3.9 Astrophysics (A-level only) — revision question pack

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

Checked against AQA 7408 section 3.9. 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.

3.9.1.1 · Astronomical telescope consisting of two converging lenses

Explanation

An astronomical telescope in normal adjustment uses a long-focal-length converging objective and a short-focal-length converging eyepiece. Parallel rays from a distant object are focused by the objective to form a real, inverted intermediate image.
This image lies in the focal plane shared by the two lenses, so the eyepiece sends parallel rays to the relaxed eye and the final image is virtual at infinity.
The lens separation is therefore $f_o+f_e$ .
Angular magnification compares the angle subtended at the eye through the telescope with that at the unaided eye; its magnitude is $M=f_o/f_e$ .
A complete ray diagram must show the intermediate image, both focal points and parallel emergent rays.

Ray path through a two-converging-lens telescope in normal adjustment.

Worked example

A telescope in normal adjustment has $f_o=1.20\,\text{m}$ and $f_e=30.0\,\text{mm}$ . Calculate its angular magnification and lens separation.

1.Convert the eyepiece focal length: $f_e=0.0300\,\text{m}$ .
2.Use $M=f_o/f_e=1.20/0.0300=40.0$ .
3.Use the normal-adjustment separation $f_o+f_e=1.20+0.0300=1.23\,\text{m}$ .

Answer: The angular magnification has magnitude $40.0$ and the lens separation is $1.23\,\text{m}$ .

Common mistakes

Don't use image-height magnification instead of the ratio of angles subtended at the eye.
Don't subtract the focal lengths even though the objective and eyepiece focal planes coincide in normal adjustment.
Don't leave the emergent rays converging, which places the final image at a finite distance rather than at infinity.

Exam tip

For a ray-diagram question, label both lenses, the shared focal plane and the real inverted intermediate image before drawing the parallel emergent rays.

Tier 1 · Easy

An astronomical telescope in normal adjustment has objective focal length $1.20\,\text{m}$ and eyepiece focal length $30.0\,\text{mm}$ . Calculate the magnitude of its angular magnification.
[1 mark]
Total for this question: 1
State where the real intermediate image is formed in an astronomical telescope in normal adjustment.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A distant crater subtends $2.0\times10^{-4}\,\text{rad}$ at the unaided eye. It is viewed through a normal-adjustment telescope with $f_o=1.500\,\text{m}$ and $f_e=25.0\,\text{mm}$ . Determine the angle subtended at the eye and the lens separation.
[3 marks]
Total for this question: 3
Explain why an astronomical telescope in normal adjustment allows relaxed viewing and has lens separation $f_o+f_e$ .
[3 marks]
Total for this question: 3
A distant double star subtends an angle of $3.50\times10^{-4}\,\text{rad}$ at the objective of a telescope. The objective focal length is $1.80\,\text{m}$ . Calculate the separation of the two images in the real intermediate image and state its orientation.
[2 marks]
Total for this question: 2

Tier 3 · Hard

A telescope is to make an object that subtends $1.50\times10^{-4}\,\text{rad}$ appear to subtend at least $4.80\times10^{-3}\,\text{rad}$ . The objective focal length is $0.960\,\text{m}$ . Determine the greatest acceptable eyepiece focal length and the corresponding maximum telescope length in normal adjustment.
[5 marks]
Total for this question: 5
A telescope in normal adjustment has angular magnification $36.0$ and lens separation $1.295\,\text{m}$ . Determine the focal length of each lens to three significant figures.
[3 marks]
Total for this question: 3
A telescope in normal adjustment is $1.260\,\text{m}$ long. A distant feature subtends $1.60\times10^{-4}\,\text{rad}$ at the objective and produces a $0.192\,\text{mm}$ image in the shared focal plane. Determine the angle subtended by the final image at the eye.
[4 marks]
Total for this question: 4
A student describes a two-converging-lens astronomical telescope in normal adjustment as follows: the objective forms an upright virtual image; the eyepiece is placed before this image; the emergent rays converge; and the angular magnification is $f_o+f_e$ . Identify and correct the errors, explaining how the corrected arrangement permits relaxed viewing.
[5 marks]
Total for this question: 5
A telescope in normal adjustment uses a $1.950\,\text{m}$ converging lens as its objective and a $52.0\,\text{mm}$ converging lens as its eyepiece. Determine its angular magnification and length. The lenses are then interchanged and the telescope is returned to normal adjustment. Determine the new angular magnification and length, then state with a reason whether interchanging the lenses is useful for viewing a distant object.
[4 marks]
Total for this question: 4

