Calculate the reflector depth and estimate the scan resolution as one wavelength. Explain why coupling gel is placed between the transducer and skin.
[6 marks]
Total for this question: 6
Ultrasound of equal incident intensity meets two boundaries. Boundary 1 separates materials with acoustic impedances $1.50\times10^6$ and $1.80\times10^6\,\text{kg m}^{-2}\text{s}^{-1}$ . Boundary 2 is a tissue-bone boundary with impedances $1.80\times10^6$ and $6.00\times10^6\,\text{kg m}^{-2}\text{s}^{-1}$ . Determine the ratio of the reflected intensities at boundary 2 and boundary 1, state which boundary gives the stronger echo, and explain the consequence for imaging tissue beyond the bone.
[6 marks]
Total for this question: 6
A boundary $6.0\,\text{cm}$ below the skin must be imaged with resolution no worse than $0.30\,\text{mm}$ . Available transducers operate at $2.5\,\text{MHz}$ and $7.5\,\text{MHz}$ . Take resolution as one wavelength, ultrasound speed as $1540\,\text{m s}^{-1}$ , and one-way attenuation as $0.50\,\text{dB}$ per centimetre per megahertz. The receiver can detect an echo only when the round-trip attenuation does not exceed $50\,\text{dB}$ . Determine which transducer meets both requirements.
[6 marks]
Total for this question: 6
A tissue layer of acoustic impedance $1.70\times10^6\,\text{kg m}^{-2}\text{s}^{-1}$ lies between tissues of impedances $1.45\times10^6$ and $2.60\times10^6\,\text{kg m}^{-2}\text{s}^{-1}$ . Ultrasound is incident from the first tissue. Ignoring attenuation and taking the transmitted intensity fraction at a boundary as $1-R$ , calculate the echo-intensity fraction from each face of the layer and determine which echo is brighter on a B-scan.
[5 marks]
Total for this question: 5
An ultrasound B-scan must image structures down to a maximum depth of $0.180\,\text{m}$ in tissue where the ultrasound speed is $1540\,\text{m s}^{-1}$ . Calculate the latest possible echo-return time and hence the greatest pulse-repetition frequency that avoids range ambiguity. Explain why a higher pulse rate could assign an echo to the wrong transmitted pulse.
[6 marks]
Total for this question: 6

3.10.4.2 · Fibre optics and endoscopy

Explanation

An optical fibre has a higher-refractive-index core surrounded by lower-index cladding. A ray travelling in the core undergoes total internal reflection when it meets the boundary from higher to lower index at an angle greater than the critical angle.
Cladding also keeps light within each fibre and reduces cross-talk.
In a flexible endoscope, a non-coherent bundle carries illumination because fibre order need not be preserved.
A coherent bundle keeps fibres in the same relative positions at both ends, so each fibre transfers one part of the image.
Flexible bundles allow internal imaging through a small opening and can also guide treatment light, reducing the need for more invasive access.

Light guided through a fibre by total internal reflection at the core-cladding boundary.

Worked example

Explain why an endoscope needs a coherent fibre bundle for imaging but may use a non-coherent bundle for illumination.

1.An image requires each fibre's position at the entrance to match its position at the exit.
2.A coherent bundle preserves that relative fibre order and therefore the spatial pattern.
3.Illumination only requires light delivery, so preserving fibre order is unnecessary.

Answer: The coherent bundle transfers the image pattern; the non-coherent bundle can supply illumination.

Common mistakes

Don't state only that light reflects, without giving both total-internal-reflection conditions.
Don't use a non-coherent bundle to transmit an image even though fibre positions are scrambled.
Don't place the cladding at a higher refractive index than the core.

Exam tip

A full total-internal-reflection statement must name travel from higher to lower refractive index and incidence above the critical angle.

Tier 1 · Easy

Name the endoscope fibre bundle that carries an image and the bundle that carries illumination into the body.
[2 marks]
Total for this question: 2
State both conditions required for total internal reflection at a core-cladding boundary.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A fibre has core refractive index $1.62$ and cladding refractive index $1.50$ . Calculate the critical angle at the core-cladding boundary and determine whether a ray incident there at $72.0^\circ$ undergoes total internal reflection.
[3 marks]
Total for this question: 3
Explain how fibre bundles let an endoscope examine and treat tissue through a small opening, and state the clinical advantage of this access.
[3 marks]
Total for this question: 3
A ray enters the core of an endoscope fibre at too large an angle to the fibre axis. Explain why the ray is not confined to the core.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A damaged endoscope has an image bundle whose fibre positions are rearranged between its two ends, and some cladding has been removed. Explain the effects on the observed image and why the illumination bundle can still work when its fibre order is random.
[5 marks]
Total for this question: 5
Two endoscope fibres have core refractive index $1.60$ . Fibre X has cladding index $1.50$ and fibre Y has cladding index $1.42$ . Determine which fibre confines a ray travelling in the core and incident on the core-cladding boundary at $66.0^\circ$ , and explain why the fibres of the image bundle must keep the same relative positions at both ends.
[5 marks]
Total for this question: 5
An endoscope objective forms a $1.8\,\text{mm}$ -wide image of a $12\,\text{mm}$ -wide tissue region on a fibre bundle. A feature must span at least two fibre-centre spacings in this image to be resolved. Bundle A is coherent with fibre-centre spacing $30\,\mu\text{m}$ . Bundle B is non-coherent with spacing $15\,\mu\text{m}$ . Determine whether either bundle can produce a resolved image of a $0.30\,\text{mm}$ -wide lesion.
[4 marks]
Total for this question: 4
An endoscope fibre has core refractive index $1.49$ and cladding refractive index $1.36$ . A ray in a plane through the axis enters its flat end from air. Determine the limiting angle for total internal reflection at the core-cladding interface and hence the largest guided angle to the fibre axis inside the core. Use refraction at the air-core face to find the corresponding largest entrance angle to the axis in air.
[4 marks]
Total for this question: 4
A ray in one image fibre undergoes 30 boundary reflections. With intact cladding, $99.2\%$ of its intensity remains after each reflection; damage reduces this to $91\%$ . Calculate the fraction of the original intensity delivered in each case and the ratio of the two outputs. Explain why intact cladding and coherent fibre ordering are separate requirements for a clear endoscope image.
[5 marks]
Total for this question: 5

3.10.4.3 · Magnetic resonance (MR) scanner

Explanation

An MR scanner uses the strong field of a superconducting magnet to align hydrogen nuclei, or protons, with spins parallel; the spinning nuclei precess about the magnetic field lines. Gradient coils vary the field with position so successive small regions of a patient cross-section can be selected and located.
A short radio-frequency pulse excites protons in a selected region and changes their spin state. As the protons de-excite after the pulse, they emit radio-frequency signals.
The emitted radio-frequency signal has the same frequency as the exciting pulse. Receiver coils detect these signals and a computer processes their position and strength to construct a visual cross-sectional image.
MR uses non-ionising radio waves and magnetic fields.
Production of the fields and detailed relaxation times are not required.

Worked example

Outline how an MR scanner obtains a cross-sectional image after the patient is placed in the main magnetic field.

1.Gradient fields select and locate successive small regions of the cross-section.
2.Short RF pulses excite aligned, precessing hydrogen nuclei in each selected region.
3.The nuclei emit RF signals as they de-excite; receiver coils detect them.
4.A computer combines signal position and strength to construct the image.

Answer: Spatial selection by gradients, RF excitation, RF detection and computer processing produce the cross-sectional image.

Common mistakes

Don't say that the static field makes stationary protons begin spinning classically.
Don't describe the detected signal as an X-ray rather than radio-frequency emission from de-exciting protons.
Don't omit the gradient coils, leaving no mechanism for locating signals within the cross-section.

Exam tip

For an outline question, keep the sequence explicit: align and precess, select, excite, detect, then process.

Tier 1 · Easy

State which nuclei provide the main signal in an MR scanner and describe their behaviour in the scanner's static magnetic field.
[2 marks]
Total for this question: 2
State the type of electromagnetic radiation used to excite protons in an MR scanner and whether this radiation is ionising.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Describe what happens to selected protons during and immediately after a short radio-frequency pulse in an MR scan.
[3 marks]
Total for this question: 3
After a radio-frequency pulse excites selected protons, state the radiation emitted as they de-excite and compare its frequency with that of the incident pulse.
[2 marks]
Total for this question: 2
A description of an MR scan claims that the main magnetic field makes stationary protons begin to spin and that the whole cross-section is scanned at once. Correct both statements.
[2 marks]
Total for this question: 2

Tier 3 · Hard

Outline how an MR scanner obtains a cross-sectional image, beginning with a patient in the main magnetic field and ending with a computer-generated image.
[5 marks]
Total for this question: 5
An MR scanner has a working main magnet, radio-frequency transmitter and receiver coils, but its gradient coils are switched off. Explain what signals can still be produced and why a cross-sectional image cannot be formed.
[5 marks]
Total for this question: 5
State the role of a short radio-frequency pulse in an MR scan, state which nuclei the pulse acts on and why they are used, and explain why a patient can safely undergo several MR scans.
[4 marks]
Total for this question: 4
In an MR scan, small region A contains $1.8$ times as many responding protons as small region B. The radio-frequency signal from B induces an emf amplitude of $0.28\,\mu\text{V}$ in a receiver coil. Assume induced emf is proportional to the number of responding protons. An amplifier has gain $2.5\times10^6$ , and the display assigns visibly different shades only when the amplifier outputs differ by at least $0.45\,\text{V}$ . Determine the output for each small region and decide whether the display resolves their signal contrast. Explain how gradient coils allow the two signals to be assigned to different positions.
[6 marks]
Total for this question: 6
Discuss this account of MR imaging: "The main magnetic field is applied only after the radio-frequency pulse. The receiver coils supply the energy that excites the selected protons. Relaxing protons emit at a higher frequency than the radio-frequency pulse, and signal amplitude alone tells the computer where each signal originated." Identify and correct the errors.
[5 marks]
Total for this question: 5

3.10.5.1 · The physics of diagnostic X-rays

Explanation

Thermionic emission releases electrons that accelerate through potential difference $V$ and decelerate at a metal target. Most energy becomes heat, so a rotating anode spreads heating.
Different electron energy losses produce the continuous X-ray spectrum; $E_{\max}=eV$ and $\lambda_{\min}=hc/(eV)$ . Characteristic lines occur when incident electrons remove inner-shell electrons and higher-shell electrons fall into vacancies.
Increasing tube current increases intensity and patient dose, while increasing voltage raises maximum photon energy and penetration.
A smaller focal spot improves sharpness but concentrates heating.
Filtration removes low-energy photons and collimation restricts the exposed region, controlling dose and image quality.

Diagnostic X-ray energy spectrum with a continuous distribution, characteristic lines and maximum energy.

Worked example

Calculate the maximum photon energy from a tube operating at $80\,\text{kV}$ , in joules and electronvolts.

1.Use $E_{\max}=eV=(1.60\times10^{-19})(80\times10^3)$ .
2.This gives $E_{\max}=1.28\times10^{-14}\,\text{J}$ .
3.An electron accelerated through $80\,\text{kV}$ gains $80\,\text{keV}$ .

Answer: $E_{\max}=1.28\times10^{-14}\,\text{J}=80\,\text{keV}$ .

Common mistakes

Don't say that increasing tube current increases maximum photon energy rather than photon number.
Don't attribute characteristic lines to arbitrary electron deceleration instead of transitions into inner-shell vacancies.
Don't claim filtration removes the highest-energy photons and thereby increases skin dose.

Exam tip

In a controls question, link current to intensity, voltage to maximum energy, focal spot to sharpness and filtration or collimation to dose.

Tier 1 · Easy

State the maximum photon energy, in $\text{keV}$ , from an X-ray tube operating at $68\,\text{kV}$ .
[1 mark]
Total for this question: 1
State the effect of increasing the tube current on the intensity of the X-ray beam.
[1 mark]
Total for this question: 1

Tier 2 · Standard

An X-ray tube uses an accelerating potential difference of $72\,\text{kV}$ . Calculate the minimum X-ray wavelength. Use $e=1.60\times10^{-19}\,\text{C}$ , $h=6.63\times10^{-34}\,\text{J s}$ and $c=3.00\times10^8\,\text{m s}^{-1}$ .
[3 marks]
Total for this question: 3
Explain how adding an aluminium filter changes an X-ray beam and reduces the dose to a patient.
[3 marks]
Total for this question: 3
A characteristic X-ray from a target has wavelength $2.75\times10^{-11}\,\text{m}$ . Calculate, in $\text{keV}$ , the separation between the two electron energy levels involved. Use $e=1.60\times10^{-19}\,\text{C}$ , $h=6.63\times10^{-34}\,\text{J s}$ and $c=3.00\times10^8\,\text{m s}^{-1}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

A rotating-anode tube operates at $95\,\text{kV}$ with a current of $3.0\,\text{mA}$ for $0.080\,\text{s}$ . It converts $1.2\%$ of the electrical energy into X-rays. Calculate the maximum photon energy and the total X-ray energy produced. Explain one benefit of rotating the anode. Use $e=1.60\times10^{-19}\,\text{C}$ .
[5 marks]
Total for this question: 5
A small focal spot reduces geometric unsharpness, but too small a spot overheats the anode. The tube potential difference and current are unchanged. Explain why reducing the focal-spot size leaves the maximum photon energy and the number of X-ray photons produced unchanged.
[3 marks]
Total for this question: 3
Two X-ray tubes use a tungsten target, and both voltages are high enough to produce tungsten's characteristic lines. Tube A operates at $80\,\text{kV}$ and $2.4\,\text{mA}$ ; tube B operates at $120\,\text{kV}$ and $1.2\,\text{mA}$ . Each tube converts $0.80\%$ of its electrical power into X-rays. For each tube, the mean photon energy is $0.40$ times its maximum photon energy. Determine the ratios of maximum photon energy and photon production rate for B relative to A, and whether their characteristic-line energies differ. Use $e=1.60\times10^{-19}\,\text{C}$ .
[5 marks]
Total for this question: 5
A target atom has K-shell and L-shell binding energies of $72.4\,\text{keV}$ and $11.3\,\text{keV}$ respectively. Calculate the wavelength of the characteristic X-ray emitted when an electron falls from the L shell into a K-shell vacancy. Determine whether a tube operating at $70.0\,\text{kV}$ can produce this line. Use $hc=1.99\times10^{-25}\,\text{J m}$ and $1\,\text{keV}=1.60\times10^{-16}\,\text{J}$ .
[5 marks]
Total for this question: 5
An X-ray tube operates at $84\,\text{kV}$ . Only $0.70\%$ of the electrical power becomes X-rays and the remainder heats a focal spot measuring $0.80\,\text{mm}$ by $1.6\,\text{mm}$ . The permitted heating power per unit area is $3.2\times10^9\,\text{W m}^{-2}$ . Calculate the greatest tube current that does not exceed this limit. State the maximum photon energy and explain why doubling the focal-spot area would permit a greater beam intensity without changing that maximum energy.
[6 marks]
Total for this question: 6

3.10.5.2 · Image detection and enhancement

Explanation

In a flat-panel detector, a scintillator converts X-rays to visible photons, photodiode pixels convert light to electrical charge, and electronic scanning transfers pixel signals to a computer. The immediate digital image can be enhanced, stored and transmitted; high sensitivity and wide dynamic range may reduce repeat exposures compared with film.
X-rays darken photographic film, while an intensifying screen converts X-rays to visible light so less dose is needed.
Fluoroscopic image intensification gives live images, although prolonged viewing can raise dose.
X-ray-opaque barium absorbs strongly and outlines low-contrast digestive structures against soft tissue.
Images arise from differential transmission and absorption, not reflected X-rays.

Worked example

Explain how a flat-panel detector converts an incident X-ray pattern into a digital image.

1.The scintillator converts absorbed X-rays into visible-light photons.
2.Photodiode pixels convert the light into electrical charge proportional to local exposure.
3.Electronic scanning reads the pixel charges and a computer constructs and enhances the image.

Answer: The scintillator-photodiode-scanning chain converts spatial X-ray intensity into digital pixel values.

Common mistakes

Don't reverse the conversion chain and make the photodiode emit X-rays.
Don't say that barium improves contrast by reflecting X-rays rather than absorbing them strongly.
Don't claim fluoroscopy reduces dose regardless of viewing time.

Exam tip

For an operation question, give the detector stages in order and identify the energy conversion at each stage.

Tier 1 · Easy

State the function of the scintillator in a flat-panel X-ray detector.
[1 mark]
Total for this question: 1
State the function of a photodiode pixel in a flat-panel X-ray detector.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Explain why a patient is given a barium suspension before an X-ray image of the digestive tract is recorded.
[3 marks]
Total for this question: 3
Explain why an intensifying screen reduces the patient dose needed to produce a photographic X-ray image, and state one disadvantage.
[3 marks]
Total for this question: 3
A proposed flat-panel detector adds the signals from every photodiode into one output before electronic scanning. Explain why this would not produce a useful image and how the detector should preserve the X-ray pattern.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A flat-panel X-ray detector has a pixel pitch of $0.25\,\text{mm}$ . Three adjacent photodiode rows fail, so the corresponding strip of the image records no data. Calculate the width of the missing strip and explain why a $0.50\,\text{mm}$ lesion lying entirely inside it would not be recorded. Explain why image enhancement cannot reconstruct the lesion and when a repeat exposure may be justified despite the additional dose.
[5 marks]
Total for this question: 5
An unenhanced digestive-tract image has very similar detector signals behind the tract and the surrounding soft tissue. Explain how a barium suspension and subsequent digital processing can make the tract easier to distinguish, and state one limitation of digital processing.
[5 marks]
Total for this question: 5
Two pixels of a flat-panel detector absorb $3.0\times10^4$ and $1.8\times10^4$ X-ray photons. Each absorbed X-ray produces $1.2\times10^3$ visible photons; $25\%$ reach the relevant photodiode and $60\%$ of these release a charge carrier. A pixel saturates at $6.0\times10^6$ charge carriers. Determine whether the exposure records a usable contrast between the pixels, and explain what digital enhancement can and cannot do to the recorded signals.
[5 marks]
Total for this question: 5
A fluoroscopic system follows a barium marker moving at up to $42\,\text{mm s}^{-1}$ . Successive displayed positions must be no more than $3.0\,\text{mm}$ apart. Each recorded frame gives the patient $0.016$ relative dose units and the examination lasts $50\,\text{s}$ . Calculate the minimum frame rate, the relative dose at this rate and the percentage dose reduction from an existing rate of $24$ frames per second. Explain why barium is used and why a lower frame rate would be unsuitable.
[6 marks]
Total for this question: 6
A photographic X-ray film requires $6.8\times10^9$ effective photon exposures to reach a specified optical density. Without an intensifying screen, each incident X-ray produces one effective exposure. An intensifying screen absorbs $57\%$ of the incident X-rays and has a supplied conversion factor of $290$ useful visible-light exposures at the film per absorbed X-ray. Calculate the number of incident X-rays required with the screen, the factor by which the incident exposure is reduced and the percentage exposure reduction. Assume patient dose is proportional to the number of incident X-rays.
[4 marks]
Total for this question: 4

3.10.5.3 · Absorption of X-rays

Explanation

For a narrow monoenergetic beam, transmitted intensity follows $I=I_0e^{-\mu x}$ , where $\mu$ is the linear attenuation coefficient and $x$ is thickness in compatible units. Attenuation is exponential, not a fixed subtraction per unit thickness.
The half-value thickness is $x_{1/2}=\ln2/\mu$ ; after $n$ half-value thicknesses, $I=I_0(1/2)^n$ .
The mass attenuation coefficient is $\mu_m=\mu/\rho$ , with SI unit $\text{m}^2\,\text{kg}^{-1}$ , allowing materials to be compared without density's direct effect.
Bone and contrast media attenuate more strongly than soft tissue, so differential transmission creates image contrast.
The equation gives transmitted, not absorbed, intensity.

Exponential X-ray attenuation and the half-value thickness.

Worked example

An absorber has $\mu=0.35\,\text{cm}^{-1}$ . Calculate its half-value thickness.

1.Use $x_{1/2}=\ln2/\mu$ .
2.Substitute $x_{1/2}=0.693/0.35=1.98\,\text{cm}$ .

Answer: The half-value thickness is $1.98\,\text{cm}$ .

Common mistakes

Don't subtract the same intensity for every centimetre instead of applying exponential attenuation.
Don't use incompatible units for $\mu$ and $x$ .
Don't report $I_0-I$ when the equation has calculated transmitted intensity $I$ .

Exam tip

Check whether the question requests transmitted intensity, absorbed intensity or half-value thickness before selecting the relation.

Tier 1 · Easy

An absorber has linear attenuation coefficient $0.28\,\text{cm}^{-1}$ . Calculate its half-value thickness.
[2 marks]
Total for this question: 2
Calculate the mass attenuation coefficient of tissue with linear attenuation coefficient $28\,\text{m}^{-1}$ and density $920\,\text{kg m}^{-3}$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

A narrow X-ray beam of intensity $6.4\,\text{W m}^{-2}$ passes through $7.0\,\text{mm}$ of tissue with $\mu=0.18\,\text{mm}^{-1}$ . Calculate the transmitted intensity.
[3 marks]
Total for this question: 3
Determine the absorber thickness that transmits $15\%$ of a narrow X-ray beam when $\mu=0.42\,\text{cm}^{-1}$ .
[3 marks]
Total for this question: 3
A narrow X-ray beam passes through $2.0\,\text{cm}$ of tissue with $\mu=0.12\,\text{cm}^{-1}$ and then $0.30\,\text{cm}$ of another material with $\mu=2.5\,\text{cm}^{-1}$ . Calculate the percentage of the incident intensity absorbed by the two layers.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An incident X-ray beam has intensity $10.0\,\text{W m}^{-2}$ . One ray crosses $18\,\text{mm}$ of soft tissue with $\mu=0.050\,\text{mm}^{-1}$ and then $4.0\,\text{mm}$ of bone with $\mu=0.32\,\text{mm}^{-1}$ . A neighbouring ray crosses $22\,\text{mm}$ of the same soft tissue only. Calculate both transmitted intensities and the ratio of the soft-tissue-only intensity to the bone-path intensity.
[5 marks]
Total for this question: 5
A $35\,\text{mm}$ sample transmits $22\%$ of an incident narrow X-ray beam. Its mass attenuation coefficient is $2.8\times10^{-2}\,\text{m}^2\,\text{kg}^{-1}$ . Determine the density of the sample.
[4 marks]
Total for this question: 4
Two filters have the same mass per unit area, $12\,\text{kg m}^{-2}$ . Filter A has mass attenuation coefficient $3.6\times10^{-2}\,\text{m}^2\,\text{kg}^{-1}$ and filter B has $1.2\times10^{-2}\,\text{m}^2\,\text{kg}^{-1}$ . Determine the percentage transmitted by each filter and the factor by which the larger transmitted intensity exceeds the smaller one.
[4 marks]
Total for this question: 4
A narrow monoenergetic X-ray beam has intensity $4.8\,\text{W m}^{-2}$ after passing through $1.5\,\text{cm}$ of a uniform absorber and $1.7\,\text{W m}^{-2}$ after passing through $5.5\,\text{cm}$ . Use the two measurements to determine the linear attenuation coefficient, the incident intensity and the half-value thickness of the absorber.
[5 marks]
Total for this question: 5
A lesion is $3.0\,\text{cm}$ thick within an $8.0\,\text{cm}$ tissue path. At photon energy L, the attenuation coefficients of the lesion and surrounding tissue are $0.40\,\text{cm}^{-1}$ and $0.26\,\text{cm}^{-1}$ . At photon energy H they are $0.19\,\text{cm}^{-1}$ and $0.13\,\text{cm}^{-1}$ . For each energy, calculate the ratio of the intensity through surrounding tissue only to that through the lesion path. The surrounding-tissue transmission must be at least $25\%$ of the incident intensity. Determine which energy meets this requirement and discuss the resulting contrast trade-off.
[4 marks]
Total for this question: 4

3.10.5.4 · CT scanner

Explanation

A CT scanner moves an X-ray tube around the patient and directs a narrow, approximately monochromatic beam through the body to an array of detectors. At many angles, detectors measure the transmitted intensity along different paths.
A computer combines these projections to reconstruct a cross-sectional visual image; successive slices may form a three-dimensional representation.
CT removes the superposition of structures in a plain radiograph and provides better localisation and useful soft-tissue information.
Its disadvantages are greater cost and usually a higher ionising-radiation dose than a simple X-ray image.
Required comparisons are limited to image resolution, cost and safety; detector construction and operation are not required.

CT acquisition using a rotating X-ray tube and detector array around a patient cross-section.

Worked example

Explain why CT can distinguish overlapping internal structures better than a plain radiograph.

1.A plain radiograph superposes attenuation from every structure along one projection.
2.CT records narrow-beam projections from many different angles.
3.Computer reconstruction assigns attenuation information to locations within a cross-sectional slice.

Answer: CT removes projection overlap by reconstructing a spatially resolved cross-section from many angular measurements.

Common mistakes

Don't describe CT as one wide-beam exposure from a fixed tube.
Don't say that CT uses non-ionising radiation.
Don't claim CT is always safer and cheaper than a plain radiograph despite its generally higher dose and cost.

Exam tip

A comparison must connect the reconstructed slice to improved localisation, then balance that benefit against cost and ionising dose.

Tier 1 · Easy

State why the X-ray tube moves around a patient during a CT scan.
[1 mark]
Total for this question: 1
State the function of the detector array in a CT scanner.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Describe how a CT scanner produces a cross-sectional image from X-rays transmitted through a patient.
[3 marks]
Total for this question: 3
Explain why a CT scanner uses a narrow, monochromatic X-ray beam.
[3 marks]
Total for this question: 3
For a CT image, explain why transmission measurements from only one projection direction would leave structures superposed.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A hospital is choosing between CT and a plain X-ray image to localise a small lung lesion partly hidden by overlapping ribs, and to confirm the position of a radiopaque feeding tube. Compare the techniques using image resolution, superposition, radiation dose, cost and availability. Recommend one technique for each task.
[6 marks]
Total for this question: 6
Discuss the suitability of a reduced-dose CT protocol that uses fewer angular projections for repeated monitoring of a large, clearly defined lesion.
[4 marks]
Total for this question: 4
A child needs three follow-up images to check that a simple forearm fracture remains correctly aligned while it heals. Compare CT with plain radiographs for this purpose in terms of image resolution, cost and safety. Give a justified recommendation.
[5 marks]
Total for this question: 5
A small soft-tissue lesion lies behind overlapping structures in a plain radiograph, and a surgeon needs its position before an operation. Compare the spatial detail, financial cost and radiation risk of taking another plain radiograph with those of using CT. Give a justified recommendation.
[5 marks]
Total for this question: 5
A CT scan covers a body length of $96\,\text{mm}$ using contiguous slices $1.5\,\text{mm}$ thick. Calculate the number of slices. The slice thickness is then halved to $0.75\,\text{mm}$ while the same body length is scanned. Calculate the new number of slices and explain why the total patient dose doubles if the dose per slice is unchanged.
[4 marks]
Total for this question: 4

3.10.6.1 · Imaging techniques

Explanation

A tracer combines a gamma-emitting radioisotope with a compound that has affinity for a chosen organ; penetrating gamma photons leave the body for external detection. Technetium-99m emits about $140\,\text{keV}$ gamma radiation, has a roughly $6\,\text{h}$ half-life and labels many compounds.
Iodine-131 is taken up by the thyroid, emits beta and gamma radiation and has a roughly $8\,\text{d}$ half-life. Indium-111 emits gamma radiation and has a roughly $2.8\,\text{d}$ half-life.
Higher gamma energy makes collimation harder because photons can penetrate the lead between adjacent collimator holes, and it lowers detection efficiency because fewer photons are fully absorbed by the detector crystal.
A molybdenum-technetium generator supplies fresh technetium-99m from longer-lived molybdenum-99.
PET detects two nearly opposite $511\,\text{keV}$ annihilation photons in coincidence to locate positron-emitting tracer activity.

Worked example

Explain why technetium-99m is suitable for diagnostic tracer imaging.

1.Its gamma radiation is penetrating enough to leave the body for external detection.
2.Its approximately $6\,\text{h}$ half-life permits imaging but limits prolonged dose.
3.It can be attached to compounds with affinity for different organs and obtained from a hospital generator.

Answer: Detectable gamma emission, a suitably short half-life, flexible labelling and generator availability make technetium-99m useful.

Common mistakes

Don't choose an alpha emitter for external imaging even though alpha radiation cannot escape tissue effectively.
Don't say the molybdenum-technetium generator creates energy rather than supplying daughter technetium-99m from molybdenum-99 decay.
Don't describe PET coincidence photons as travelling in the same direction instead of nearly opposite directions.

Exam tip

Justify a tracer by linking radiation type, half-life, gamma energy and organ-specific labelling to the intended scan.

Tier 1 · Easy

State why a medical tracer used for external imaging should emit gamma radiation.
[1 mark]
Total for this question: 1
Give the organ in which an iodine-131 tracer is preferentially taken up.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Justify the use of technetium-99m for imaging an organ. Refer to its $6\,\text{h}$ half-life, its $140\,\text{keV}$ gamma emission and its chemical use.
[3 marks]
Total for this question: 3
Explain the importance of a molybdenum-technetium generator in a hospital imaging department.
[3 marks]
Total for this question: 3
Iodine-131 emits $364\,\text{keV}$ gamma photons, whereas technetium-99m emits $140\,\text{keV}$ gamma photons. Explain two consequences of this difference for gamma-camera imaging.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Explain how a PET scan can map regions of high glucose uptake after a patient receives a positron-emitting glucose analogue. Include the nuclear event, the detection method and how position is inferred.
[5 marks]
Total for this question: 5
A labelled white-cell scan records tracer distribution for $30\,\text{h}$ . Technetium-99m emits $140\,\text{keV}$ gamma photons and has a $6.0\,\text{h}$ half-life. Indium-111 emits $171\,\text{keV}$ and $245\,\text{keV}$ gamma photons and has a $67\,\text{h}$ half-life. Both can label the cells. Determine the more suitable isotope.
[5 marks]
Total for this question: 5
A PET reconstruction program places every annihilation at the midpoint between the two detectors that record a coincident pair. Explain why this rule is not valid for a scanner without photon time-of-flight information. Describe how coincident detections are used correctly to locate a small region of high tracer uptake. The number of detector elements is fixed; explain why making each element narrower could improve spatial resolution at the expense of count rate.
[5 marks]
Total for this question: 5
Two opposite PET detectors have individual count rates $3.2\times10^4\,\text{s}^{-1}$ and $4.5\times10^4\,\text{s}^{-1}$ . The accidental-coincidence rate is $R_a=2R_1R_2\Delta t$ . Calculate $R_a$ for coincidence windows of $15\,\text{ns}$ and $6.0\,\text{ns}$ , and the percentage reduction. Explain the benefit and one limitation of narrowing the window.
[5 marks]
Total for this question: 5
Two gamma-emitting compounds have the same total activity, photon energy and half-life. The target organ has mass $0.24\,\text{kg}$ . Compound X places $15\%$ of its activity in the organ and distributes the remainder uniformly through $40\,\text{kg}$ of other tissue. Compound Y places $8.0\%$ in the organ and distributes its remainder through the same other-tissue mass. For equal tissue masses viewed with the same detection efficiency, calculate the organ-to-background count-rate ratio for each compound and evaluate which is the better tracer.
[6 marks]
Total for this question: 6

3.10.6.2 · Half-life

Explanation

Physical half-life $T_P$ is the time for half the unstable nuclei to decay and is a property of the radionuclide. Biological half-life $T_B$ is the time for biological processes to remove half the tracer from an organ or body, ignoring radioactive decay.
Effective half-life $T_E$ describes the combined reduction caused by both processes.
The rates add, so $1/T_E=1/T_B+1/T_P$ ; consequently $T_E$ must be shorter than both component half-lives.
All values must use the same time unit before reciprocals are combined.
The equation concerns the amount or activity remaining in the body, not a change in the radionuclide's intrinsic physical half-life.

Worked example

A tracer has $T_P=12\,\text{h}$ and $T_B=18\,\text{h}$ . Calculate $T_E$ .

1.Use $1/T_E=1/18+1/12=5/36\,\text{h}^{-1}$ .
2.Take the reciprocal: $T_E=36/5=7.2\,\text{h}$ .
3.Check that $7.2\,\text{h}$ is shorter than both $12\,\text{h}$ and $18\,\text{h}$ .

Answer: The effective half-life is $7.2\,\text{h}$ .

Common mistakes

Don't add $T_P$ and $T_B$ directly instead of adding their reciprocals.
Don't use hours for one half-life and days for the other.
Don't obtain an effective half-life longer than a component half-life and fail to reject the result.

Exam tip

Use the shorter-than-both check immediately after evaluating an effective half-life.

Tier 1 · Easy

Define the biological half-life of a tracer in an organ.
[2 marks]
Total for this question: 2
Define the physical half-life of a radionuclide.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A tracer has physical half-life $18\,\text{h}$ and biological half-life $30\,\text{h}$ . Calculate its effective half-life.
[3 marks]
Total for this question: 3
A student claims that a tracer with physical half-life $12\,\text{h}$ and biological half-life $30\,\text{h}$ has an effective half-life of $20\,\text{h}$ . Explain why this result must be rejected without calculating the effective half-life.
[3 marks]
Total for this question: 3
A tracer has physical half-life $8.0\,\text{h}$ and effective half-life $5.0\,\text{h}$ . Calculate, as a percentage, the ratio of the amount remaining after $10\,\text{h}$ to the amount that would remain if only radioactive decay occurred.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A radionuclide has physical half-life $12.0\,\text{h}$ . Measurements in a patient give an effective half-life of $7.20\,\text{h}$ . Calculate the biological half-life and the percentage of the initial activity remaining in the patient after $21.6\,\text{h}$ .
[5 marks]
Total for this question: 5
A tracer has physical half-life $20\,\text{h}$ and biological half-life $30\,\text{h}$ . Determine the time for its activity in the body to fall to $18\%$ of the initial value.
[5 marks]
Total for this question: 5
After $8.0\,\text{h}$ , radioactive decay alone would leave $70\%$ of a tracer, but measurements in a patient show that only $35\%$ remains. Assuming radioactive decay and biological removal act independently, determine the physical, biological and effective half-lives.
[4 marks]
Total for this question: 4
A patient receives equal tracer administrations of $4.8\,\text{MBq}$ at $0$ , $6.0\,\text{h}$ and $12.0\,\text{h}$ . The tracer has an effective half-life of $9.0\,\text{h}$ . Calculate the total activity in the patient immediately after the third administration and the additional time for this total to fall below $2.0\,\text{MBq}$ , assuming no further tracer is given.
[5 marks]
Total for this question: 5
The same radionuclide is present in organs A and B, so its physical half-life is $14\,\text{h}$ in both. Its biological half-life is $21\,\text{h}$ in A and $7.0\,\text{h}$ in B. Initially the activities are $2.0\,\text{MBq}$ in A and $5.0\,\text{MBq}$ in B. Calculate each effective half-life, then determine when the two organ activities become equal and their common activity at that time.
[6 marks]
Total for this question: 6

3.10.6.3 · Gamma camera

Explanation

A gamma camera's lead collimator absorbs photons travelling in unsuitable directions, so accepted photons retain directional information; narrower holes improve spatial resolution but reduce count rate. A scintillation crystal converts each absorbed gamma photon into a visible-light flash and a light guide distributes it to photomultiplier tubes.
At each photocathode, light ejects photoelectrons.
Successive dynodes held at increasing potentials produce secondary emission, multiplying the electron number into a large anode pulse.
Relative photomultiplier outputs locate the flash, while pulse size estimates photon energy so scattered photons can be rejected.
A computer accumulates many accepted event positions to form the tracer-distribution image.

Gamma-camera chain from directional collimation to scintillation and photomultiplier detection.

Worked example

Explain the trade-off produced by making gamma-camera collimator holes narrower.

1.Narrow holes accept a smaller range of photon directions.
2.The accepted direction is better defined, improving spatial resolution.
3.Fewer photons pass through, reducing count rate and sensitivity.

Answer: Narrower holes improve spatial resolution but reduce sensitivity and require longer counting.

Common mistakes

Don't say the collimator focuses gamma photons like a glass lens.
Don't place the photocathode before the scintillation crystal and make it absorb gamma photons directly.
Don't claim narrower holes increase both resolution and count rate.

Exam tip

An operation answer should follow one photon through collimator, crystal, photocathode, dynodes and anode before explaining image formation.

Tier 1 · Easy

State the function of the lead collimator in a gamma camera.
[1 mark]
Total for this question: 1
State the function of the scintillation crystal in a gamma camera.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Describe how a gamma photon entering a gamma camera produces an electrical pulse at the output of a photomultiplier tube.
[3 marks]
Total for this question: 3
Explain why pulse-height selection improves the image produced by a gamma camera.
[3 marks]
Total for this question: 3
Two identical, equally calibrated neighbouring photomultiplier tubes receive light from each scintillation. For event P their pulse sizes are $8.0$ and $2.0$ units; for event Q they are $5.0$ and $5.0$ units. Assume the two tubes together collect all the light from each scintillation. Deduce the relative positions of P and Q and whether the same gamma energy was deposited in each event.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Explain how a gamma camera determines the position of tracer activity and improves the quality of the recorded image. Include the roles of the collimator, photomultiplier array and pulse-height selection.
[5 marks]
Total for this question: 5
A photomultiplier releases three photoelectrons at its photocathode. Each of its ten dynodes multiplies the electron number by $4.0$ . Calculate the charge reaching the anode and explain why successive dynodes are held at increasing positive potentials. Use $e=1.60\times10^{-19}\,\text{C}$ .
[4 marks]
Total for this question: 4
A gamma camera with collimator holes of diameter $d$ has spatial resolution $10\,\text{mm}$ and records $900$ counts per second for $20\,\text{s}$ . The hole length is unchanged and accepted count rate is proportional to $d^2$ . Reducing $d$ by half also halves the resolution distance. Determine whether the modified camera can resolve two tracer regions $7.0\,\text{mm}$ apart and the counting time needed to collect the original number of counts.
[5 marks]
Total for this question: 5
During one acquisition, a target region records $5600$ accepted events and an equal-area background region records $4800$ . For two independent counts, use uncertainty in their difference $=\sqrt{N_1+N_2}$ . The significance of the excess is the excess divided by its uncertainty. Calculate the significance of the excess. A narrower pulse-height window retains $72\%$ of the target excess but only $25\%$ of the background events in each region. Calculate the new significance and evaluate the change.
[4 marks]
Total for this question: 4
A gamma-camera operator makes three changes at once: removes the lead collimator, widens the accepted pulse-height window to include low-energy events, and uses a scintillation crystal that produces only half as much visible light per absorbed gamma photon. Evaluate how each change affects the count rate, spatial information, image quality and detector output.
[6 marks]
Total for this question: 6

3.10.6.4 · Use of high-energy X-rays

Explanation

External radiotherapy directs high-energy X-rays into a tumour.
Their penetration delivers energy at depth, where ionisation damages DNA and prevents successful cell division.
Several beam directions can converge on the tumour: the target receives the combined dose, while each region of healthy tissue lies in fewer paths and receives less dose.
Collimation shapes each field, shielding protects selected regions, and imaging plus computer planning locate the tumour and choose beam directions.
Healthy tissue cannot receive zero dose, so treatment planning balances tumour control against damage to surrounding organs.

External X-ray beams from several directions converge on a tumour.