3.9.1.2 · Reflecting telescopes

Explanation

A Cassegrain reflecting telescope uses a large parabolic concave primary mirror and a small convex secondary mirror.
Parallel incident rays reflect from the primary and begin to converge; before reaching the primary focus, they strike the secondary and are reflected back through a central hole in the primary towards the eyepiece.
A parabolic primary brings on-axis parallel rays to one focus, avoiding the spherical aberration produced by a spherical mirror.
Reflectors also avoid chromatic aberration because reflection does not depend on refractive index.
Compared with a large refracting objective, a mirror can be supported across its back and made with a larger practical aperture, although the secondary obstructs part of the incoming beam.

Cassegrain ray path from the parabolic primary to the convex secondary and eyepiece.

Worked example

Explain two reasons why a large Cassegrain reflector is usually preferred to a refractor of the same aperture.

1.Reflection does not separate wavelengths by refractive index, so the reflector has no chromatic aberration.
2.A parabolic primary focuses on-axis parallel rays together, avoiding the spherical aberration of a spherical surface.
3.The primary mirror can be supported across its back, whereas a large lens is supported at its edge and is more liable to distort.

Answer: The reflector controls both named aberrations and allows a mechanically practical large aperture.

Common mistakes

Don't draw the convex secondary as a lens and let the rays pass through it.
Don't claim that every mirror avoids spherical aberration; a spherical mirror still gives spherical aberration.
Don't state that a reflector has no disadvantages, ignoring the obstruction caused by the secondary mirror.

Exam tip

In a compare question, link each design feature to its consequence: parabolic surface to reduced spherical aberration, and reflection to no chromatic aberration.

Tier 1 · Easy

State one aberration that a reflecting telescope avoids because its primary element is a mirror rather than a lens.
[1 mark]
Total for this question: 1
State the shape of the primary mirror in a Cassegrain reflecting telescope.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Describe the path of initially parallel rays through a Cassegrain reflecting telescope up to the eyepiece.
[3 marks]
Total for this question: 3
A designer increases the diameter of the convex secondary mirror in a Cassegrain telescope while leaving the primary unchanged. Explain the effect on the detected image.
[3 marks]
Total for this question: 3
Explain why a large primary mirror can retain its shape more effectively than a refracting objective lens of the same diameter.
[2 marks]
Total for this question: 2

Tier 3 · Hard

A research group must choose between a large refracting telescope and a Cassegrain reflector of the same aperture. Discuss why the reflector is usually preferred, including aberrations and construction.
[5 marks]
Total for this question: 5
Parallel rays strike a spherical concave primary mirror at different distances from its principal axis. Explain why replacing it with a parabolic primary improves the image.
[3 marks]
Total for this question: 3
A team compares one relative merit of a Cassegrain reflector with a refractor. Each has an outer objective diameter of $4.00\,\text{m}$ . The reflector's secondary mirror has diameter $0.800\,\text{m}$ and blocks the corresponding part of the incoming beam. The refracting objective transmits $90.0\%$ of the incident radiation. Determine the ratio of the detected signal from the reflector to that from the refractor. Deduce which design detects the greater signal. Ignore all other losses.
[3 marks]
Total for this question: 3
A technician claims that, in a Cassegrain telescope, incoming parallel rays first pass through the central hole, then reflect from the convex secondary and finally from the primary before reaching the eyepiece. Identify the errors and give the correct light path. Explain why the primary surface is parabolic rather than spherical.
[5 marks]
Total for this question: 5
A Cassegrain reflector is to have the same clear collecting area as an unobstructed circular aperture of diameter $3.60\,\text{m}$ . Its convex secondary has diameter $0.200D$ , where $D$ is the primary diameter. Determine $D$ and the percentage of the incoming beam area obstructed by the secondary. State one further optical condition needed for the reflector to form a sharp on-axis image.
[4 marks]
Total for this question: 4

3.9.1.3 · Single dish radio telescopes, I-R, U-V and X-ray telescopes

Explanation

A single-dish radio telescope resembles an optical reflector: a concave dish collects waves and directs them to a receiver near the focus, and its collecting power is proportional to $D^2$ .
Radio wavelengths are much longer than visible wavelengths, so $\theta\approx\lambda/D$ means that a much larger dish is needed for comparable angular resolution.
Radio telescopes can operate through cloud and in daylight, but require sites with little human-made radio interference.
Infrared observations favour high, dry sites and cooled detectors because atmospheric water vapour and thermal radiation interfere.
Most ultraviolet and X-ray astronomy is performed above the atmosphere, usually on satellites, because those wavelengths are strongly absorbed before reaching the ground.