Worked example

Explain how using three beam directions reduces harm compared with delivering the entire tumour dose through one path.

1.Each beam contributes only part of the prescribed dose.
2.All three overlap at the tumour, so their doses add at the target.
3.Most surrounding tissue lies in only one beam path and receives a smaller dose.

Answer: Beam convergence maintains tumour dose while spreading the unavoidable healthy-tissue dose across different regions.

Common mistakes

Don't claim that surrounding healthy tissue receives no radiation.
Don't name collimation without explaining that it restricts the tissue inside the treatment field.
Don't list shielding without explaining that it limits exposure outside the planned field.

Exam tip

Use comparative dose language: full combined dose at the tumour, smaller dose in each healthy-tissue path.

Tier 1 · Easy

State why high-energy X-rays are used to treat a tumour deep inside the body.
[1 mark]
Total for this question: 1
Give one way shielding is used to limit exposure of healthy cells during external X-ray treatment.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Explain how directing X-ray beams at a tumour from several angles can reduce damage to healthy tissue.
[3 marks]
Total for this question: 3
Explain how imaging and computer planning can limit the exposure of healthy cells during external X-ray treatment.
[3 marks]
Total for this question: 3
A radiotherapy planning grid has equal-area cells of $0.50\,\text{cm}^2$ . A tumour covers $20$ cells. An unshaped rectangular field contains $48$ cells, whereas a tumour-matched field contains $26$ cells and still covers the whole tumour. Calculate and compare the healthy-tissue area inside the two fields, then explain why field shaping reduces off-target exposure.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A tumour of mass $0.18\,\text{kg}$ receives $1.8\,\text{Gy}$ in each of $20$ treatment sessions. Calculate the total energy absorbed by the tumour. Explain two methods, other than reducing the prescribed tumour dose, that limit exposure of healthy cells. Use $1\,\text{Gy}=1\,\text{J kg}^{-1}$ .
[5 marks]
Total for this question: 5
Four equal X-ray beams approach a tumour from different directions. A healthy region lies in the path of only one beam and receives $30\%$ of the energy absorbed per kilogram that this beam delivers at the tumour. Calculate the healthy region's absorbed energy per kilogram as a percentage of the tumour's total, and compare it with using one beam under the same conditions.
[5 marks]
Total for this question: 5
Two X-ray beam plans deliver the same total dose to a tumour. In plan A, three beams provide $40\%$ , $35\%$ and $25\%$ of the tumour dose; a nearby organ receives $20\%$ , $8.0\%$ and $0\%$ of those respective contributions. In plan B, three unequal beams provide $50\%$ , $30\%$ and $20\%$ of the tumour dose; the organ receives $12\%$ , $4.0\%$ and $0\%$ of those respective contributions. Calculate the organ dose for each plan as a percentage of the tumour dose and determine the safer plan.
[5 marks]
Total for this question: 5
A patient says that high-energy X-rays used to treat a deep tumour can be made to affect tumour cells only. Evaluate this statement. Explain why high-energy X-rays are used, how they can stop tumour growth and why treatment can also harm healthy cells.
[5 marks]
Total for this question: 5
A tumour cell absorbs $4.6\times10^{-11}\,\text{J}$ from high-energy X-rays. The mean energy transferred per ionisation is $32\,\text{eV}$ . Estimate the number of ionisations produced. Explain how the resulting cell damage can prevent tumour-cell division and why the same absorbed energy would be harmful to a healthy cell. Use $e=1.60\times10^{-19}\,\text{C}$ .
[5 marks]
Total for this question: 5

3.10.6.5 · Use of radioactive implants

Explanation

Brachytherapy places sealed radioactive sources inside or very close to a tumour, producing a high local dose while reducing irradiation of distant tissue. Beta-emitting implants are suitable because beta particles are ionising but have a short range in tissue; a more penetrating gamma source would expose more healthy tissue beyond the target.
Several small sources can distribute dose through an irregular tumour. Activity, half-life, source position and treatment time determine the delivered dose, and temporary implants can be removed after the planned exposure.
Staff exposure is limited by short handling times, distance, shielding and remote handling tools.
The benefit must be linked to localisation of dose, not merely to beta being less penetrating.
When the activity falls appreciably during a treatment, the number of decays is $N_0(1-e^{-\lambda t})$ with $N_0=A_0/\lambda$ , not $A_0t$ .

Worked example

Explain why a beta-emitting implant may be preferred to a gamma-emitting implant for a small local tumour.

1.Beta radiation is ionising and can damage tumour-cell DNA.
2.Its short range deposits most energy close to the implanted source.
3.Gamma radiation would travel farther and irradiate more healthy tissue outside the target.

Answer: The beta implant concentrates an effective ionising dose within the local tumour while limiting distant exposure.

Common mistakes

Don't state only that beta is less penetrating without linking range to localised tumour dose.
Don't place the implant outside the patient and describe it as external-beam therapy.
Don't ignore source half-life and treatment time when discussing delivered dose.

Exam tip

For a justify question, connect radiation range directly to the required target size and protection of surrounding tissue.

Tier 1 · Easy

State why a beta emitter is suitable for a radioactive implant placed inside a tumour.
[1 mark]
Total for this question: 1
State one advantage of distributing several small beta-emitting implants through an irregularly shaped tumour.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Explain why a beta-emitting implant may expose less healthy tissue than an external high-energy X-ray beam used to deliver the same tumour dose.
[3 marks]
Total for this question: 3
Explain why a temporary implant is removed once the prescribed dose has been delivered.
[2 marks]
Total for this question: 2
Implant A has activity $4.0\,\text{MBq}$ and is used for $30\,\text{min}$ ; implant B has activity $2.0\,\text{MBq}$ and is used for $60\,\text{min}$ . Their half-lives are much longer than the treatment times. Beta particles from either implant have range $4.0\,\text{mm}$ in tissue. Determine whether the implants produce the same number of decays and whether either can irradiate tumour cells $6.0\,\text{mm}$ from the source.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A temporary beta-emitting implant has initial activity $9.6\,\text{MBq}$ and physical half-life $48\,\text{h}$ . Calculate its activity after $120\,\text{h}$ . Explain how the position and radiation type of the implant help protect healthy tissue.
[5 marks]
Total for this question: 5
A beta-emitting implant has constant activity $2.4\,\text{MBq}$ during a $1.8\times10^3\,\text{s}$ treatment. The mean beta energy is $0.52\,\text{MeV}$ and $65\%$ of the emitted energy is absorbed by a $0.018\,\text{kg}$ tumour. Calculate the tumour's absorbed dose. Use $e=1.60\times10^{-19}\,\text{C}$ and $1\,\text{Gy}=1\,\text{J kg}^{-1}$ .
[5 marks]
Total for this question: 5
A beta-emitting implant must give a $0.423\,\text{Gy}$ absorbed dose to a $0.020\,\text{kg}$ tumour in $9.0\,\text{h}$ . The implant half-life is $6.0\,\text{h}$ , the mean energy per decay is $1.12\times10^{-13}\,\text{J}$ , and the tumour absorbs $75\%$ of the emitted energy. Calculate the initial activity of the implant, allowing for its activity decrease. Explain why multiplying the initial activity by the treatment time would give an incorrect number of decays. Use $1\,\text{Gy}=1\,\text{J kg}^{-1}$ .
[4 marks]
Total for this question: 4
Three identical beta-emitting implants release a total of $0.090\,\text{J}$ during treatment. A correctly positioned implant deposits $82\%$ of its energy in a $0.015\,\text{kg}$ tumour, with the remaining $18\%$ absorbed in one shared $0.040\,\text{kg}$ region of nearby healthy tissue. One implant is misplaced and instead deposits $25\%$ in the tumour and $75\%$ in the healthy tissue. Calculate both absorbed doses for this arrangement and compare them with the doses when all three implants are correctly positioned. Use $1\,\text{Gy}=1\,\text{J kg}^{-1}$ .
[6 marks]
Total for this question: 6
Calculate the minimum number and suitable positions of identical beta-emitting implants along the centre line of an $18\,\text{mm}$ by $8.0\,\text{mm}$ rectangular tumour, measured from one short edge. Beta particles have an effective range of $5.0\,\text{mm}$ in the tissue, and every point of the tumour must be within range of a source. The total activity must be $8.1\,\text{MBq}$ ; calculate the activity of each source and explain why using several sources helps localise the dose.
[6 marks]
Total for this question: 6

3.10.6.6 · Imaging comparisons

Explanation

Plain X-rays are fast, available and resolve bone well, but use ionising radiation, superpose structures and give poor soft-tissue contrast. CT improves localisation through cross-sectional images but generally costs more and gives a higher ionising dose.
Ultrasound is portable, inexpensive, real-time and non-ionising, but is operator-dependent and transmits poorly through bone or gas.
MR gives excellent soft-tissue contrast without ionising radiation but is slower, costly and less convenient for some patients.
Radionuclide and PET scans reveal physiological function, yet require internal ionising tracers and usually have lower spatial resolution.
A valid comparison applies resolution, safety and convenience to the specified organ and diagnostic purpose.

Worked example

Compare ultrasound and CT for repeated imaging of a moving fetus.

1.Ultrasound is non-ionising and provides real-time images, supporting repeated monitoring.
2.CT uses ionising X-rays and generally delivers a greater dose.
3.Although CT may provide different anatomical detail, its radiation risk makes it unsuitable for routine repeated fetal imaging.

Answer: Ultrasound is preferred because it gives convenient real-time imaging without ionising dose.

Common mistakes

Don't declare one technique universally best without relating it to the tissue or clinical need.
Don't call MR an ionising X-ray technique.
Don't compare image quality vaguely without distinguishing spatial resolution, soft-tissue contrast and functional information.

Exam tip

Structure comparisons as paired consequences for the stated case: resolution or contrast, convenience, then ionising-radiation risk.

Tier 1 · Easy

Select a suitable imaging technique for repeated bedside monitoring of moving soft tissue when ionising radiation must be avoided.
[1 mark]
Total for this question: 1
Give a suitable imaging technique for a quick, low-cost examination of a suspected broken wrist.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Compare CT and MR for repeated imaging of a brain tumour. Refer to soft-tissue contrast, patient dose and convenience.
[3 marks]
Total for this question: 3
Compare MR and ultrasound for imaging a suspected deep knee-ligament tear.
[3 marks]
Total for this question: 3
Explain why a radionuclide gamma-camera scan may reveal reduced kidney function when a CT image shows no structural abnormality.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A patient may have a kidney stone. Evaluate ultrasound, CT and MR for locating the stone, using spatial resolution, soft-tissue or stone contrast, radiation dose, availability and convenience. Reach a justified recommendation.
[6 marks]
Total for this question: 6
Evaluate PET and MR for investigating whether a soft-tissue tumour has returned after treatment. Give a justified recommendation for the main imaging technique.
[5 marks]
Total for this question: 5
A patient needs both a structural image and a map of activity for a suspected $9\,\text{mm}$ liver lesion. Total scanning time must not exceed $60\,\text{min}$ , and the stated ionising-dose limit is $1.0$ relative dose unit. Ultrasound gives $4\,\text{mm}$ resolution in $20\,\text{min}$ with zero relative dose units; MR gives $2\,\text{mm}$ in $50\,\text{min}$ with zero; CT gives $1\,\text{mm}$ in $8\,\text{min}$ at $1.15$ times the dose limit; PET maps activity at $6\,\text{mm}$ resolution in $35\,\text{min}$ at $0.85$ times the limit. Determine the pair of techniques that meets every requirement and justify why the other structural techniques cannot replace the chosen one.
[5 marks]
Total for this question: 5
A hospital proposes replacing CT with either ultrasound or MR for every patient with a suspected bleed inside the skull, because both alternatives are non-ionising. Evaluate this proposal in terms of the information required, image resolution, safety and practical convenience.
[6 marks]
Total for this question: 6
A high-resolution CT dataset of an organ has already been recorded. A team claims that software processing can turn it into a map of metabolic activity, convert it into a live record of motion and remove the radiation dose already delivered. Evaluate each claim and explain when PET or ultrasound would supply genuinely new information.
[6 marks]
Total for this question: 6

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

3.10.1.1 · Physics of vision

Tier 1 · Easy

Mark scheme for 3.10.1.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Rod cells; the image has no colour or has low spatial resolution.	Low light levels stimulate the more sensitive rod cells. Rods do not provide colour vision, and the convergence of several rods onto one nerve pathway gives low spatial resolution; either image property earns the second marking point.	2
02.1	Cone cells; neighbouring cones show little convergence and usually send signals along separate pathways.	Cones operate best in bright light and provide colour vision. Their pathways show little convergence, so signals from neighbouring retinal positions remain separate.	2

Tier 2 · Standard

Mark scheme for 3.10.1.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Several rod cells can share one nerve pathway, so signals from the two illuminated rods may be combined and the brain cannot distinguish their positions; sensitivity is increased but spatial resolution is reduced.	Identify convergence: several rods connect to a common sensory pathway. Their signals are combined, which helps a small light stimulus reach the detection threshold. Because the same pathway does not identify which individual rod responded, the two source positions cannot be resolved.	3
02.1	The ciliary muscles contract so the lens becomes more curved. Its power increases and its focal length decreases, allowing the more divergent rays from the nearer letter to form a real image on the retina.	Moving the object closer makes the rays reaching the eye more divergent. Award an equivalent accommodation description using ciliary-muscle contraction, increased lens curvature, increased power or decreased focal length when it is linked correctly. The additional refraction brings the rays to a focus at the fixed retinal surface.	3
03.1	The rays should refract at the cornea, which provides most of the eye's refraction, and refract further at the lens. They should converge on the retina to form a real, inverted image.	Correct the optical sequence: the air-cornea boundary provides most of the refraction, and the lens provides further adjustable refraction. The emerging rays must actually converge at the retinal surface, so the image there is real. A converging eye system forms an inverted retinal image, not an upright virtual one.	4

Tier 3 · Hard

Mark scheme for 3.10.1.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	In bright light cones respond, so the grid is seen in colour and with high spatial resolution because cone pathways show little convergence. In dim light rods dominate, so colour is lost and fine lines are less easily resolved because many rods share a pathway, although this convergence makes the eye more sensitive to weak light.	For bright light, award the linked chain cones, colour vision, and high resolution from separate or weakly convergent pathways. For dim light, award rods, no colour, and reduced resolution from convergence. Complete the comparison by linking that convergence to greater sensitivity because signals from several rods can combine.	5
02.1	Slightly away from the centre of vision, rod cells dominate. Many rods converge onto one pathway, allowing weak signals to combine and making the faint star easier to detect. Direct vision uses the cone-rich central region; cone pathways have little convergence, so the positions of the two bright images remain separate and spatial resolution is greater.	For the faint object, link peripheral rod vision to convergence and summation, which increase sensitivity. For the close pair, link direct cone-rich vision to low convergence, so signals from neighbouring image positions are not combined. This preserves positional information and allows the pair to be resolved.	5
03.1	The large patch can still be identified by colour because the cone types and their responses are unchanged. The fine grid has poor spatial resolution or appears scrambled because cone signals no longer preserve retinal position. The faint grey patch is detected normally because low-light vision uses the unaffected, convergent rod pathways.	Separate the information encoded by receptor type from that encoded by pathway position. Preserved cone types retain the relative responses needed for the colour of a large uniform area. Randomised cone positions destroy the spatial mapping needed to reproduce fine detail. In very low light rods dominate; because their sensitivity and convergent pathways are unchanged, detection of the faint grey patch is not impaired by this cone-pathway fault.	5
04.1	In X, each pathway receives only $0.11\,\text{mV}$ , so neither point is detected. In X the separate cone pathways would resolve the two points if the light reached threshold. In Y the faint light is detected: the nine illuminated rods supply $9(0.11)=0.99\,\text{mV}$ . With both images present the pathway receives at most $12(0.11)=1.32\,\text{mV}$ and still gives one output. The images therefore activate the same convergent pathway, so they are perceived as one point and are not resolved.	For X, compare the single-cone input directly with threshold: $0.11<0.90\,\text{mV}$ , so neither separate pathway fires. For Y, nine rods of the group are illuminated, giving $9(0.11)=0.99\,\text{mV}>0.90\,\text{mV}$ . Rod convergence therefore increases sensitivity. Because both images feed one pathway, the output contains no separate positional signals, so spatial resolution is lost.	6
05.1	The threshold intensity at $650\,\text{nm}$ is $3.2\times10^{-8}\,\text{W m}^{-2}$ . Rods dominate at these very low intensities; they do not distinguish colour and many converge on each nerve pathway, reducing spatial resolution. At much greater illumination, cones respond, providing colour vision and higher spatial resolution because their pathways show little convergence.	Threshold sensitivity is inversely proportional to the intensity needed for detection, so $S_{510}/S_{650}=I_{650}/I_{510}=16$ . Therefore $I_{650}=16(2.0\times10^{-9})=3.2\times10^{-8}\,\text{W m}^{-2}$ . Near threshold, the response is rod-dominated: rods do not encode colour and their convergent pathways combine positional information. Increased illumination activates cones; the different cone responses support colour vision, while little pathway convergence preserves fine spatial detail.	5

3.10.1.2 · Defects of vision and their correction using lenses

Tier 1 · Easy

Mark scheme for 3.10.1.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	A diverging (concave) lens with negative power.	A myopic eye is too powerful, so parallel rays would focus before the retina. A diverging lens reduces the convergence before the rays enter the eye; its focal length and power are negative.	2
02.1	A cylindrical lens component is used; the axis gives the orientation of that cylindrical correction.	Astigmatism involves unequal focusing in different planes. A cylindrical component supplies direction-dependent power, and its axis from $0^\circ$ to $180^\circ$ specifies the orientation of the correction.	2

Tier 2 · Standard

Mark scheme for 3.10.1.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$v=+0.267\,\text{m}$ and $\|m\|=0.667$	The focal length is $f=1/P=1/6.25=0.160\,\text{m}$ . Hence $1/v=1/f-1/u=6.25-1/0.400=3.75\,\text{m}^{-1}$ , so $v=+0.267\,\text{m}$ . The magnification magnitude is $\|m\|=\|v/u\|=0.2667/0.400=0.667$ .	3
02.1	$f=-0.303\,\text{m}$ ; it is a diverging lens for myopia and forms a virtual image at the eye's far point.	Use $P=1/f$ , so $f=1/P=1/(-3.30)=-0.30303\,\text{m}=-0.303\,\text{m}$ to $3$ significant figures. The negative power identifies a diverging lens. It corrects myopia by making light from a distant object appear to come from a virtual image at the eye's far point.	3
03.1	The spherical component acts in every plane. The cylindrical component supplies additional correction in the plane perpendicular to its axis. The axis gives the orientation along which the cylindrical component has zero power.	Interpret the prescription rather than combining its numerical values. The sphere provides the same correction in all planes. The cylinder adds direction-dependent correction in the plane perpendicular to the stated axis, while the axis specifies the orientation with no cylindrical power.	2