Worked example

An optical telescope of diameter $0.200\,\text{m}$ observes at $550\,\text{nm}$ . Estimate the radio-dish diameter needed for the same angular resolution at $0.210\,\text{m}$ .

1.For equal resolution, set $\lambda_{\text{opt}}/D_{\text{opt}}=\lambda_{\text{radio}}/D_{\text{radio}}$ .
2.Rearrange to $D_{\text{radio}}=D_{\text{opt}}\lambda_{\text{radio}}/\lambda_{\text{opt}}$ .
3.Substitute to obtain $D_{\text{radio}}=0.200(0.210)/(550\times10^{-9})=7.64\times10^4\,\text{m}$ .

Answer: The required diameter is approximately $7.64\times10^4\,\text{m}$ , showing why radio dishes need very large apertures.

Common mistakes

Don't claim that a radio dish has optical-telescope resolution at the same diameter despite its much longer wavelength.
Don't place an X-ray telescope at ground level even though the atmosphere absorbs astronomical X-rays.
Don't say that every non-visible telescope uses a glass objective lens.

Exam tip

A comparison must address structure, observing position, use, resolving power and collecting power rather than naming wavelength bands alone.

Tier 1 · Easy

State why most X-ray telescopes used for astronomy are placed on satellites.
[1 mark]
Total for this question: 1
State the function of the concave dish in a single-dish radio telescope.
[1 mark]
Total for this question: 1

Tier 2 · Standard

An optical telescope of diameter $0.200\,\text{m}$ observes at $550\,\text{nm}$ . Estimate the diameter of a radio dish observing at $0.210\,\text{m}$ that would have the same Rayleigh angular resolution.
[3 marks]
Total for this question: 3
Compare the structure of a single-dish radio telescope with that of a Cassegrain reflecting telescope. State one similarity and one difference.
[2 marks]
Total for this question: 2
A single-dish radio telescope has diameter $25.0\,\text{m}$ and observes at wavelength $0.150\,\text{m}$ . An optical reflector has diameter $2.50\,\text{m}$ and observes at $600\,\text{nm}$ . Compare their collecting powers and minimum angular resolutions. Assume neither aperture is obstructed and use $\theta\approx\lambda/D$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

Discuss suitable observing locations for radio, infrared, ultraviolet and X-ray astronomy, and explain why a radio dish generally has a much larger diameter than an optical telescope.
[6 marks]
Total for this question: 6
Explain why an infrared observatory uses cooled equipment at a high, dry ground site.
[4 marks]
Total for this question: 4
A single-dish radio telescope has diameter $100\,\text{m}$ and observes at wavelength $0.0500\,\text{m}$ . An optical reflector has diameter $2.00\,\text{m}$ and observes at $500\,\text{nm}$ . Two sources have angular separation $4.00\times10^{-4}\,\text{rad}$ . Determine whether each telescope can resolve the sources, compare their collecting powers, and explain one suitable siting condition for each instrument. Assume unobstructed apertures and use $\theta\approx\lambda/D$ .
[6 marks]
Total for this question: 6
Each of two single-dish radio telescopes has diameter $76.0\,\text{m}$ and observes at wavelength $0.180\,\text{m}$ . An optical reflecting telescope has diameter $2.40\,\text{m}$ and observes at wavelength $520\,\text{nm}$ . Compare the minimum angular resolutions of one radio telescope and the optical telescope. The two radio telescopes are to be linked as an interferometer with baseline $4.00\times10^3\,\text{m}$ . Determine the minimum angular resolution of the linked array and state, with a reason, which single instrument it now outperforms. Use $\theta\approx\lambda/D$ , taking the baseline as the effective diameter for the linked array.
[5 marks]
Total for this question: 5
A proposal suggests converting a ground-based single-dish radio telescope into an ultraviolet telescope simply by replacing the radio receiver at the focus with an ultraviolet detector. Discuss the proposal with reference to telescope structure, the observing position and angular resolution.
[5 marks]
Total for this question: 5

3.9.1.4 · Advantages of large diameter telescopes

Explanation

A larger telescope diameter improves both resolution and sensitivity. The Rayleigh criterion gives the minimum angular separation as $\theta\approx\lambda/D$ , where $\theta$ is in radians, so increasing $D$ allows two closer sources to be resolved at the same wavelength.
Collecting power is proportional to the aperture area and therefore to $D^2$ , so a larger telescope detects fainter sources in a given exposure. A CCD has greater quantum efficiency than the eye: it detects a greater proportion of the incident photons.
It can also accumulate charge during a long exposure and gives a permanent digital record that is convenient to process.
Detector resolution depends on pixel size, while the telescope aperture still sets the diffraction limit.
No knowledge of CCD internal structure is required.