Tier 3 · Hard

Mark scheme for 3.10.1.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$+2.8\,\text{D}$	The object is real, so $u=+0.25\,\text{m}$ . The required image is virtual, so $v=-0.80\,\text{m}$ . Then $1/f=1/u+1/v=1/0.25+1/(-0.80)=4.00-1.25=2.75\,\text{m}^{-1}$ . Therefore $P=1/f=+2.75\,\text{D}$ , which is $+2.8\,\text{D}$ to $2$ significant figures. The positive sign confirms that a converging lens is required.	5
02.1	$P=-1.1\,\text{D}$ ; the phone produces a virtual image $0.28\,\text{m}$ in front of the eye, within its clear-focus range.	For the distant sign, the lens must form a virtual image at the far point: $u=\infty$ and $v=-0.90\,\text{m}$ . Hence $P=1/u+1/v=-1/0.90=-1.11\,\text{D}=-1.1\,\text{D}$ to $2$ significant figures. For the phone, $1/v=P-1/u=-1.11-1/0.40=-3.61\,\text{m}^{-1}$ , so $v=-0.277\,\text{m}=-0.28\,\text{m}$ . This virtual image lies between $0.25\,\text{m}$ and $0.90\,\text{m}$ in front of the eye, so the same lens allows the phone to be focused.	5
03.1	M requires $-1.54\,\text{D}$ , a diverging lens. H requires $+2.22\,\text{D}$ , a converging lens.	For M, a distant object has $u=\infty$ and must form a virtual image at the far point, so $v=-0.65\,\text{m}$ . Hence $P=1/u+1/v=0-1/0.65=-1.538\,\text{D}$ , giving $-1.54\,\text{D}$ and therefore a diverging lens. For H, $u=+0.30\,\text{m}$ and the required virtual image is at the near point, so $v=-0.90\,\text{m}$ . Thus $P=1/0.30+1/(-0.90)=2.222\,\text{D}$ , giving $+2.22\,\text{D}$ and therefore a converging lens.	5
04.1	The closest object is $0.285\,\text{m}$ from the lens and the magnification magnitude is $3.86$ . The converging lens forms an enlarged virtual image at the eye's unaided near point, where the hypermetropic eye can focus it.	At the closest clear position the lens must form a virtual image at the unaided near point, so $v=-1.10\,\text{m}$ . Using $P=1/u+1/v$ , $1/u=2.60+1/1.10=3.50909\,\text{m}^{-1}$ and $u=0.284974\,\text{m}=0.285\,\text{m}$ . The magnification magnitude is $\|m\|=\|v/u\|=1.10/0.284974=3.86$ . The positive-power lens adds convergence, allowing a nearer object to produce a virtual image at a distance the eye can focus.	5
05.1	The contact-lens power is $-2.17\,\text{D}$ and the spectacle-lens power is $-2.25\,\text{D}$ . The spectacle lens sits $16\,\text{mm}$ nearer the far point, so it must form the virtual image only $0.444\,\text{m}$ away instead of $0.460\,\text{m}$ ; the shorter image distance needs a slightly larger magnitude of negative power.	For a distant object, $u=\infty$ and $P=1/v$ . The contact lens is at the cornea, so $v=-0.460\,\text{m}$ and $P=-1/0.460=-2.17391\,\text{D}=-2.17\,\text{D}$ . The far point is $0.460-0.016=0.444\,\text{m}$ in front of the spectacle lens, so $v=-0.444\,\text{m}$ and $P=-1/0.444=-2.25225\,\text{D}=-2.25\,\text{D}$ . Both lenses are diverging; the smaller lens-to-far-point distance for the spectacles ( $0.444\,\text{m}$ against $0.460\,\text{m}$ ) requires the slightly larger magnitude of power.	5

3.10.2.1 · Ear as a sound detection system

Tier 1 · Easy

Mark scheme for 3.10.2.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	The tympanic membrane (eardrum) and the ossicles.	Sound pressure variations act on the tympanic membrane. Its vibration is transmitted through the malleus, incus and stapes, collectively called the ossicles.	2
02.1	$1.4\times10^{-6}\,\text{N}$	Use $F=pA$ . Therefore $F=0.030(4.8\times10^{-5})=1.44\times10^{-6}\,\text{N}=1.4\times10^{-6}\,\text{N}$ to $2$ significant figures.	2

Tier 2 · Standard

Mark scheme for 3.10.2.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The ossicles provide a lever force gain, and this force acts on the oval window, whose area is much smaller than the tympanic membrane; since pressure is force divided by area, the pressure increases.	First identify the lever action of the ossicles and link it to an increased force. Then compare areas: the oval window is smaller than the tympanic membrane. Using $p=F/A$ , a larger force acting over a smaller area gives a greater pressure in the inner-ear fluid.	3
02.1	The frequency is unchanged, the displacement amplitude at the oval window is smaller, and the pressure is greater.	The ossicles transmit a forced vibration, so the oval window vibrates at the driving frequency. Its displacement amplitude is reduced, but the ossicle lever action and the smaller oval-window area increase the pressure delivered to the inner-ear fluid.	3
03.1	The tympanic membrane vibrates and the ossicles transmit the vibration to the oval window. The oval window sets up a pressure wave in the inner-ear fluid, which stimulates receptors producing electrical impulses in the auditory nerve.	Follow the three credited stages in order: vibration of the tympanic membrane is passed through the ossicles to the oval window; the oval window produces a pressure wave in the inner-ear fluid; receptors are stimulated and produce electrical impulses in the auditory nerve.	3

Tier 3 · Hard

Mark scheme for 3.10.2.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.29\,\text{Pa}$	The force on the tympanic membrane is $F_1=p_1A_1=0.012(5.5\times10^{-5})=6.6\times10^{-7}\,\text{N}$ . The ossicle output force is $F_2=1.4F_1=9.24\times10^{-7}\,\text{N}$ . Therefore $p_2=F_2/A_2=(9.24\times10^{-7})/(3.2\times10^{-6})=0.28875\,\text{Pa}=0.29\,\text{Pa}$ to $2$ significant figures.	5
02.1	The required ossicle force gain is $1.6$ . The area reduction contributes more because the tympanic-to-oval-window area ratio is $18.8$ , much greater than the lever gain of $1.6$ .	The tympanic force is $F_1=p_1A_1=0.020(6.4\times10^{-5})=1.28\times10^{-6}\,\text{N}$ . The required oval-window force is $F_2=p_2A_2=0.62(3.4\times10^{-6})=2.11\times10^{-6}\,\text{N}$ . Hence the required force gain is $F_2/F_1=2.108/1.28=1.646875=1.6$ to $2$ significant figures. The area ratio is $A_1/A_2=(6.4\times10^{-5})/(3.4\times10^{-6})=18.8$ , so the area reduction contributes much more than the lever gain.	4
03.1	Without the ossicles, vibration of the tympanic membrane is not transmitted effectively to the oval window. The oval window therefore produces a much smaller pressure wave in the inner-ear fluid, so the receptors produce far fewer electrical impulses in the auditory nerve. The tuning fork vibrates the skull and inner ear directly, bypassing the missing ossicles, so some receptor stimulation remains.	For air conduction, trace the missing middle-ear link: the tympanic membrane still vibrates, but without the ossicles it does not drive the oval window effectively. The reduced inner-ear pressure wave gives less receptor stimulation and fewer auditory-nerve impulses. For bone conduction, vibration passes through the skull to the inner ear, bypassing the tympanic membrane-to-oval-window route and giving the distinct fourth credit point.	4
04.1	The force amplitude is restored, but the larger piston area produces a smaller fluid pressure amplitude. The weaker pressure wave stimulates the inner-ear receptors less, so they produce fewer electrical impulses in the auditory nerve and the sound is perceived as quieter than with a normal oval window.	Use the relationship qualitatively: for the same force amplitude, spreading the force over a larger area reduces pressure. The oversized piston therefore produces a lower-amplitude pressure wave in the inner-ear fluid. This gives less receptor stimulation, fewer auditory-nerve impulses and a quieter perceived sound.	4
05.1	The fluid pressure amplitude is $0.51\,\text{Pa}$ and the fluid vibrates at $1.3\,\text{kHz}$ . The pressure wave stimulates inner-ear receptors, which convert the mechanical disturbance into electrical impulses in the auditory nerve.	The force delivered to the fluid is $0.72(1.9\times10^{-6})=1.368\times10^{-6}\,\text{N}$ . Hence $p=F/A=(1.368\times10^{-6})/(2.7\times10^{-6})=0.50667\,\text{Pa}=0.51\,\text{Pa}$ to $2$ significant figures. A forced vibration retains the driving frequency, so the fluid wave is at $1.3\,\text{kHz}$ . This mechanical pressure wave stimulates sensory receptors, producing electrical impulses in the auditory nerve.	5

3.10.2.2 · Sensitivity and frequency response

Tier 1 · Easy

Mark scheme for 3.10.2.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$40\,\text{dB}$	Use $L=10\log_{10}(I/I_0)$ . Thus $L=10\log_{10}((1.0\times10^{-8})/(1.0\times10^{-12}))=10\log_{10}(10^4)=40\,\text{dB}$ .	2
02.1	$8.0\times10^{-4}\,\text{W m}^{-2}$	For an isotropic source, $I=P/(4\pi r^2)$ . Therefore $I=0.040/(4\pi(2.0)^2)=7.96\times10^{-4}\,\text{W m}^{-2}$ , which is $8.0\times10^{-4}\,\text{W m}^{-2}$ to $2$ significant figures.	2

Tier 2 · Standard

Mark scheme for 3.10.2.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$I=2.0\times10^{-4}\,\text{W m}^{-2}$ and $P=4.0\times10^{-7}\,\text{W}$ .	Rearrange $L=10\log_{10}(I/I_0)$ to give $I=I_0 10^{L/10}$ . Hence $I=(1.0\times10^{-12})10^{83/10}=1.9952623\times10^{-4}\,\text{W m}^{-2}=2.0\times10^{-4}\,\text{W m}^{-2}$ to $2$ significant figures. Using the unrounded intensity, $P=IA=(1.9952623\times10^{-4})(2.0\times10^{-3})=3.9905246\times10^{-7}\,\text{W}=4.0\times10^{-7}\,\text{W}$ to $2$ significant figures.	3
02.1	A decrease of $12\,\text{dB}$ (accept $12.0\,\text{dB}$ ).	Intensity follows the inverse-square law, so $I_2/I_1=(2.5/10.0)^2=1/16$ . Hence $\Delta L=10\log_{10}(I_2/I_1)=10\log_{10}(1/16)=-12.04\,\text{dB}$ . The level therefore decreases by $12\,\text{dB}$ to $2$ significant figures; $12.0\,\text{dB}$ is also acceptable.	3
03.1	$58\,\text{dBA}$ at $125\,\text{Hz}$ and $75\,\text{dBA}$ at $2.0\,\text{kHz}$ ; the larger weighted reading represents the ear's greater sensitivity to the $2.0\,\text{kHz}$ tone despite equal physical intensity levels.	Apply the supplied weighting to each unweighted level: $74-16=58\,\text{dBA}$ and $74+1=75\,\text{dBA}$ . The $17\,\text{dBA}$ difference is not a physical-intensity difference; it results from frequency weighting that approximates the greater human sensitivity at $2.0\,\text{kHz}$ .	3

Tier 3 · Hard

Mark scheme for 3.10.2.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$14\,\text{m}$ ; normal hearing is less sensitive at $12\,\text{kHz}$ than near $3\,\text{kHz}$ , so the higher-frequency tone has a greater equal-loudness threshold.	First find the intensity: $I=I_0 10^{L/10}=(1.0\times10^{-12})10^{8.5}=3.16\times10^{-4}\,\text{W m}^{-2}$ . For an isotropic source, $I=P/(4\pi r^2)$ , so $r=\sqrt{P/(4\pi I)}=\sqrt{0.80/(4\pi(3.16\times10^{-4}))}=14.2\,\text{m}$ , giving $14\,\text{m}$ to $2$ significant figures. Equal-loudness curves have their minimum near a few kilohertz; at $12\,\text{kHz}$ a larger level is required for the same perceived loudness.	5
02.1	The $200\,\text{Hz}$ intensity is $20$ times the $1.0\,\text{kHz}$ intensity. Since $P=IA$ , the incident-power ratio for equal areas is also $20$ . The ear is less sensitive at $200\,\text{Hz}$ .	The level difference is $53-40=13\,\text{dB}$ . Therefore $I_{200}/I_{1000}=10^{13/10}=19.95=20$ to $2$ significant figures. Since $P=IA$ , equal receiving areas give the same power ratio. Requiring $20$ times the intensity for equal loudness shows that hearing is less sensitive at $200\,\text{Hz}$ than at $1.0\,\text{kHz}$ .	4
03.1	The protected levels are $81\,\text{dB}$ at $250\,\text{Hz}$ and $61\,\text{dB}$ at $4.0\,\text{kHz}$ . The protected $250\,\text{Hz}$ tone is equivalent to a $4.0\,\text{kHz}$ tone about $5\,\text{dB}$ above the protected $4.0\,\text{kHz}$ level, so it is perceived as louder.	For the low tone, $L'=92+10\log_{10}(0.080)=81.03\,\text{dB}$ . For the high tone, $L'=84+10\log_{10}(0.0050)=60.99\,\text{dB}$ . A level of $81\,\text{dB}$ at $250\,\text{Hz}$ has the same loudness as about $81-15=66\,\text{dB}$ at $4.0\,\text{kHz}$ . This exceeds the actual protected $4.0\,\text{kHz}$ level by about $5\,\text{dB}$ , so the $250\,\text{Hz}$ tone is perceived as louder.	4
04.1	The combined level is $80\,\text{dB}$ , giving $72\,\text{dBA}$ after weighting. It does not exceed the $75\,\text{dBA}$ limit.	Add intensities, not decibel readings. Relative to the same reference intensity, $I/I_0=10^{73/10}+10^{79/10}=1.995\times10^7+7.943\times10^7=9.938\times10^7$ . Thus $L=10\log_{10}(9.938\times10^7)=79.973\,\text{dB}=80\,\text{dB}$ . Apply the supplied weighting: $L_A=79.973-8=71.973\,\text{dBA}=72\,\text{dBA}$ , which is below $75\,\text{dBA}$ .	5
05.1	Speaker B requires an acoustic power of $1.1\times10^{-2}\,\text{W}$ (accept $1.07\times10^{-2}\,\text{W}$ ).	Equal loudness requires $L_A-L_B=14\,\text{dB}$ , so $I_A/I_B=10^{14/10}=25.1189$ . Now $I_A=0.12/(4\pi(6.0)^2)$ and $I_B=P_B/(4\pi(9.0)^2)$ . Therefore $P_B=0.12(9.0/6.0)^2/25.1189=0.0107489\,\text{W}=1.1\times10^{-2}\,\text{W}$ to $2$ significant figures. The smaller high-frequency intensity is sufficient because the ear is more sensitive at $2.5\,\text{kHz}$ than at $160\,\text{Hz}$ .	4

3.10.2.3 · Defects of hearing

Tier 1 · Easy

Mark scheme for 3.10.2.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	At high frequencies.	Age-related hearing loss does not raise the threshold equally at all frequencies. The characteristic change is a larger loss towards the high-frequency end of an equal-loudness or threshold graph.	1
02.1	Prolonged exposure to excessive noise.	A localised loss or notch near $4\,\text{kHz}$ is characteristic of deterioration associated with prolonged excessive-noise exposure.	1

Tier 2 · Standard

Mark scheme for 3.10.2.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$30\,\text{dB}$ and a factor of $1.0\times10^3$	The hearing loss is the vertical threshold difference: $38-8=30\,\text{dB}$ . Using $\Delta L=10\log_{10}(I_2/I_1)$ gives $I_2/I_1=10^{30/10}=10^3$ . The impaired threshold intensity is therefore $1000$ times the normal value.	3
02.1	$26\,\text{dB}$ ; the impaired curve is displaced upwards by $26\,\text{dB}$ at this frequency.	Hearing loss is the level difference, so $\Delta L=10\log_{10}(I_{\text{impaired}}/I_{\text{normal}})=10\log_{10}(400)=26.02\,\text{dB}$ . This is $26\,\text{dB}$ to $2$ significant figures. The impaired ear therefore needs a level $26\,\text{dB}$ higher for the same loudness, so its equal-loudness curve lies $26\,\text{dB}$ above the normal curve at that frequency.	3
03.1	The impaired equal-loudness curve is raised by different amounts at different frequencies, with the greatest upward shift at high frequency. A uniform level increase raises all frequency components equally, so it does not compensate for the larger high-frequency loss and the relative loudness balance remains abnormal.	Age-related deterioration is frequency dependent rather than a single threshold shift. The high-frequency part of the equal-loudness curve moves upward more, meaning that a larger added level is required there. Adding the same level everywhere leaves the difference between low- and high-frequency hearing losses uncompensated.	3

Tier 3 · Hard

Mark scheme for 3.10.2.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	A is consistent with age-related deterioration because the loss rises strongly at high frequency. B is consistent with excessive-noise exposure because its largest loss is near $4\,\text{kHz}$ . Each equal-loudness curve is shifted upward most in its affected range, so greater intensity is needed there and sounds at those frequencies are perceived as quieter.	Compare the frequency dependence rather than the total loss. A's largest value is at $10\,\text{kHz}$ , matching age-related high-frequency loss. B has a pronounced maximum at $4\,\text{kHz}$ , matching excessive-noise exposure. A positive hearing loss raises the required level on the equal-loudness curve; the larger the loss, the larger the upward displacement and the quieter an unchanged sound is perceived.	5
02.1	$50\,\text{dB}$ ; the listener's threshold curve has a narrow upward notch of about $50\,\text{dB}$ near $4.0\,\text{kHz}$ but is close to normal at neighbouring frequencies. This is consistent with prolonged exposure to excessive noise.	The threshold ratio is $(2.0\times10^{-7})/(2.0\times10^{-12})=1.0\times10^5$ . Therefore the hearing loss is $10\log_{10}(10^5)=50\,\text{dB}$ . This produces a local upward displacement of about $50\,\text{dB}$ in the threshold curve at $4.0\,\text{kHz}$ , while the neighbouring parts remain close to normal. A pronounced local notch around $4\,\text{kHz}$ indicates excessive-noise deterioration rather than the broad high-frequency loss typical of ageing.	4
03.1	For the normal listener, $I_{4.0\,\text{kHz}}/I_{0.50\,\text{kHz}}=0.20$ . For the listener with hearing damage, the ratio is $100$ . The reversal shows a large loss of sensitivity near $4.0\,\text{kHz}$ relative to $0.50\,\text{kHz}$ .	Use $I_2/I_1=10^{(L_2-L_1)/10}$ . For the normal curve, $L_{4.0}-L_{0.50}=31-38=-7\,\text{dB}$ , so the ratio is $10^{-0.7}=0.1995\approx0.20$ . For the damaged-hearing curve, the difference is $73-53=20\,\text{dB}$ , giving $10^2=100$ . Equal loudness therefore changes from needing less intensity at $4.0\,\text{kHz}$ to needing much more, showing frequency-dependent hearing loss around $4.0\,\text{kHz}$ .	4
04.1	The aided listener requires external levels of $48\,\text{dB}$ at $0.75\,\text{kHz}$ and $46\,\text{dB}$ at $3.5\,\text{kHz}$ . These remain $2\,\text{dB}$ and $7\,\text{dB}$ above the normal requirements, so the aid improves both frequencies but under-corrects the high-frequency loss more strongly and does not fully restore the normal balance.	Subtract the aid gain from the impaired level required at each frequency. The external requirements are $58-10=48\,\text{dB}$ and $76-30=46\,\text{dB}$ . Compare with the normal curve: the residual losses are $48-46=2\,\text{dB}$ and $46-39=7\,\text{dB}$ . Because the residual shift is frequency dependent and larger at $3.5\,\text{kHz}$ , the aid has not restored the original equal-loudness relationship.	5
05.1	The combined threshold is $3.8\times10^{-11}\,\text{W m}^{-2}$ and the hearing loss is $21\,\text{dB}$ (accept $3.78\times10^{-11}\,\text{W m}^{-2}$ and $21.0\,\text{dB}$ ). The curve has a broad upward displacement at high frequency from ageing, with an additional local upward notch near $4.0\,\text{kHz}$ from noise exposure.	The combined intensity factor is $18(7.0)=126$ . Therefore $I_{\text{threshold}}=126(3.0\times10^{-13})=3.78\times10^{-11}\,\text{W m}^{-2}$ . The level increase is $10\log_{10}(126)=21.0037\,\text{dB}$ , giving $21\,\text{dB}$ . The broad age-related loss raises the high-frequency region, while multiplying by the extra local factor produces a superimposed notch around $4.0\,\text{kHz}$ .	5

3.10.3.1 · Simple ECG machines and the normal ECG waveform

Tier 1 · Easy

Mark scheme for 3.10.3.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	The QRS complex is ventricular depolarisation and the T wave is ventricular repolarisation.	Match each named feature to the electrical activity of the ventricles: rapid ventricular depolarisation produces QRS, and the later recovery or repolarisation produces T.	2
02.1	Atrial depolarisation; potential difference.	The P wave is produced by depolarisation of the atria. Skin electrodes measure a potential difference arising from the heart's electrical activity, so the trace is not a record of blood pressure.	2

Tier 2 · Standard

Mark scheme for 3.10.3.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.192\,\text{s}$ ; it lies within the normal range.	Use $t=x/v$ for the horizontal paper scale. The PR interval is $4.8/25=0.192\,\text{s}$ . Since $0.12\,\text{s}\leq0.192\,\text{s}\leq0.20\,\text{s}$ , the measured conduction interval is within the supplied normal range.	3
02.1	$18\,\text{mm}$	The cardiac period is $T=60/84=0.7143\,\text{s}$ . The paper distance for one period is $vT=25(0.7143)=17.86\,\text{mm}$ . This gives an R-peak separation of $18\,\text{mm}$ to $2$ significant figures.	3
03.1	The P wave and QRS complex occur at their original times, while the T wave occurs later after the QRS complex. Successive R peaks keep the same separation because the heart period is unchanged.	The P wave represents atrial depolarisation, so it is unaffected. The QRS complex represents ventricular depolarisation, so its timing is also unchanged. The T wave represents ventricular repolarisation and is therefore delayed. The R-R separation measures the cardiac period; the stated unchanged period keeps that separation constant.	3

Tier 3 · Hard

Mark scheme for 3.10.3.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Clean the skin and use low-resistance conductive gel, then secure non-reactive electrodes so movement does not change the contact. Use shielded leads kept away from mains sources and a differential low-noise amplifier. The amplifier needs high gain because the cardiac potential differences are only of order millivolts.	Link each fault to a remedy. Cleaning and conductive gel lower and stabilise skin-electrode contact resistance. Securing the electrodes and keeping the patient still reduces movement artefact. Shielded leads and separation from mains wiring reduce induced electrical noise. A differential, low-noise amplifier rejects common interference, while high gain makes the millivolt ECG large enough to record.	5
02.1	The voltage gain is $2.0\times10^3$ . The percentage uncertainties are $2.5\%$ for one interval and $0.50\%$ for five intervals, so timing several intervals reduces the percentage uncertainty.	Convert the input to volts: $0.80\,\text{mV}=0.80\times10^{-3}\,\text{V}$ . The gain is $1.6/(0.80\times10^{-3})=2.0\times10^3$ . For one interval, the percentage uncertainty is $(0.02/0.80)\times100=2.5\%$ . For five intervals it is $(0.02/4.00)\times100=0.50\%$ . The same absolute timing uncertainty is a smaller fraction of the longer total time, so averaging over several intervals gives a more precise period and heart rate.	5
03.1	The heart-signal output is $+0.96\,\text{V}$ and the identical interference cancels. The unequal extra interference at B produces an output error of $-0.16\,\text{V}$ . Conductive gel lowers and stabilises the skin-electrode contact resistance, reducing unequal pickup and contact changes.	The differential heart signal is $0.70-(-0.50)=1.20\,\text{mV}$ , so the output is $800(1.20\times10^{-3})=0.96\,\text{V}$ . The common $0.180\,\text{V}$ appears in both inputs and subtracts to zero. An additional $+0.20\,\text{mV}$ at B changes $V_A-V_B$ by $-0.20\,\text{mV}$ , so the output error is $800(-0.20\times10^{-3})=-0.16\,\text{V}$ . Low, stable contact resistance from conductive gel reduces differences between electrode contacts and movement-related changes.	4
04.1	The attempted outputs are $0.20\,\text{V}$ , $1.5\,\text{V}$ and $0.35\,\text{V}$ . The QRS complex clips at the recorder limit. The greatest suitable gain is $7.1\times10^2$ (the limiting gain is $714$ , or $714.29$ to $2$ d.p.).	Convert millivolts to volts and multiply by gain: $1100(0.18\times10^{-3})=0.198\,\text{V}$ , $1100(1.4\times10^{-3})=1.54\,\text{V}$ and $1100(0.32\times10^{-3})=0.352\,\text{V}$ . Only QRS exceeds $1.0\,\text{V}$ , so its peak amplitude is clipped and cannot be measured correctly. The largest input sets the gain limit: $G_{\max}=1.0/(1.4\times10^{-3})=714.286$ , so a gain no greater than about $714$ is required.	5
05.1	The mean heart rate is $85.7\,\text{beats min}^{-1}$ (accept $86\,\text{beats min}^{-1}$ ), the QRS amplitude is $1.30\,\text{mV}$ , and the following broad wave is the T wave, representing ventricular repolarisation.	Six R peaks contain five R-R intervals. Their total time is $105/30=3.500\,\text{s}$ , so the mean period is $3.500/5=0.7000\,\text{s}$ and the rate is $60/0.7000=85.714\,\text{beats min}^{-1}$ . The voltage calibration gives $V_{QRS}=(18.2/14.0)(1.00\,\text{mV})=1.300\,\text{mV}$ . On a normal trace the broad feature after QRS is T, produced by ventricular repolarisation.	6

3.10.4.1 · Ultrasound imaging

Tier 1 · Easy

Mark scheme for 3.10.4.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Use ultrasound: it produces real-time images and is non-ionising, so repeated imaging does not give the fetus an ionising-radiation dose.	Ultrasound echoes can be processed continuously to display fetal motion in real time. Ultrasound is non-ionising, so it avoids the cumulative ionising-radiation risk that would make repeated X-ray imaging unsuitable.	2
02.1	Electrical energy to ultrasound for transmission, and returning ultrasound to electrical energy for detection.	An alternating potential difference makes the crystal deform and emit an ultrasound pulse. A returning echo deforms the crystal and produces a detected potential difference.	2

Tier 2 · Standard

Mark scheme for 3.10.4.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.179\%$	The impedances are $Z_A=\rho_Ac_A=1000(1500)=1.500\times10^6\,\text{kg m}^{-2}\text{s}^{-1}$ and $Z_B=1060(1540)=1.6324\times10^6\,\text{kg m}^{-2}\text{s}^{-1}$ . Thus $I_r/I_i=((Z_B-Z_A)/(Z_B+Z_A))^2=((1.6324-1.500)/(1.6324+1.500))^2=1.787\times10^{-3}$ . Multiplying by $100$ gives $0.179\%$ .	4
02.1	$28\,\text{mm}$ (accept $27.7\,\text{mm}$ ).	The extra round-trip time through the layer is $(88-52)\,\mu\text{s}=36\times10^{-6}\,\text{s}$ . Hence the thickness is $d=c\Delta t/2=1540(36\times10^{-6})/2=0.02772\,\text{m}=27.72\,\text{mm}$ . This is $28\,\text{mm}$ to $2$ significant figures; accept $27.7\,\text{mm}$ .	3
03.1	At each transducer position a short pulse is emitted and returning echoes are detected. Echo delay gives boundary depth using the out-and-back travel time, while echo amplitude sets the brightness of the displayed spot. Combining the depth and brightness data from successive transducer positions constructs a two-dimensional image.	The transducer supplies one line of pulse-echo data at each position. The computer converts each return time using $d=ct/2$ , so the echo is placed at the correct depth. Its amplitude controls the displayed brightness. Placing successive lines beside one another at their transducer positions builds the two-dimensional B-scan.	4

Tier 3 · Hard

Mark scheme for 3.10.4.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.100\,\text{m}$ depth and $0.39\,\text{mm}$ resolution; the gel removes air and reduces the acoustic-impedance mismatch, so less intensity is reflected at the skin.	The pulse makes a round trip, so $d=ct/2=1540(130\times10^{-6})/2=0.1001\,\text{m}$ , giving $0.100\,\text{m}$ . Its wavelength is $\lambda=c/f=1540/(4.0\times10^6)=3.85\times10^{-4}\,\text{m}=0.385\,\text{mm}$ , so a one-wavelength resolution estimate is $0.39\,\text{mm}$ . Air has a very different acoustic impedance from tissue and would reflect most incident ultrasound; gel displaces the air and provides a better impedance match, increasing transmission into the body.	6
02.1	The reflected-intensity ratio is $35$ . Boundary 2 gives the stronger echo; its large tissue-bone impedance mismatch reflects a much greater fraction of the ultrasound, leaving less transmitted intensity and making tissue beyond the bone difficult to image.	Use $R=((Z_2-Z_1)/(Z_2+Z_1))^2$ . For boundary 1, $R_1=((1.80-1.50)/(1.80+1.50))^2=8.264\times10^{-3}$ . For boundary 2, $R_2=((6.00-1.80)/(6.00+1.80))^2=0.28994$ . Since the incident intensities are equal, $I_{r2}/I_{r1}=R_2/R_1=35.08$ , giving $35$ to $2$ significant figures. Boundary 2 therefore gives the stronger echo. Its larger reflection coefficient means less ultrasound is transmitted through the bone, so echoes from structures beyond it are weak.	6
03.1	Only the $7.5\,\text{MHz}$ transducer: its resolution is $0.205\,\text{mm}$ and its round-trip attenuation is $45\,\text{dB}$ .	At $2.5\,\text{MHz}$ , $\lambda=c/f=1540/(2.5\times10^6)=6.16\times10^{-4}\,\text{m}=0.616\,\text{mm}$ , so the resolution requirement is not met. Its round-trip attenuation is $2(0.50)(6.0)(2.5)=15\,\text{dB}$ . At $7.5\,\text{MHz}$ , $\lambda=1540/(7.5\times10^6)=2.05\times10^{-4}\,\text{m}=0.205\,\text{mm}$ , and the round-trip attenuation is $2(0.50)(6.0)(7.5)=45\,\text{dB}$ . This is below $50\,\text{dB}$ while the wavelength is below $0.30\,\text{mm}$ , so only $7.5\,\text{MHz}$ satisfies both constraints.	6
04.1	The near-face echo is $6.30\times10^{-3}$ of the incident intensity. The far-face echo returning to the first tissue is $4.33\times10^{-2}$ of the incident intensity, $6.87$ times stronger, so the far face is brighter (accept $4.32\times10^{-2}$ to $4.33\times10^{-2}$ and $6.86$ to $6.88$ ).	For the near face, $R_1=((1.70-1.45)/(1.70+1.45))^2=0.00629882$ . For the far face, $R_2=((2.60-1.70)/(2.60+1.70))^2=0.0438075$ . The pulse crosses the near face twice, so its returning fraction is $(1-R_1)R_2(1-R_1)=(0.993701)^2(0.0438075)=0.0432573$ . The ratio $0.0432573/0.00629882=6.86753$ , so the far-boundary echo has the greater intensity and is displayed more brightly.	5
05.1	The latest echo returns after $2.34\times10^{-4}\,\text{s}$ ( $234\,\mu\text{s}$ ), so the greatest pulse-repetition frequency is $4.28\times10^3\,\text{Hz}$ (about $4.3\,\text{kHz}$ ). If pulses are sent faster, a deep echo can return after a newer pulse has been transmitted and be assigned to that newer pulse, making the calculated depth ambiguous.	The latest echo makes a round trip of $2d=0.360\,\text{m}$ , so $t_{\max}=2d/c=0.360/1540=2.33766\times10^{-4}\,\text{s}=234\,\mu\text{s}$ . The interval between pulses must be at least this long. Therefore $f_{\max}=1/t_{\max}=1540/(2(0.180))=4277.78\,\text{Hz}=4.28\,\text{kHz}$ . At a higher repetition frequency, a new pulse is transmitted before the deepest echo returns, so the receiver could associate that late echo with the newer pulse and calculate an incorrect depth.	6

3.10.4.2 · Fibre optics and endoscopy

Tier 1 · Easy

Mark scheme for 3.10.4.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	The coherent bundle carries the image, and the non-coherent (incoherent) bundle carries illumination.	Image information requires the fibre positions to be preserved, so it uses a coherent bundle. Illumination only needs light delivery and therefore uses a non-coherent or incoherent bundle.	2
02.1	Light must travel from the higher-index core towards the lower-index cladding, and its incidence angle must exceed the critical angle.	Total internal reflection is possible only for travel from higher to lower refractive index. It occurs when the incidence angle at that boundary is greater than the critical angle.	2

Tier 2 · Standard

Mark scheme for 3.10.4.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$67.8^\circ$ ; total internal reflection occurs.	For light travelling from core to cladding, $\sin c=n_{\text{cladding}}/n_{\text{core}}=1.50/1.62$ . Hence $c=\sin^{-1}(1.50/1.62)=67.8^\circ$ . The incidence angle $72.0^\circ$ is greater than $c$ , and the light is travelling from higher to lower refractive index, so total internal reflection occurs.	3
02.1	A coherent bundle returns an image of the internal tissue, while another fibre bundle can deliver illumination or treatment light to the target. The narrow bundles pass through a small opening, so access is less invasive and causes less tissue damage than a large surgical opening.	Credit the separate functions: an ordered coherent bundle carries the image back to the clinician, while fibres can carry light into the body for illumination or treatment. Because both bundles are narrow, they can reach the target through a small opening. This gives less invasive access, reducing damage to surrounding tissue.	3
03.1	The large angle to the fibre axis makes the angle of incidence at the core-cladding boundary too small. It is then less than the critical angle, so total internal reflection does not occur and the ray refracts into the cladding.	For a meridional ray, increasing the angle to the fibre axis decreases the incidence angle at the side wall. Confinement requires incidence from the higher-index core at an angle greater than the critical angle. Here that condition fails, so the ray is transmitted into the lower-index cladding instead of being totally internally reflected.	3

Tier 3 · Hard

Mark scheme for 3.10.4.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Rearranging the image fibres destroys the one-to-one spatial mapping, so light from each object point emerges at the wrong position. Missing cladding allows light to cross between neighbouring fibres or escape, reducing contrast and blurring the image. Random order does not prevent the illumination bundle working because it only delivers light into the body and does not encode spatial information.	A coherent image bundle must preserve the relative position of every fibre; losing that order scrambles the image. Core-cladding total internal reflection confines each ray, so missing cladding permits leakage and cross-talk, producing dimmer or blurred regions. In the illumination bundle, only the total delivered light matters, so a non-coherent arrangement remains suitable.	5
02.1	Only fibre Y confines the ray by total internal reflection. Its critical angle is $62.6^\circ$ , whereas X has critical angle $69.6^\circ$ . The $66.0^\circ$ incidence exceeds the critical angle only for Y. The relative fibre positions must be preserved so light from each object point emerges at the corresponding image position rather than producing a scrambled image.	For X, $c_X=\sin^{-1}(1.50/1.60)=69.6^\circ$ , so $66.0^\circ<c_X$ and total internal reflection does not occur. For Y, $c_Y=\sin^{-1}(1.42/1.60)=62.6^\circ$ , so $66.0^\circ>c_Y$ and the ray is confined. Total internal reflection transports light within each fibre, but the complete image is reproduced only if the fibres preserve their relative ordering from one end of the bundle to the other.	5
03.1	Bundle A cannot resolve the lesion because its $45\,\mu\text{m}$ image is below the $60\,\mu\text{m}$ criterion. Bundle B samples finely enough but cannot form an image because it is non-coherent.	The image scale is $1.8/12=0.150$ , so the lesion image is $0.150(0.30\,\text{mm})=0.045\,\text{mm}=45\,\mu\text{m}$ wide. Bundle A needs $2(30)=60\,\mu\text{m}$ , so it fails the sampling criterion. Bundle B needs only $2(15)=30\,\mu\text{m}$ and therefore samples finely enough. At least one mark requires the separate coherence verdict: because B is non-coherent, it cannot preserve relative positions and cannot form an image.	4
04.1	The critical angle is $65.9^\circ$ , the largest internal angle to the axis is $24.1^\circ$ , and the largest entrance angle in air is $37.5^\circ$ (accept $37.4^\circ$ to $37.5^\circ$ ).	At the core-cladding boundary, $c=\sin^{-1}(1.36/1.49)=65.888^\circ$ . The side-wall incidence angle and the ray angle to the axis sum to $90^\circ$ , so the limiting internal angle is $90-65.888=24.112^\circ$ . At the flat end, Snell's law gives $1.00\sin\theta_{\text{air}}=1.49\sin(24.112^\circ)$ , so $\theta_{\text{air}}=37.495^\circ$ . Entrance angles below this limit lead to incidence above the critical angle and total internal reflection.	4
05.1	The delivered fractions are $0.786$ with intact cladding and $0.059$ when damaged, so the intact fibre delivers about $13$ times as much intensity. Cladding confines light and limits leakage or cross-talk, whereas coherent ordering preserves the spatial mapping from object points to image points.	Successive retained fractions multiply. With intact cladding, $I/I_0=0.992^{30}=0.785869$ . With damage, $I/I_0=0.91^{30}=0.0590530$ . Their ratio is $0.785869/0.0590530=13.3079$ . Good confinement keeps each signal bright and within its own fibre, but it does not determine where that fibre ends; a coherent bundle must also keep relative fibre positions unchanged so the emerging pattern reproduces the image.	5

3.10.4.3 · Magnetic resonance (MR) scanner

Tier 1 · Easy

Mark scheme for 3.10.4.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Hydrogen nuclei (protons); their spins align and precess about the magnetic field direction.	The abundant hydrogen nuclei in body tissue supply the signal. In the strong static field, their spin states become aligned and the proton magnetic moments precess about the field lines.	2
02.1	Radio-frequency radiation is used; it is non-ionising.	A short radio-frequency pulse supplies energy to change proton spin states. Radio waves are non-ionising, unlike X-rays or gamma radiation.	2

Tier 2 · Standard

Mark scheme for 3.10.4.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The pulse excites the protons and changes their spin state. When the pulse stops, they relax or de-excite and emit radio-frequency signals that can be detected.	During the pulse, radio-frequency energy at the appropriate frequency is absorbed and selected protons change to an excited spin state. Once excitation ends, the protons return towards their original state. This relaxation or de-excitation releases an RF signal for the receiver coils.	3
02.1	The protons emit radio-frequency photons with the same frequency as the incident pulse.	De-excitation releases photons in the radio-frequency part of the electromagnetic spectrum. Their frequency is the same as that of the incident radio-frequency pulse.	2
03.1	The protons already possess spin; the main field aligns their spin axes rather than starting the spin. Gradient fields select successive small regions of the cross-section.	Award one mark for correcting the classical-spin misconception: proton spin already exists and the main field aligns the spin axes. Award one mark for the spatial-selection step: gradient fields select successive small regions rather than the whole cross-section being scanned at once.	2