Worked example

A telescope diameter increases from $0.50\,\text{m}$ to $2.0\,\text{m}$ at the same wavelength. Compare its minimum resolvable angle and collecting power.

1.The diameter increases by a factor of $2.0/0.50=4$ .
2.Since $\theta\propto1/D$ , the minimum resolvable angle becomes one quarter as large.
3.Since collecting power is proportional to $D^2$ , it increases by $4^2=16$ .

Answer: The minimum resolvable angle is quartered and the collecting power is multiplied by $16$ .

Common mistakes

Don't state that collecting power is proportional to diameter instead of diameter squared.
Don't say that a larger minimum angular resolution is better; better resolution means a smaller resolvable angle.
Don't attribute unlimited resolution to smaller CCD pixels and ignore the aperture's diffraction limit.

Exam tip

For a diameter-ratio question, apply the ratio once to $\theta\propto D^{-1}$ and square it for collecting power.

Tier 1 · Easy

A telescope aperture is increased from $0.30\,\text{m}$ to $0.60\,\text{m}$ . Determine the factor by which its collecting power increases.
[1 mark]
Total for this question: 1
State what is meant by the quantum efficiency of a telescope detector.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Calculate the Rayleigh minimum angular resolution of a $2.50\,\text{m}$ telescope at wavelength $500\,\text{nm}$ . State the effect of increasing the diameter.
[3 marks]
Total for this question: 3
A ground-based optical telescope observes visible light at $600\,\text{nm}$ through a $1.20\,\text{m}$ aperture. A satellite-borne ultraviolet telescope observes at $240\,\text{nm}$ through a $0.800\,\text{m}$ aperture. Determine the ratio of the ultraviolet minimum resolvable angle to the visible-light value.
[2 marks]
Total for this question: 2
A telescope has aperture diameter $0.800\,\text{m}$ . A source supplies $1.50\times10^4$ photons per square metre per second at the aperture. Calculate the number recorded in a $12.0\,\text{s}$ exposure by a detector with quantum efficiency $0.650$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

Two telescopes have diameters $4.00\,\text{m}$ and $10.0\,\text{m}$ and observe at $600\,\text{nm}$ . A pair of faint stars is separated by $8.0\times10^{-8}\,\text{rad}$ . Determine which telescope can resolve the pair, compare their collecting powers, and explain one advantage of recording with a CCD rather than the eye.
[6 marks]
Total for this question: 6
Compare the exposure time and diffraction-limited angular resolution obtained when a $1.50\,\text{m}$ telescope with detector quantum efficiency $0.18$ is replaced by a $3.00\,\text{m}$ telescope with efficiency $0.72$ . The original exposure is $2.40\times10^2\,\text{s}$ and both observations record the same photon count at the same wavelength; report the new exposure time to two significant figures.
[5 marks]
Total for this question: 5
A telescope observing at $550\,\text{nm}$ must resolve a binary with angular separation $1.10\times10^{-7}\,\text{rad}$ . The photon flux at its aperture is $2.00\times10^2\,\text{m}^{-2}\text{s}^{-1}$ , its detector quantum efficiency is $0.400$ , and it must record at least $5.00\times10^5$ photons in $300\,\text{s}$ . Determine the minimum aperture diameter that satisfies both requirements. Use $\theta\approx\lambda/D$ .
[5 marks]
Total for this question: 5
A telescope has aperture diameter $2.40\,\text{m}$ , focal length $18.0\,\text{m}$ and detector pixels of width $8.10\,\mu\text{m}$ . It observes at $510\,\text{nm}$ . Estimate the angular size represented by one pixel and the diffraction-limited angle. Take the one-pixel angle as the detector's minimum separable angle and hence identify the limiting resolution. Repeat the conclusion when the pixels are replaced by $3.00\,\mu\text{m}$ pixels. Use $\theta\approx\lambda/D$ and the small-angle approximation.
[5 marks]
Total for this question: 5
A telescope of aperture diameter $1.10\,\text{m}$ receives $2.40\times10^4$ photons per square metre per second from a star. Its detector records $2.85\times10^5$ photons in $18.0\,\text{s}$ . Determine the detector's quantum efficiency. The aperture is then masked to diameter $0.700\,\text{m}$ . Calculate the exposure needed for the same recorded count and determine the factor by which the minimum angular resolution worsens at unchanged wavelength.
[5 marks]
Total for this question: 5

3.9.2.1 · Classification by luminosity

Explanation

Apparent magnitude $m$ describes how bright a star appears from Earth on the Hipparcos scale. The order is reversed: a smaller or more negative magnitude means a brighter object, while the dimmest stars visible to the unaided eye have $m\approx+6$ .
The scale is logarithmic and subjective.
A difference of one magnitude corresponds to an intensity ratio of $2.51$ , so a magnitude difference $\Delta m$ corresponds to an intensity ratio $2.51^{\lvert\Delta m\rvert}$ , with the lower-magnitude object being more intense.
Apparent magnitude alone does not give luminosity because distance changes the detected intensity.
Examiners expect the direction of the comparison as well as the calculated ratio.

Worked example

Star A has apparent magnitude $1.0$ and star B has apparent magnitude $6.0$ . Calculate how many times brighter A appears than B.

1.The magnitude difference is $6.0-1.0=5.0$ .
2.Use the intensity ratio $2.51^5=99.7$ .
3.Star A has the lower apparent magnitude, so A is the brighter star.

Answer: Star A appears approximately $100$ times brighter than star B.

Common mistakes

Don't call the magnitude- $6$ star brighter because $6$ is numerically greater than $1$ .
Don't multiply the magnitude difference by $2.51$ instead of raising $2.51$ to that difference.
Don't treat apparent magnitude as a direct measure of the star's intrinsic luminosity.

Exam tip

After finding an intensity ratio, state explicitly that the object with the lower apparent magnitude appears brighter.

Tier 1 · Easy

Two stars have apparent magnitudes $+1.0$ and $+4.0$ . State which appears brighter.
[1 mark]
Total for this question: 1
State the approximate apparent magnitude of the dimmest star visible to the unaided eye under dark conditions.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Star P has apparent magnitude $2.00$ and star Q has apparent magnitude $5.00$ . Calculate the ratio $I_P/I_Q$ .
[3 marks]
Total for this question: 3
Two stars have an apparent-intensity ratio of $60.0$ . Determine their apparent-magnitude difference to three significant figures.
[2 marks]
Total for this question: 2
Two unresolved stars have apparent magnitudes $4.00$ and $5.50$ . Determine the apparent magnitude of their combined light.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Interstellar dust reduces the received intensity from a star by a factor of $20.0$ . Before the dimming, its apparent magnitude was $2.00$ . Determine its new apparent magnitude and whether it should remain visible to the unaided eye under dark conditions.
[5 marks]
Total for this question: 5
An unresolved cluster has apparent magnitude $5.00$ . Each identical star in the cluster would have apparent magnitude $6.50$ if observed separately. Estimate the number of stars in the cluster, giving a whole-number result.
[4 marks]
Total for this question: 4
A star has apparent magnitude $4.40$ above Earth's atmosphere. Atmospheric dust transmits $45.0\%$ of its intensity, and a thin cloud transmits a further fraction $x$ . The dimmest star visible to the unaided eye has apparent magnitude $6.00$ . Determine the minimum value of $x$ for the star to remain visible.
[4 marks]
Total for this question: 4
An unresolved eclipsing binary has apparent magnitude $3.20$ outside eclipse. During a total eclipse only star A is visible, and the apparent magnitude is $3.75$ . Determine the apparent magnitude of star B when observed alone and the percentage of the out-of-eclipse intensity supplied by star B.
[5 marks]
Total for this question: 5
A nova has apparent magnitude $7.30$ before an outburst. At maximum brightness its received intensity is $320$ times the original value. Later, its intensity has fallen to one twenty-fifth of the maximum value. Determine the apparent magnitude at maximum and at the later time, and state at which of the three stages it is visible to the unaided eye under dark conditions.
[4 marks]
Total for this question: 4

3.9.2.2 · Absolute magnitude, M

Explanation

Absolute magnitude $M$ is the apparent magnitude a star would have if placed $10\,\text{pc}$ from Earth, so it permits intrinsic luminosities to be compared without the effect of differing distances. The distance modulus is $m-M=5\log_{10}(d/10)$ , where $d$ must be in parsecs.
A more negative value of $M$ represents a more luminous star.
One parsec is approximately $3.26$ light years; both are units of distance, although a light year is defined using the distance light travels in one year.
In calculations, the logarithm is base ten and the distance must not be entered in metres.
A valid conclusion distinguishes apparent brightness $m$ from intrinsic brightness $M$ .