Tier 3 · Hard

Mark scheme for 3.10.4.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The main field aligns proton spins, which precess about the field. Gradient coils select and locate successive small regions of the cross-section. Short RF pulses excite protons in each selected region; as they relax, the protons emit RF signals. Receiver coils detect the signals and a computer uses their position and strength to construct the image.	Give the process in causal order: alignment and precession in the static field; position-dependent gradient fields selecting a slice or small region; an RF pulse changing proton spin states; relaxation or de-excitation producing RF emissions; and detection followed by computer processing into a spatial image. No calculation of relaxation times is needed.	5
02.1	The main field can still align hydrogen nuclei and make them precess. An RF pulse can excite the protons, which emit detectable RF signals as they de-excite. Without gradients, successive small regions are not selected or located, so the receiver signals lack position information and the computer cannot construct a cross-sectional image.	Separate signal production from image location. The static field, RF excitation, de-excitation and RF detection still operate, so a combined signal may be recorded. Gradient coils normally impose a position-dependent field to select and encode regions. With no gradient, the computer cannot map signal strength to coordinates in the patient.	5
03.1	The pulse transfers energy to selected protons and excites them into a higher-energy spin state. As they relax, they emit the radio-frequency signal detected by the receiver coils. The pulse acts on hydrogen nuclei (protons), which are used because they are abundant in the water and fat of body tissue. The magnetic fields and radio-frequency radiation are non-ionising, so several scans do not carry a cumulative risk of tissue ionisation.	Credit the MR-internal chain: the short pulse supplies energy and changes selected proton spin states; relaxation produces the RF response detected for the image; the nuclei involved are hydrogen nuclei (protons), abundant in body tissue; and the fields and radio waves used are non-ionising, so several scans do not accumulate a risk from tissue ionisation.	4
04.1	The outputs are $0.70\,\text{V}$ for B and $1.26\,\text{V}$ for A. Their difference is $0.56\,\text{V}$ , so it exceeds the $0.45\,\text{V}$ requirement and the display resolves the contrast. Gradient coils make the magnetic field position dependent, allowing the small regions to be selected or encoded separately so the computer can map signal strength to position.	For B, $V_{\text{out}}=(2.5\times10^6)(0.28\times10^{-6})=0.700\,\text{V}$ . The coil emf for A is $1.8(0.28)=0.504\,\mu\text{V}$ , so its output is $(2.5\times10^6)(0.504\times10^{-6})=1.260\,\text{V}$ . The difference $1.260-0.700=0.560\,\text{V}$ is greater than the required $0.45\,\text{V}$ , so the two shades are distinguishable. Gradient fields vary with position, selecting or encoding successive small regions; the computer combines that position information with the detected strengths to construct the cross-sectional image.	6
05.1	The main magnetic field must be applied first to align the proton spin axes. An RF transmitter supplies the short pulse that excites selected protons; receiver coils instead detect the signal as the protons relax. The detected signal has the same frequency as the exciting pulse. Gradient fields provide position information, while signal amplitude provides intensity information, so the computer combines both to construct the cross-section.	Correct each causal step. The static main field is established before RF excitation and aligns the proton spin axes. The RF transmitter, not the receiver coils, transfers energy to selected protons; the receiver coils detect the response during relaxation. De-excitation produces RF emission at the same frequency as the exciting pulse. Gradient fields encode position, and the computer combines that spatial information with signal amplitude to reconstruct the cross-sectional image.	5

3.10.5.1 · The physics of diagnostic X-rays

Tier 1 · Easy

Mark scheme for 3.10.5.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$68\,\text{keV}$	An electron accelerated through $68\,\text{kV}$ gains $68\,\text{keV}$ . If it loses all this energy in one interaction, $E_{\max}=68\,\text{keV}$ .	1
02.1	The X-ray beam intensity increases.	A larger current sends more electrons to the target each second, so more X-ray photons are produced per second and the beam intensity increases.	1

Tier 2 · Standard

Mark scheme for 3.10.5.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.7\times10^{-11}\,\text{m}$	Convert the voltage: $V=72\times10^3\,\text{V}$ . The maximum photon energy is $E_{\max}=eV=(1.60\times10^{-19})(72\times10^3)=1.152\times10^{-14}\,\text{J}$ . Hence $\lambda_{\min}=hc/E_{\max}=(6.63\times10^{-34})(3.00\times10^8)/(1.152\times10^{-14})=1.7266\times10^{-11}\,\text{m}$ , which is $1.7\times10^{-11}\,\text{m}$ to two significant figures.	3
02.1	The filter preferentially absorbs low-energy X-ray photons, increasing the mean energy of the transmitted beam and reducing the skin dose from photons that would be absorbed without contributing to the image.	Low-energy photons are more likely to be absorbed near the patient's surface and are unlikely to reach the detector. Aluminium filtration removes many of these photons before they enter the patient. The remaining beam has a greater mean photon energy, and the avoidable skin dose is reduced.	3
03.1	$45.2\,\text{keV}$	The photon energy equals the level separation. First, $E=hc/\lambda=(6.63\times10^{-34})(3.00\times10^8)/(2.75\times10^{-11})=7.2327\times10^{-15}\,\text{J}$ . Since $1\,\text{keV}=1.60\times10^{-16}\,\text{J}$ , the separation is $(7.2327\times10^{-15})/(1.60\times10^{-16})=45.2\,\text{keV}$ .	3

Tier 3 · Hard

Mark scheme for 3.10.5.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.5\times10^{-14}\,\text{J}$ and $0.27\,\text{J}$ ; rotation spreads heating over the target	The maximum photon energy is $E_{\max}=eV=(1.60\times10^{-19})(95\times10^3)=1.52\times10^{-14}\,\text{J}=1.5\times10^{-14}\,\text{J}$ to two significant figures. The electrical energy supplied is $VIt=(95\times10^3)(3.0\times10^{-3})(0.080)=22.8\,\text{J}$ . Therefore the X-ray energy is $0.012\times22.8=0.2736\,\text{J}=0.27\,\text{J}$ . Rotating the anode continually changes the impact area, spreading the large thermal energy and reducing local overheating or target damage.	5
02.1	The maximum photon energy is unchanged because the potential difference is unchanged. The same current supplies the same number of electrons per second, so approximately the same number of X-ray photons is produced per second.	The maximum X-ray photon energy is $E_{\max}=eV$ , so it depends on the tube potential difference rather than the focal-spot size. A fixed current means the same charge, and therefore the same number of electrons, reaches the anode each second. Reducing the impact area does not change that electron rate, so the photon count per second is approximately unchanged.	3
03.1	Tube B has $1.5$ times the maximum photon energy, half the photon production rate, and characteristic lines at the same energies as tube A.	Since $E_{\max}=eV$ , $E_{\max,B}/E_{\max,A}=120/80=1.5$ . The X-ray power is $P_X=0.0080VI$ , while $E_{\text{mean}}=0.40E_{\max}$ . Substituting $E_{\max}=eV$ gives $\dot N=P_X/E_{\text{mean}}=0.0080I/(0.40e)$ , so $\dot N_B/\dot N_A=I_B/I_A=1.2/2.4=0.50$ ; explicitly, the rates are $1.5\times10^{14}\,\text{s}^{-1}$ and $3.0\times10^{14}\,\text{s}^{-1}$ . Characteristic energies are fixed by tungsten's electron-level separations, so changing the voltage does not move those lines once both voltages exceed the relevant thresholds.	5
04.1	$2.04\times10^{-11}\,\text{m}$ (accept $2.03$ to $2.04\times10^{-11}\,\text{m}$ ); the $70.0\,\text{kV}$ tube cannot create the K-shell vacancy, because at least $72.4\,\text{kV}$ is required	The photon energy is the difference between the binding energies: $E_\gamma=72.4-11.3=61.1\,\text{keV}$ . In joules this is $(61.1)(1.60\times10^{-16})=9.776\times10^{-15}\,\text{J}$ . Hence $\lambda=hc/E_\gamma=(1.99\times10^{-25})/(9.776\times10^{-15})=2.0356\times10^{-11}\,\text{m}$ , giving $2.04\times10^{-11}\,\text{m}$ to three significant figures. An incident electron must supply at least the $72.4\,\text{keV}$ K-shell binding energy, requiring a minimum tube potential difference of $72.4\,\text{kV}$ . Therefore $70.0\,\text{kV}$ is insufficient even though it exceeds the energy of the emitted photon.	5
05.1	$49\,\text{mA}$ and $84\,\text{keV}$ ; doubling the area permits twice the current and hence approximately twice the photon rate, while $E_{\max}=eV$ is unchanged	The focal-spot area is $(0.80\times10^{-3})(1.6\times10^{-3})=1.28\times10^{-6}\,\text{m}^2$ . Its maximum heating power is $(3.2\times10^9)(1.28\times10^{-6})=4096\,\text{W}$ . Since the heating fraction is $1-0.0070=0.993$ , $0.993VI=4096$ , so $I=4096/[0.993(84\times10^3)]=4.9106\times10^{-2}\,\text{A}=49\,\text{mA}$ to two significant figures. The maximum photon energy is $eV=84\,\text{keV}$ . Doubling the spot area doubles the allowed heating power and therefore the allowed current. A larger current supplies more electrons per second and increases photon intensity, but the unchanged potential difference leaves the energy gained by each electron, and hence $E_{\max}$ , unchanged.	6

3.10.5.2 · Image detection and enhancement

Tier 1 · Easy

Mark scheme for 3.10.5.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	It converts incident X-ray photons into visible-light photons.	The marking point is the energy conversion: the scintillator absorbs X-rays and emits visible light, which the photodiode pixels can detect.	1
02.1	It converts visible light into an electrical signal.	Visible photons from the scintillator produce charge in the photodiode, giving an electrical signal for that pixel.	1

Tier 2 · Standard

Mark scheme for 3.10.5.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Barium absorbs X-rays strongly, so less radiation reaches the detector behind the digestive tract and its outline has greater contrast with surrounding soft tissue.	Barium is X-ray opaque, so it has greater attenuation than the surrounding soft tissues. Regions behind the barium therefore give a different detector signal because fewer X-rays are transmitted. This increases image contrast and makes the shape of the digestive tract visible.	3
02.1	One absorbed X-ray photon produces many visible photons, and the film is more sensitive to visible light than to X-rays, so fewer incident X-ray photons are needed; spreading of the light reduces image sharpness.	The screen absorbs an X-ray photon and emits many visible photons. Photographic film responds much more strongly to this visible light than it does directly to X-rays, so the required film darkening is obtained with fewer incident X-ray photons and a smaller patient dose. The visible light spreads before reaching the film, making boundaries less sharp.	3
03.1	Adding all signals loses their positions, so different X-ray patterns could give the same total output. Each photodiode signal must remain linked to its pixel as electronic scanning reads the array, allowing a computer to assign pixel brightness and reconstruct the spatial image.	A scintillator and photodiode array convert the local X-ray intensity into a separate electrical charge at each position. Summing first retains only total exposure and discards the mapping between signal and position. The array must instead be addressed pixel by pixel during electronic scanning, so the computer can place each value at the corresponding image location and enhance the resulting digital pattern.	3

Tier 3 · Hard

Mark scheme for 3.10.5.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The missing strip is $0.75\,\text{mm}$ wide. The $0.50\,\text{mm}$ lesion fits entirely within the strip, so none of its detector data is recorded. The image therefore has a $0.75\,\text{mm}$ strip with no anatomical information. Enhancement cannot reconstruct spatial data that the failed rows never recorded. A repeat may be justified if the missing strip could conceal clinically necessary detail and the diagnostic benefit outweighs the extra ionising-radiation dose.	The failed width is $3(0.25\,\text{mm})=0.75\,\text{mm}$ . Since the $0.50\,\text{mm}$ lesion lies wholly inside this wider strip, every photodiode position that would record it has failed, so the lesion produces no recorded spatial data. Digital enhancement can remap or interpolate recorded neighbouring values, but it cannot determine the unrecorded anatomy inside the strip and therefore cannot reconstruct the lesion. Repeating the exposure adds ionising-radiation dose, so it is justified only when the missing anatomy could affect diagnosis and the expected diagnostic benefit outweighs that added risk.	5
02.1	Barium attenuates X-rays more strongly than soft tissue, creating a larger transmitted-intensity and detector-signal difference. Digital processing can rescale that signal range into a larger displayed brightness difference, but it cannot recover detail that was not recorded because the signals were indistinguishable or dominated by noise.	Barium is much more X-ray opaque than the surrounding soft tissues. Less radiation is therefore transmitted through a barium-filled region, so its detector signal differs more strongly from that of nearby tissue and the physical image contrast increases. After detection, a computer can map the recorded signal range onto a wider range of displayed pixel brightnesses. This makes an existing signal difference easier to see, but processing cannot recreate spatial information when the original detector signals contain no usable difference or are masked by noise.	5
03.1	The pixels store $5.4\times10^6$ and $3.2\times10^6$ charge carriers, so neither saturates and their signal ratio is $1.7$ ; enhancement can enlarge the displayed brightness difference but cannot restore detail lost through saturation, noise or light spreading.	Each absorbed X-ray produces $(1.2\times10^3)(0.25)(0.60)=180$ charge carriers. The pixel signals are therefore $(3.0\times10^4)(180)=5.40\times10^6$ and $(1.8\times10^4)(180)=3.24\times10^6$ charge carriers. Both are below the $6.0\times10^6$ saturation limit, and their ratio is $5.40/3.24=1.67$ , so a physical signal difference has been recorded. Digital processing can map that difference onto a wider displayed brightness range, but it cannot recreate information that the detector failed to distinguish because of saturation, noise or loss of spatial detail.	5
04.1	$14$ frames per second, $11$ relative dose units and a $42\%$ dose reduction (accept $43\%$ ); barium increases differential X-ray absorption, while a lower frame rate would allow the marker to move more than $3.0\,\text{mm}$ between images	The time between frames must be at most $3.0/42=0.07143\,\text{s}$ , so the minimum frame rate is $42/3.0=14\,\text{s}^{-1}$ . In $50\,\text{s}$ this records $(14)(50)=700$ frames and gives $(700)(0.016)=11.2$ relative dose units, or $11$ units to two significant figures. At $24\,\text{s}^{-1}$ the relative dose is $(24)(50)(0.016)=19.2$ units. The reduction is $(19.2-11.2)/19.2=0.4167$ , or $42\%$ (accept $43\%$ ). Barium attenuates X-rays more strongly than neighbouring soft tissue, increasing the transmitted-intensity difference and making the marker visible. Below $14\,\text{s}^{-1}$ it can move more than $3.0\,\text{mm}$ between frames, so clinically relevant motion could be missed even though dose would fall further.	6
05.1	$4.1\times10^7$ incident X-rays; the incident exposure and patient dose are reduced by a factor of $1.7\times10^2$ , corresponding to a $99.4\%$ reduction	Each incident X-ray produces $(0.57)(290)=165.3$ useful film exposures when the screen is used. The required incident number is therefore $(6.8\times10^9)/165.3=4.114\times10^7$ , or $4.1\times10^7$ to two significant figures. Relative to the $6.8\times10^9$ incident X-rays needed without the screen, the reduction factor is $(6.8\times10^9)/(4.114\times10^7)=165.3$ , or $1.7\times10^2$ . The percentage reduction is $[1-(1/165.3)]\times100=99.395\%$ , giving $99.4\%$ . Because dose is proportional to incident X-ray number, the patient dose falls by the same factor and percentage.	4

3.10.5.3 · Absorption of X-rays

Tier 1 · Easy

Mark scheme for 3.10.5.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.5\,\text{cm}$	Use $x_{1/2}=\ln 2/\mu=0.693/0.28=2.48\,\text{cm}$ , which is $2.5\,\text{cm}$ to two significant figures.	2
02.1	$3.0\times10^{-2}\,\text{m}^2\,\text{kg}^{-1}$	Use $\mu_m=\mu/\rho=28/920=0.03043\,\text{m}^2\,\text{kg}^{-1}$ , which is $3.0\times10^{-2}\,\text{m}^2\,\text{kg}^{-1}$ to two significant figures.	2

Tier 2 · Standard

Mark scheme for 3.10.5.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.8\,\text{W m}^{-2}$	Use $I=I_0e^{-\mu x}$ . The exponent is $-\mu x=-(0.18)(7.0)=-1.26$ . Therefore $I=6.4e^{-1.26}=1.82\,\text{W m}^{-2}=1.8\,\text{W m}^{-2}$ .	3
02.1	$4.5\,\text{cm}$	Use $I/I_0=e^{-\mu x}=0.15$ . Taking natural logarithms gives $x=-\ln(0.15)/0.42=4.517\,\text{cm}$ , so the required thickness is $4.5\,\text{cm}$ to two significant figures.	3
03.1	$63\%$	The attenuation exponents add: $\mu_1x_1+\mu_2x_2=(0.12)(2.0)+(2.5)(0.30)=0.99$ . Hence $I/I_0=e^{-0.99}=0.3716$ , so the absorbed fraction is $1-0.3716=0.6284$ . Therefore $63\%$ of the incident intensity is absorbed to two significant figures.	3

Tier 3 · Hard

Mark scheme for 3.10.5.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.1\,\text{W m}^{-2}$ , $3.3\,\text{W m}^{-2}$ and a ratio of $2.9$	Attenuation exponents add for successive materials. For the bone path, $\mu x=(0.050)(18)+(0.32)(4.0)=0.90+1.28=2.18$ , so $I_{\text{bone}}=10.0e^{-2.18}=1.130\,\text{W m}^{-2}=1.1\,\text{W m}^{-2}$ . For soft tissue only, $\mu x=(0.050)(22)=1.10$ , so $I_{\text{soft}}=10.0e^{-1.10}=3.329\,\text{W m}^{-2}=3.3\,\text{W m}^{-2}$ . Using unrounded values, the ratio is $I_{\text{soft}}/I_{\text{bone}}=2.945=2.9$ to two significant figures.	5
02.1	$1.5\times10^3\,\text{kg m}^{-3}$	Convert the thickness: $x=0.035\,\text{m}$ . From $I/I_0=e^{-\mu x}$ , $\mu=-\ln(0.22)/0.035=43.26\,\text{m}^{-1}$ . Since $\mu_m=\mu/\rho$ , $\rho=\mu/\mu_m=43.26/(2.8\times10^{-2})=1.545\times10^3\,\text{kg m}^{-3}$ , which is $1.5\times10^3\,\text{kg m}^{-3}$ to two significant figures.	4
03.1	Filter A transmits $65\%$ and filter B transmits $87\%$ ; B transmits $1.3$ times the intensity transmitted by A.	Since $\mu_m=\mu/\rho$ , the exponent $\mu x$ can be written as $\mu_m\rho x$ , where $\rho x$ is the mass per unit area. For A, $I_A/I_0=e^{-(3.6\times10^{-2})(12)}=e^{-0.432}=0.6492$ , or $65\%$ . For B, $I_B/I_0=e^{-(1.2\times10^{-2})(12)}=e^{-0.144}=0.8659$ , or $87\%$ . The factor is $I_B/I_A=0.8659/0.6492=1.334$ , which is $1.3$ to two significant figures.	4
04.1	$\mu=0.26\,\text{cm}^{-1}$ , $I_0=7.1\,\text{W m}^{-2}$ and $x_{1/2}=2.7\,\text{cm}$	Dividing the two forms of $I=I_0e^{-\mu x}$ eliminates $I_0$ : $4.8/1.7=e^{\mu(5.5-1.5)}$ . Hence $\mu=\ln(4.8/1.7)/4.0=0.25950\,\text{cm}^{-1}$ , giving $0.26\,\text{cm}^{-1}$ . Using the first reading, $I_0=4.8e^{(0.25950)(1.5)}=7.0842\,\text{W m}^{-2}$ , or $7.1\,\text{W m}^{-2}$ . Finally, $x_{1/2}=\ln2/\mu=0.69315/0.25950=2.6711\,\text{cm}$ , which is $2.7\,\text{cm}$ to two significant figures.	5
05.1	The surrounding-to-lesion intensity ratios are $1.5$ at L and $1.2$ at H. Surrounding tissue transmits $12\%$ at L and $35\%$ at H, so only H meets the $25\%$ requirement, although it produces less image contrast.	The two paths differ only where $3.0\,\text{cm}$ of surrounding tissue is replaced by lesion. Thus $I_{\text{sur}}/I_{\text{les}}=e^{(\mu_{\text{les}}-\mu_{\text{sur}})(3.0)}$ . At L this is $e^{(0.40-0.26)(3.0)}=1.5220$ , giving $1.5$ to two significant figures; at H it is $e^{(0.19-0.13)(3.0)}=1.1972$ , giving $1.2$ . The all-surrounding transmissions are $e^{-(0.26)(8.0)}=0.1249$ , or $12\%$ , and $e^{-(0.13)(8.0)}=0.3535$ , or $35\%$ . Energy H is the only one above the transmission floor. Its smaller attenuation-coefficient difference gives weaker lesion contrast, so the increased transmission is obtained at the cost of a smaller detector-signal ratio.	4

3.10.5.4 · CT scanner

Tier 1 · Easy

Mark scheme for 3.10.5.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	To obtain attenuation measurements through the body from many directions.	Movement changes the beam direction, providing the many projections that the computer needs to reconstruct a cross-section.	1
02.1	It measures the X-ray intensity transmitted along each path through the patient.	The detectors convert the transmitted beam measurements into signals for computer reconstruction.	1

Tier 2 · Standard

Mark scheme for 3.10.5.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	A narrow beam and detector array collect transmitted intensities at many angles as the tube moves, and a computer reconstructs these projection data into a cross-sectional image.	The moving tube directs a narrow X-ray beam through the selected section from a succession of angles. The detector array measures the transmitted intensity for many paths. A computer processes this set of projections to calculate the attenuation distribution and display a cross-sectional image.	3
02.1	A narrow beam defines a small path for spatial resolution, while a monochromatic beam has one photon energy so changes in transmission can be related consistently to attenuation.	Restricting the beam width limits the region contributing to each detector reading, helping the reconstruction distinguish nearby structures. A monochromatic beam has a single photon energy and therefore a consistent attenuation coefficient for a given material. Variations in measured transmission can then be assigned more reliably to the tissues along each narrow path.	3
03.1	Structures on the same X-ray path contribute to one combined attenuation measurement, so their separate positions along that direction cannot be identified. Measurements from many angles provide different projections that allow a computer to locate the structures separately within a cross-sectional image.	A single projection records only the total attenuation along each path through the patient. Different arrangements of structures along one path can therefore give the same reading and remain superposed. Rotating the source makes the X-rays cross each region along different paths, and combining these projections lets the reconstruction assign attenuation to separate positions within the slice.	3

Tier 3 · Hard

Mark scheme for 3.10.5.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Use CT to localise the lung lesion because cross-sectional images remove rib superposition and give better depth information; use a plain X-ray for the feeding tube because its radiopaque position is shown adequately at lower dose, lower cost and greater availability.	CT reconstructs slices, so ribs at different depths do not remain superimposed on the lesion and its three-dimensional location is clearer. That extra information can justify CT's higher ionising dose, cost and lower availability for the lesion. A plain radiograph is quick, cheap, widely available and gives a lower dose. Because a radiopaque tube already has strong contrast and only its overall position must be checked, the projection image is sufficient. Therefore use CT for the obscured lesion and a plain X-ray for the tube check.	6
02.1	Fewer projections can reduce ionising dose, but provide less angular information and may reduce image resolution; the protocol may be suitable if the large lesion remains visible and the lower repeated dose outweighs the risk of missing a small change.	Each angular projection requires X-rays to pass through the patient, so collecting fewer projections can lower the total ionising exposure. The computer then has less directional information, so fine detail and image resolution may deteriorate. A large lesion with clear boundaries may still be monitored adequately, but the protocol must retain enough resolution to detect a clinically important change. Its suitability depends on whether that limitation is outweighed by the safety advantage of a smaller accumulated dose from repeated scans.	4
03.1	CT gives higher-resolution cross-sectional detail but costs more and generally gives a larger ionising dose. Plain radiographs should be used because they can show the alignment of a simple fracture adequately at lower cost and lower dose, which is especially important for repeated imaging of a child.	CT reconstructs slices and can resolve fine detail without superposition, so it would provide more anatomical information. It is more expensive and normally exposes the patient to a greater X-ray dose than a plain radiograph. A simple forearm fracture has strong bone contrast, and the clinical target is only its overall alignment, so the extra CT resolution is unlikely to justify its added cost and dose across three examinations. Recommend plain radiographs unless they fail to show a feature needed for treatment.	5
04.1	A plain radiograph is cheaper and normally gives a lower ionising dose, but overlapping structures remain superposed and may still conceal the lesion. CT costs more and usually gives a greater dose, but its reconstructed cross-sectional images provide better resolution and localisation without superposition. Use CT because the exact position is needed for surgery and the radiograph has not supplied it, while keeping the CT exposure as low as practicable.	A plain radiograph records attenuation along each complete X-ray path, so structures at different depths are superposed. Repeating the same type of projection may therefore leave the small lesion hidden even though the examination is relatively inexpensive and uses a lower dose. CT combines projections from many angles to reconstruct slices, removing superposition and locating the lesion with better spatial detail. This costs more and normally exposes the patient to a greater ionising dose. In this case the extra information has a stated surgical purpose and was not available from the radiograph, so the benefit of CT can justify its added cost and radiation risk.	5
05.1	$64$ slices initially and $128$ slices after halving the thickness; twice as many slices are exposed, so an unchanged dose per slice gives twice the total dose	The original number of contiguous slices is $96/1.5=64$ . Halving the slice thickness gives $96/0.75=128$ slices, which is twice the original number. Total dose is the dose per slice multiplied by the number of slices. Since the dose per slice is unchanged while the slice count doubles, the total patient dose also doubles.	4

3.10.6.1 · Imaging techniques

Tier 1 · Easy

Mark scheme for 3.10.6.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Gamma radiation can pass out of the body and be detected externally.	The required property is penetration: enough gamma photons must escape the tissue to reach the detector outside the patient.	1
02.1	The thyroid gland	The thyroid takes up iodine to make thyroid hormones, so labelled iodine becomes concentrated there.	1

Tier 2 · Standard

Mark scheme for 3.10.6.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Its half-life is long enough for preparation and imaging but short enough to limit dose; $140\,\text{keV}$ gamma photons can leave the body for detection; and technetium-99m can be attached to compounds that concentrate in a chosen organ.	Award one linked justification for each stated property. A $6\,\text{h}$ half-life provides usable activity during the scan but causes activity to fall soon afterwards. The $140\,\text{keV}$ gamma radiation is penetrating enough for external detection while avoiding an unnecessarily high photon energy. Labelling a compound with organ affinity makes the recorded count distribution show that organ's function or structure.	3
02.1	Longer-lived molybdenum-99 continually produces technetium-99m, allowing the hospital to obtain fresh short-lived tracer on site without frequent direct deliveries of technetium-99m.	Technetium-99m has a short half-life, so a transported stock would lose activity rapidly. The longer-lived molybdenum-99 parent remains in the generator and decays to technetium-99m. Staff can extract fresh daughter isotope when required, providing a practical local supply while retaining the dose advantage of the short-lived tracer.	3
03.1	The more penetrating $364\,\text{keV}$ photons are more likely to pass through the lead between adjacent collimator holes or through the scintillation crystal. Passage through the lead between adjacent collimator holes weakens directional information and reduces spatial resolution, while incomplete absorption reduces detection efficiency; $140\,\text{keV}$ photons are easier to collimate and detect.	Higher-energy gamma radiation is more penetrating. Some iodine-131 photons can cross lead between adjacent collimator holes instead of following the selected direction, so the recorded event may be assigned to an incorrect line and the image is less sharply resolved. They are also less likely to deposit all their energy in a finite crystal. Technetium-99m's lower-energy photons are therefore more readily stopped by the collimator where required and absorbed by the detector.	3

Tier 3 · Hard

Mark scheme for 3.10.6.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The labelled glucose accumulates in active tissue; emitted positrons annihilate with electrons to produce two opposite $511\,\text{keV}$ photons; coincidence detectors identify a line of response, and many such events are reconstructed into an uptake map.	The glucose analogue is carried into tissues, so regions with greater uptake contain more tracer. A radionuclide in the tracer emits a positron. After losing energy, the positron annihilates with an electron and produces two $511\,\text{keV}$ gamma photons travelling in approximately opposite directions. Detectors on opposite sides register near-simultaneous photons as a coincidence, locating the event somewhere along the joining line. A computer combines many lines of response to reconstruct the spatial distribution of uptake.	5
02.1	Indium-111 is more suitable because useful activity remains throughout the $30\,\text{h}$ observation and its gamma photons can leave the body, although its longer residence gives a more prolonged dose.	The $30\,\text{h}$ observation lasts five technetium-99m half-lives, leaving only $(1/2)^5=1/32$ of its initial activity. It lasts $30/67=0.448$ indium-111 half-lives, leaving $(1/2)^{0.448}=0.733$ of the initial activity. Both isotopes emit penetrating gamma radiation and both can be attached to the required cells, but indium-111 maintains a detectable distribution for the full observation. Its longer effective residence in the patient prolongs irradiation, so the extra dose is justified only because the scan requires extended tracking.	5
03.1	A coincident pair shows only that the annihilation occurred somewhere on the line joining the nearly opposite detectors, not necessarily at its midpoint. Many lines of response are reconstructed, and their concentration locates high uptake. Smaller detector elements define narrower lines and improve spatial resolution, but intercept fewer photons and reduce count rate.	Electron-positron annihilation produces two $511\,\text{keV}$ photons moving in nearly opposite directions. Detecting them within a coincidence window identifies the joining line of response. Without a measured arrival-time difference, no position along that line is selected, so a midpoint assignment invents information. The computer accumulates many lines; the region through which many of them pass is displayed as high uptake. Reducing detector-element width narrows the range of possible lines and improves spatial resolution, while the smaller collecting area lowers the number of accepted pairs per second.	5
04.1	$43\,\text{s}^{-1}$ and $17\,\text{s}^{-1}$ , a $60\%$ reduction; the narrower window rejects more unrelated detections but may also reject genuine pairs if detector timing uncertainty is comparable with the window	For $15\,\text{ns}$ , $R_a=2(3.2\times10^4)(4.5\times10^4)(15\times10^{-9})=43.2\,\text{s}^{-1}$ , giving $43\,\text{s}^{-1}$ to two significant figures. For $6.0\,\text{ns}$ , $R_a=2(3.2\times10^4)(4.5\times10^4)(6.0\times10^{-9})=17.28\,\text{s}^{-1}$ , or $17\,\text{s}^{-1}$ . The reduction is $(43.2-17.28)/43.2=0.600$ , or $60\%$ . A smaller coincidence interval makes two unrelated photons less likely to be paired, reducing false lines of response. If the interval becomes narrower than the spread in arrival-time measurements, some true opposite-photon pairs will fall outside it and detection efficiency will decrease.	5
05.1	The organ-to-background ratios are $29$ for X and $14$ for Y, so X is the better tracer because its greater organ affinity produces stronger image contrast under otherwise equal conditions.	For X, the activity concentration in the organ is proportional to $0.15/0.24=0.625$ per kilogram, while the background concentration is $0.85/40=0.02125$ per kilogram. Equal viewed masses therefore give ratio $0.625/0.02125=29.412$ , or $29$ to two significant figures. For Y, the ratio is $(0.080/0.24)/(0.92/40)=14.493$ , or $14$ . Because photon energy, physical half-life, total activity and detector response are stated to be equal, the larger count ratio directly represents better contrast between the target and background. X is preferable because its labelled compound has greater affinity for the organ; its penetrating gamma photons can then carry that more concentrated distribution to the external detector.	6

3.10.6.2 · Half-life

Tier 1 · Easy

Mark scheme for 3.10.6.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	The time for biological processes alone to remove half of the tracer from the organ.	The definition must identify removal by biological processes and a reduction to one half; radioactive decay is excluded from this definition.	2
02.1	The time for the number of undecayed nuclei, or the activity, to fall to half its initial value.	Physical half-life describes radioactive decay of the radionuclide itself. The definition requires a fall to one half and is independent of biological removal.	2

Tier 2 · Standard

Mark scheme for 3.10.6.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$11\,\text{h}$	Use $1/T_E=1/T_B+1/T_P=1/30+1/18=0.0889\,\text{h}^{-1}$ . Therefore $T_E=1/0.0889=11.25\,\text{h}$ , which is $11\,\text{h}$ to two significant figures.	3
02.1	Radioactive decay and biological removal act together, so the total rate of loss is greater than either rate alone. The effective half-life must therefore be shorter than both component half-lives, whereas $20\,\text{h}$ is longer than the $12\,\text{h}$ physical half-life.	The reciprocal relation adds the two positive removal rates: $1/T_E=1/T_B+1/T_P$ . Their sum is greater than either separate rate, so its reciprocal $T_E$ must be smaller than both $T_B$ and $T_P$ . The claimed $20\,\text{h}$ exceeds $T_P=12\,\text{h}$ and is impossible.	3
03.1	$59\%$	With both processes acting, the remaining fraction is $(1/2)^{10/5.0}=0.250$ . With radioactive decay alone it would be $(1/2)^{10/8.0}=0.4204$ . The required fraction is $0.250/0.4204=0.5946$ , so $59\%$ remains relative to the physical-decay-only amount.	3

Tier 3 · Hard

Mark scheme for 3.10.6.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$18.0\,\text{h}$ and $12.5\%$	Rearrange the reciprocal relation: $1/T_B=1/T_E-1/T_P=1/7.20-1/12.0=0.0556\,\text{h}^{-1}$ , so $T_B=18.0\,\text{h}$ . The elapsed time is $21.6/7.20=3.00$ effective half-lives. The remaining activity fraction is therefore $(1/2)^3=0.125$ , giving $12.5\%$ .	5
02.1	$30\,\text{h}$	First find the effective half-life: $1/T_E=1/20+1/30=1/12\,\text{h}^{-1}$ , so $T_E=12\,\text{h}$ . Radioactive decrease gives $0.18=(1/2)^{t/T_E}$ . Taking logarithms, $t=12\ln(0.18)/\ln(0.5)=29.69\,\text{h}$ , which is $30\,\text{h}$ to two significant figures.	5
03.1	$T_P=16\,\text{h}$ , $T_B=8.0\,\text{h}$ and $T_E=5.3\,\text{h}$	For physical decay, $0.70=(1/2)^{8.0/T_P}$ , so $T_P=8.0\ln(0.5)/\ln(0.70)=15.55\,\text{h}=16\,\text{h}$ . The biological process supplies the additional factor $0.35/0.70=0.50$ in $8.0\,\text{h}$ , so $T_B=8.0\,\text{h}$ . Finally, $1/T_E=1/15.55+1/8.0$ , giving $T_E=5.282\,\text{h}=5.3\,\text{h}$ to two significant figures.	4
04.1	$9.7\,\text{MBq}$ immediately after the third administration and $21\,\text{h}$ more to fall below $2.0\,\text{MBq}$ (accept $20.5\,\text{h}$ )	At $12.0\,\text{h}$ , the three contributions are $4.8(1/2)^{12/9}$ , $4.8(1/2)^{6/9}$ and $4.8\,\text{MBq}$ . Their sum is $4.8[0.39685+0.62996+1]=9.7287\,\text{MBq}$ , giving $9.7\,\text{MBq}$ to two significant figures. If $t$ is measured from the third administration, $2.0=9.7287(1/2)^{t/9.0}$ . Therefore $t=9.0\ln(2.0/9.7287)/\ln(0.5)=20.54\,\text{h}$ , so the total falls below the threshold after $21\,\text{h}$ to two significant figures. The effective half-life is used because both physical decay and biological removal reduce the activity in the patient.	5
05.1	$T_{E,A}=8.4\,\text{h}$ , $T_{E,B}=4.7\,\text{h}$ ; the activities become equal after $14\,\text{h}$ at approximately $0.64\,\text{MBq}$ each (accept $0.62$ to $0.64\,\text{MBq}$ )	For A, $1/T_{E,A}=1/14+1/21$ , so $T_{E,A}=8.4\,\text{h}$ . For B, $1/T_{E,B}=1/14+1/7.0$ , so $T_{E,B}=4.6667\,\text{h}=4.7\,\text{h}$ . Equality requires $2.0(1/2)^{t/8.4}=5.0(1/2)^{t/4.6667}$ . Hence $2.5=2^{t(1/4.6667-1/8.4)}$ and $t=13.880\,\text{h}=14\,\text{h}$ . Substitution without rounding gives $A=2.0(1/2)^{13.880/8.4}=0.6362\,\text{MBq}$ , or $0.64\,\text{MBq}$ . Although B starts higher, its shorter biological and effective half-lives make its activity decrease more rapidly.	6

3.10.6.3 · Gamma camera

Tier 1 · Easy

Mark scheme for 3.10.6.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	It absorbs gamma photons arriving from unsuitable directions so that detected photons retain directional information.	The collimator selects photon directions before the scintillator; it does not focus gamma rays with refraction.	1
02.1	It converts absorbed gamma-photon energy into visible light.	A gamma interaction in the crystal produces a flash of visible photons for the photomultiplier array.	1

Tier 2 · Standard

Mark scheme for 3.10.6.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The photon produces a light flash in the scintillator, the photocathode emits photoelectrons, and successive dynodes multiply them to give an anode pulse.	The gamma photon deposits energy in the scintillation crystal, which emits visible photons. Light reaching a photomultiplier photocathode releases electrons by the photoelectric effect. These electrons accelerate to successive dynodes and cause secondary emission at each stage, so a much larger electron pulse is collected at the anode.	3
02.1	Pulse height indicates deposited photon energy, so lower-energy scattered events can be rejected because their directions no longer identify the source position accurately.	The summed photomultiplier pulse is related to the energy deposited in the scintillator. Gamma photons scattered in the patient generally reach the camera with reduced energy and altered direction. An energy window rejects their lower pulse heights, preventing those events from being assigned to misleading positions and improving image contrast or resolution.	3
03.1	P occurred nearer the first photomultiplier, Q occurred midway between them, and the equal total pulse of $10$ units means the same gamma energy was deposited in each event.	The larger share of event P's light reaches the first photomultiplier, so P is nearer that tube. Equal signals for Q place it midway between the two tubes. Position is inferred from the relative pulse sizes, while deposited energy is related to their sum; both events have total pulse size $10$ units and therefore correspond to the same deposited energy.	3

Tier 3 · Hard

Mark scheme for 3.10.6.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The collimator selects photon directions; relative signals from nearby photomultipliers locate each scintillation; pulse height estimates photon energy so scattered events are rejected; accumulating accepted events produces the image, with a resolution-sensitivity trade-off set by the collimator.	The lead collimator accepts only a restricted range of directions, linking a detected photon to a line from the patient. A scintillation spreads light to several photomultipliers, and the relative sizes of their pulses allow the computer to estimate the interaction position. The total pulse height is related to deposited photon energy, so an energy window rejects lower-energy scattered photons that would be placed incorrectly. The computer accumulates the accepted positions to form the count image. Smaller collimator apertures sharpen directional selection and improve resolution, but fewer photons pass, so sensitivity falls.	5
02.1	$5.0\times10^{-13}\,\text{C}$ ; each successive dynode is more positive, so electrons are accelerated and arrive with enough energy to release several secondary electrons	After ten dynodes the electron number is $N=3(4.0)^{10}=3.145728\times10^6$ . The collected charge is $Q=Ne=(3.145728\times10^6)(1.60\times10^{-19})=5.033\times10^{-13}\,\text{C}$ , which is $5.0\times10^{-13}\,\text{C}$ to two significant figures. Each later dynode is more positive so that electrons accelerate towards it and arrive with enough kinetic energy to release several secondary electrons, sustaining the multiplication.	4
03.1	The new resolution distance is $5.0\,\text{mm}$ , so the $7.0\,\text{mm}$ separation can be resolved; the count rate falls to $225\,\text{s}^{-1}$ and $80\,\text{s}$ is needed.	Halving the hole diameter reduces the resolution distance from $10\,\text{mm}$ to $5.0\,\text{mm}$ . Since $7.0\,\text{mm}>5.0\,\text{mm}$ , the two regions can be distinguished. The original count is $(900)(20)=1.80\times10^4$ . The new rate is $900(1/2)^2=225\,\text{s}^{-1}$ , so the required time is $(1.80\times10^4)/225=80\,\text{s}$ . Narrower holes improve directional selection but reduce sensitivity, hence the longer acquisition.	5
04.1	The significance increases from $7.8$ to $11$ standard uncertainties (accept $10.6$ ) even though fewer events remain, because the narrower energy window rejects a greater fraction of background than of the tracer excess.	Initially the excess is $5600-4800=800$ events and its uncertainty is $\sqrt{5600+4800}=101.98$ , so the significance is $800/101.98=7.8446$ , or $7.8$ . Treat the target count as $4800$ background events plus an $800$ event excess. After narrowing the window, the background in each region is $0.25(4800)=1200$ , while the retained excess is $0.72(800)=576$ . The target and background counts are therefore $1776$ and $1200$ . The new significance is $576/\sqrt{1776+1200}=10.559$ , or $11$ to two significant figures. Pulse-height selection has improved the evidence for localisation because lower-energy scattered events form much of the rejected background, despite the reduced total count.	4
05.1	Removing the collimator raises the count rate but destroys directional information and spatial resolution. A wider energy window accepts more scattered photons, raising counts but adding events at incorrect positions and reducing contrast. The weaker scintillation gives fewer photoelectrons and smaller photomultiplier pulses, so genuine events are more likely to fall below the detection threshold and photon-energy discrimination becomes less reliable.	Without the lead collimator, photons from a wide range of directions reach the scintillator. More events are detected, but their source directions are unknown, so accumulating them cannot produce a sharply located tracer distribution. Widening the pulse-height window also increases the recorded rate, but low-energy photons are commonly those scattered in the patient; scattering changes direction, so these events are placed misleadingly and degrade image contrast or resolution. If each absorbed gamma photon produces fewer visible photons, fewer photoelectrons are released at the photocathodes and the resulting anode pulse is smaller. Some genuine events may then fall below the counting threshold, while the greater relative effect of pulse fluctuations makes it harder to separate full-energy events from scattered events. The change therefore reduces useful sensitivity and weakens energy selection rather than improving the image.	6

3.10.6.4 · Use of high-energy X-rays

Tier 1 · Easy

Mark scheme for 3.10.6.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	They are sufficiently penetrating to reach and ionise the deep tumour.	The required link is high photon energy to penetration to the depth of the tumour.	1
02.1	Shielding is placed so that it absorbs X-rays travelling towards healthy tissue outside the treatment field.	The shielding attenuates radiation outside the planned target region, reducing the dose received by healthy cells.	1

Tier 2 · Standard

Mark scheme for 3.10.6.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	All beams overlap at the tumour so their doses add there, whereas each healthy region is crossed by fewer beams and receives a smaller dose.	Aim each beam so that it passes through the tumour. At the tumour the energy deposited by all beam directions adds to the therapeutic dose. Away from the tumour, the entry and exit paths are spread through different healthy regions, so any one region receives only part of the total dose.	3
02.1	Imaging locates the tumour and nearby healthy organs, and computer planning selects beam directions and field shapes that overlap at the tumour while avoiding healthy tissue as far as possible.	Imaging provides the positions and boundaries of the target and nearby sensitive structures. A computer uses this spatial information to choose beam directions and collimated field shapes. The beams are arranged to combine at the tumour while reducing the number or intensity of paths through healthy cells.	3
03.1	The unshaped field contains $14\,\text{cm}^2$ of healthy tissue. The tumour-matched field contains $3.0\,\text{cm}^2$ of healthy tissue. This is $11\,\text{cm}^2$ less healthy tissue, a reduction of about $79\%$ . Field shaping excludes healthy cells from the beam cross-section while preserving tumour coverage, so less healthy tissue absorbs off-target X-rays.	The unshaped field includes $48-20=28$ healthy-tissue cells, giving area $28(0.50)=14\,\text{cm}^2$ . The matched field includes $26-20=6$ healthy-tissue cells, giving $6(0.50)=3.0\,\text{cm}^2$ . The decrease is $14-3.0=11\,\text{cm}^2$ , or $(11/14)\times100=78.6\%$ , about $79\%$ . Matching the field more closely to the tumour removes healthy tissue from the beam cross-section while retaining tumour coverage, so fewer healthy cells receive ionising radiation.	4

Tier 3 · Hard

Mark scheme for 3.10.6.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$6.5\,\text{J}$ ; suitable methods include converging beams from different directions, collimation or shielding, and accurate imaging and planning	Since $1\,\text{Gy}=1\,\text{J kg}^{-1}$ , the energy absorbed per session is $E=Dm=(1.8)(0.18)=0.324\,\text{J}$ . Over $20$ sessions, $E_{\text{total}}=20(0.324)=6.48\,\text{J}=6.5\,\text{J}$ . Healthy-tissue exposure can be limited by using several beam directions that overlap at the tumour but spread entry dose, and by collimating or shielding the beam to exclude other tissue. Accurate imaging and computer planning improve targeting so that less healthy tissue lies within the treatment fields.	5
02.1	With four beams the healthy region receives $7.5\%$ of the tumour total; with one beam it would receive $30\%$ , so using four directions reduces this exposure by a factor of four.	Each equal beam contributes $100\%/4=25\%$ of the tumour's total absorbed energy per kilogram. The healthy region receives $0.30\times25\%=7.5\%$ of the tumour total because only one beam crosses it. A single beam would contribute the whole tumour total, so the same healthy region would receive $0.30\times100\%=30\%$ . The four-direction plan therefore reduces this healthy-tissue exposure by a factor of $30/7.5=4$ .	5
03.1	The organ receives $10.8\%$ of the tumour dose with plan A and $7.2\%$ with plan B, so plan B gives the smaller organ dose.	For plan A, the organ dose is $(0.40)(0.20)+(0.35)(0.080)+(0.25)(0)=0.108$ , or $10.8\%$ of the tumour dose. For plan B, the organ dose is $(0.50)(0.12)+(0.30)(0.040)+(0.20)(0)=0.072$ , or $7.2\%$ . Both plans preserve the prescribed tumour dose, but plan B reduces the organ dose by $3.6$ percentage points and is therefore safer for that organ.	5
04.1	High-energy X-rays are sufficiently penetrating to reach a deep tumour. They produce ionisation in tumour cells, causing cell damage that can prevent successful division. The effect is not selective: X-rays also pass through and ionise healthy tissue, so healthy cells can be damaged and their exposure cannot be reduced to zero. The treatment is justified only when the expected tumour control outweighs this unavoidable risk.	A deep target requires penetrating radiation so that useful energy reaches the tumour instead of being absorbed mainly near the surface. High-energy X-rays are ionising: the ionisation can damage tumour cells and prevent them from dividing successfully. Nothing in the interaction identifies a cell as malignant, so the same radiation can ionise and damage healthy cells on the beam path or beyond the tumour. Treatment therefore cannot make healthy-cell exposure exactly zero. Its use depends on delivering enough energy to control the tumour while keeping the unavoidable healthy-tissue risk as small as practicable.	5
05.1	$9.0\times10^6$ ionisations; ionisation can damage a cell and prevent successful division, but the same process can damage healthy cells, so the absorbed energy must be localised to the tumour as far as possible	The energy transferred per ionisation is $(32)(1.60\times10^{-19})=5.12\times10^{-18}\,\text{J}$ . The number produced is $(4.6\times10^{-11})/(5.12\times10^{-18})=8.9844\times10^6$ , or $9.0\times10^6$ to two significant figures. Ionisation causes cell damage, so the tumour cell may be unable to divide successfully. Ionisation is not selective: the same absorbed energy can damage a healthy cell. Accurate treatment planning is therefore required to concentrate the absorbed energy in the target and limit healthy-cell exposure.	5

3.10.6.5 · Use of radioactive implants

Tier 1 · Easy

Mark scheme for 3.10.6.5 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Beta radiation has a short range in tissue, so its ionising dose is concentrated near the implant.	Link the limited penetration of beta particles to a high local dose and reduced dose to distant healthy tissue.	1
02.1	The sources can give a more even local dose throughout the tumour.	Several source positions can match the tumour shape and reduce regions that are too far from a short-range beta source.	1

Tier 2 · Standard

Mark scheme for 3.10.6.5 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The implant is beside the tumour and beta particles have a short range, so most energy is deposited locally; an external X-ray beam must cross healthy tissue before and after reaching the tumour.	The source is positioned within or adjacent to the target, avoiding a long entry path. Beta particles are strongly ionising over a limited range, so their energy is deposited close to the implant. High-energy X-rays are penetrating and an external beam necessarily irradiates some healthy tissue along its entry and exit paths.	3
02.1	Removing the implant stops further irradiation of the tumour and avoids an unnecessary additional dose to nearby healthy tissue.	Once the planned dose has been delivered, leaving the beta source in place would continue to ionise tissue. Removing it ends the exposure and limits damage to healthy cells near the tumour.	2
03.1	Each implant produces $7.2\times10^9$ decays, but neither can directly irradiate cells $6.0\,\text{mm}$ away because this exceeds the $4.0\,\text{mm}$ beta range.	For A, $N=At=(4.0\times10^6)(30\times60)=7.2\times10^9$ . For B, $N=(2.0\times10^6)(60\times60)=7.2\times10^9$ , so the emitted particle totals are equal. Cells $6.0\,\text{mm}$ away lie beyond the stated $4.0\,\text{mm}$ range, showing why source positions must be distributed through a larger tumour.	3

Tier 3 · Hard

Mark scheme for 3.10.6.5 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.7\,\text{MBq}$ ; placement in or beside the tumour and the short range of beta radiation localise the dose	The elapsed number of half-lives is $120/48=2.5$ . Hence $A=9.6(1/2)^{2.5}=1.70\,\text{MBq}=1.7\,\text{MBq}$ . Placing the source within or close to the tumour avoids irradiating a long path of healthy tissue before the radiation reaches its target. Beta particles have a short range in tissue, so most ionisation occurs near the implant and relatively little dose reaches distant cells. The source can also be removed when the planned treatment time is complete.	5
02.1	$1.3\times10^{-2}\,\text{Gy}$	The number of decays is $N=At=(2.4\times10^6)(1.8\times10^3)=4.32\times10^9$ . The mean energy per decay is $(0.52\times10^6)(1.60\times10^{-19})=8.32\times10^{-14}\,\text{J}$ . The tumour absorbs $0.65(4.32\times10^9)(8.32\times10^{-14})=2.336\times10^{-4}\,\text{J}$ . Hence $D=E/m=(2.336\times10^{-4})/0.018=1.298\times10^{-2}\,\text{Gy}$ , which is $1.3\times10^{-2}\,\text{Gy}$ to two significant figures.	5
03.1	$5.0\,\text{MBq}$ The activity decreases during the treatment, so the initial activity is not the activity throughout and $A_0t$ overestimates the number of decays.	The required absorbed energy is $E=Dm=(0.423)(0.020)=8.46\times10^{-3}\,\text{J}$ . The required number of decays is $N=(8.46\times10^{-3})/[0.75(1.12\times10^{-13})]=1.007\times10^{11}$ . The decay constant is $\lambda=\ln2/(6.0\times3600)=3.209\times10^{-5}\,\text{s}^{-1}$ and $t=9.0\times3600=3.24\times10^4\,\text{s}$ . From $N=(A_0/\lambda)(1-e^{-\lambda t})$ , $A_0=N\lambda/(1-e^{-\lambda t})=5.00\times10^6\,\text{Bq}=5.0\,\text{MBq}$ to two significant figures. Multiplying $A_0$ by $t$ would treat the activity as constant at its initial value, although radioactive decay makes it decrease throughout the treatment.	4
04.1	With one misplaced implant, the tumour receives $3.8\,\text{Gy}$ and the healthy tissue $0.83\,\text{Gy}$ ; with all three correctly positioned, the doses are $4.9\,\text{Gy}$ and $0.41\,\text{Gy}$ respectively (accept $0.40\,\text{Gy}$ )	Each implant emits $0.090/3=0.030\,\text{J}$ . With two correct sources and one misplaced source, the tumour absorbs $2(0.82)(0.030)+(0.25)(0.030)=0.0567\,\text{J}$ , so $D_{\text{tumour}}=0.0567/0.015=3.78\,\text{Gy}=3.8\,\text{Gy}$ . Healthy tissue absorbs $2(0.18)(0.030)+(0.75)(0.030)=0.0333\,\text{J}$ , giving $0.0333/0.040=0.8325\,\text{Gy}=0.83\,\text{Gy}$ . If all three are correctly positioned, the tumour dose is $3(0.82)(0.030)/0.015=4.92\,\text{Gy}=4.9\,\text{Gy}$ and the healthy-tissue dose is $3(0.18)(0.030)/0.040=0.405\,\text{Gy}=0.41\,\text{Gy}$ . Accurate placement both increases the intended local tumour dose and reduces irradiation of nearby healthy tissue.	6
05.1	Use three implants at $3.0\,\text{mm}$ , $9.0\,\text{mm}$ and $15\,\text{mm}$ from one short edge, each with activity $2.7\,\text{MBq}$	The furthest points from the centre line are $4.0\,\text{mm}$ away. For a corner to lie within the $5.0\,\text{mm}$ beta range, the greatest allowed displacement along the centre line is $\sqrt{5.0^2-4.0^2}=3.0\,\text{mm}$ . One source can therefore cover a $6.0\,\text{mm}$ length of the tumour, so an $18\,\text{mm}$ length requires at least $18/6.0=3$ sources. Positions $3.0$ , $9.0$ and $15\,\text{mm}$ make the end corners and the midpoints between sources exactly $5.0\,\text{mm}$ away or less. Equal activities give $8.1/3=2.7\,\text{MBq}$ per source. Distributing short-range beta sources prevents distant parts of the tumour being underdosed while limiting irradiation beyond its boundary.	6

3.10.6.6 · Imaging comparisons

Tier 1 · Easy

Mark scheme for 3.10.6.6 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Ultrasound imaging	Ultrasound is portable, produces real-time images and does not use ionising radiation, so it meets all three requirements.	1
02.1	Plain X-ray imaging	A plain X-ray image is quick and widely available, and bone gives strong X-ray contrast, so it suits the stated purpose despite the small ionising dose.	1

Tier 2 · Standard

Mark scheme for 3.10.6.6 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	MR generally gives better soft-tissue contrast and no ionising dose, making it preferable for repeated scans, but CT is usually faster and more available while exposing the patient to ionising radiation.	MR distinguishes soft tissues well, so tumour boundaries can have high contrast. It uses magnetic fields and RF signals rather than ionising radiation, which is advantageous for repeated imaging. CT is often quicker and more readily available, but each scan contributes an X-ray dose; the justified preference is therefore MR unless speed or access is the overriding constraint.	3
02.1	MR gives higher-resolution soft-tissue detail of the deep ligament without ionising radiation, but is slower, costlier and less convenient; ultrasound is non-ionising, portable and can show movement in real time, but has lower resolution and is more operator-dependent.	MR distinguishes soft tissues with good spatial detail, so it can show the position and extent of a deep ligament injury. It does not use ionising radiation, but the scanner is expensive, less available and takes longer. Ultrasound is also non-ionising and is convenient for a portable, real-time examination. Its image quality depends more strongly on the operator and acoustic access, and its resolution of a deep ligament is generally lower than MR.	3
03.1	A labelled compound is taken up or cleared according to kidney function, so the gamma-camera count distribution can show reduced physiological activity. CT reconstructs X-ray attenuation and mainly shows anatomy, so normally shaped tissue can appear unremarkable despite impaired function.	The tracer is chosen for its biological behaviour, and emitted gamma photons allow its changing distribution to be detected outside the body. Reduced uptake or clearance changes the recorded count pattern, providing functional information. CT instead maps the spatial distribution of X-ray attenuation, giving high-resolution structural information but not necessarily detecting an early functional change.	3

Tier 3 · Hard

Mark scheme for 3.10.6.6 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Ultrasound is a reasonable first choice because it is available, convenient and non-ionising; CT is preferred if a more precise high-resolution location is needed despite its larger dose; MR is less suitable because stone-like calcified material gives poor contrast and the scan is costly and less available.	Ultrasound is portable or widely accessible, quick and non-ionising, so it is suitable as an initial investigation, although its spatial resolution and image quality are lower and depend on the acoustic path. CT gives high-resolution cross-sectional localisation and strong contrast for dense stone material without superposition, but it uses a comparatively large ionising dose and is more costly. MR provides strong soft-tissue contrast without ionising radiation, yet calcified stone material is poorly shown, scans take longer and access is more limited. Therefore choose ultrasound first when it can answer the question, and use CT when the clinical need for precise localisation outweighs the radiation risk; MR is not the best choice for this target.	6
02.1	PET is the better main technique for deciding whether active tumour has returned because it can reveal abnormal metabolic uptake. It uses an internal ionising tracer and has lower spatial resolution, while MR gives detailed soft-tissue anatomy without ionising radiation and is useful for complementary localisation.	A PET tracer accumulates according to physiological activity, so unusually high uptake can provide direct evidence of active recurrence. PET requires an internally administered radioactive tracer, gives an ionising dose and generally has lower spatial resolution. MR provides strong soft-tissue contrast and detailed anatomical position without ionising radiation, making repeated structural imaging safer, but anatomy alone may not establish whether tissue is metabolically active. Recommend PET as the main test for the stated question of active recurrence, with MR used when precise anatomical localisation is also required.	5
03.1	Use ultrasound with PET: together they take $55\,\text{min}$ , give $0.85$ times the dose limit and provide structural and functional information at resolutions smaller than $9\,\text{mm}$ . MR plus PET takes $85\,\text{min}$ , while CT alone exceeds the dose limit and CT plus PET gives twice the limit.	PET is required for the activity map and resolves $6\,\text{mm}$ , smaller than the $9\,\text{mm}$ lesion. Pairing it with ultrasound gives total time $35+20=55\,\text{min}$ and relative dose $0.85+0=0.85$ of the limit; ultrasound's $4\,\text{mm}$ structural resolution is also sufficient. MR plus PET would take $50+35=85\,\text{min}$ and fail the time limit. CT alone gives $1.15$ times the limit, and CT plus PET gives $1.15+0.85=2.00$ times the limit. Ultrasound with PET is therefore the only pair satisfying information, resolution, time and safety constraints.	5
04.1	The proposal is not justified as a universal rule. Ultrasound is non-ionising and portable, but bone blocks the acoustic path, so it is unsuitable for imaging an adult brain through the skull. MR is also non-ionising and can give high-resolution soft-tissue images, but it is slower and may be unsuitable for a patient with some metallic implants or who cannot remain still. CT uses ionising X-rays and therefore carries a radiation risk, but it is fast and gives cross-sectional images that can locate the bleed without superposition. Choose the method that supplies the required information promptly while keeping radiation risk proportionate.	Avoiding ionising radiation is a genuine safety advantage, but a technique is useful only if it can obtain the required information. Ultrasound waves do not pass effectively through the skull, so ultrasound cannot replace CT for this purpose despite its portability and absence of ionising dose. MR can provide high-resolution soft-tissue information without ionising radiation, but the examination usually takes longer and may be impractical for some patients, which matters when a bleed must be found quickly. CT exposes the patient to ionising X-rays, yet it rapidly reconstructs cross-sectional images and avoids the superposition present in a plain radiograph. The decision therefore rests on resolution, diagnostic usefulness, convenience and safety. Cost may affect availability, but it is not needed to justify the clinical choice and carries no independent marking point here.	6
05.1	Processing can enhance recorded CT attenuation values but cannot invent metabolic or time-dependent information that was never measured, and it cannot undo the ionising dose. PET can add a tracer-uptake map of physiological activity, while ultrasound can add real-time motion without ionising radiation when there is a suitable acoustic path.	CT stores a spatial map reconstructed from X-ray attenuation measurements. Display processing can rescale contrast or emphasise boundaries already present, but attenuation alone does not record tracer uptake or metabolism, so it cannot be converted into a functional map. One completed CT acquisition also contains no sequence of later times, so software cannot make genuine live motion from the single dataset. The exposure occurred when the X-rays crossed the patient and cannot be removed afterwards. PET would add physiological information because a radioactive tracer accumulates according to activity, although it adds ionising dose and generally has lower spatial resolution. Ultrasound would add a real-time, non-ionising view of motion and is portable, but its usefulness depends on operator skill and an acoustic path not blocked by bone or gas. These methods complement rather than retrospectively transform the CT data.	6