Worked example

A star has apparent magnitude $m=7.5$ and is $100\,\text{pc}$ away. Calculate its absolute magnitude.

1.Use $m-M=5\log_{10}(d/10)$ .
2.Substitute $d=100\,\text{pc}$ : $7.5-M=5\log_{10}(10)=5$ .
3.Rearrange to $M=7.5-5=2.5$ .

Answer: The star's absolute magnitude is $M=+2.5$ .

Common mistakes

Don't substitute distance in metres into a relation that requires parsecs.
Don't use a natural logarithm instead of $\log_{10}$ .
Don't say that a more positive absolute magnitude means a greater intrinsic luminosity.

Exam tip

Write the distance modulus with $d/10$ before substitution; this makes both the parsec unit and the sign of $M$ easier to track.

Tier 1 · Easy

A star is exactly $10\,\text{pc}$ from Earth and has apparent magnitude $4.3$ . State its absolute magnitude.
[1 mark]
Total for this question: 1
Two stars have absolute magnitudes $-2.0$ and $+3.0$ . State which star is more luminous.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A star has apparent magnitude $8.4$ and absolute magnitude $3.4$ . Calculate its distance from Earth in parsecs.
[3 marks]
Total for this question: 3
A star is $24.0\,\text{ly}$ from Earth. Calculate its distance in parsecs and explain why its apparent magnitude is not equal to its absolute magnitude. Use $1\,\text{pc}=3.26\,\text{ly}$ .
[2 marks]
Total for this question: 2
A star has absolute magnitude $+1.80$ and is $65.2\,\text{ly}$ from Earth. Determine its apparent magnitude. Use $1\,\text{pc}=3.26\,\text{ly}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

A supergiant has apparent magnitude $12.0$ and absolute magnitude $-1.0$ . Determine its distance in parsecs and in light years. Use $1\,\text{pc}=3.26\,\text{ly}$ .
[5 marks]
Total for this question: 5
Stars A and B each have apparent magnitude $m=8.00$ . Star A is $40.0\,\text{pc}$ away and star B is $4.00\times10^2\,\text{pc}$ away. Determine the absolute magnitude of each star, then use the inverse-square law to determine the luminosity ratio $L_B/L_A$ .
[4 marks]
Total for this question: 4
A star has absolute magnitude $-2.00$ and observed apparent magnitude $10.00$ . Interstellar dust adds $1.50$ to its apparent magnitude. Determine the star's distance and the factor by which its distance would be overestimated if the dust were ignored.
[5 marks]
Total for this question: 5
A star is independently known to be $320\,\text{pc}$ from Earth and has absolute magnitude $-0.80$ . Its observed apparent magnitude is $8.00$ . Determine the apparent magnitude expected without interstellar absorption, the additional magnitude due to absorption, and the fraction of the star's intensity transmitted.
[5 marks]
Total for this question: 5
Star A has absolute magnitude $-3.20$ and is $80.0\,\text{pc}$ from Earth. Star B has absolute magnitude $+1.30$ but the same apparent magnitude as A. Determine the distance of B and the luminosity ratio $L_A/L_B$ .
[5 marks]
Total for this question: 5

3.9.2.3 · Classification by temperature, black-body radiation

Explanation

A star may be modelled as a black body. Its continuous spectrum has a single peak; increasing temperature moves the peak to shorter wavelength and increases the area under the curve.
Wien's displacement law, $\lambda_{\max}T=2.9\times10^{-3}\,\text{m K}$ , estimates surface temperature from the peak wavelength. Stefan's law is $P=\sigma AT^4$ and, for a spherical star, $A=4\pi R^2$ , so $P\propto R^2T^4$ .
Received intensity follows $I=P/(4\pi d^2)$ when emission is isotropic and absorption is negligible.
Calculations require wavelength in metres and temperature in kelvin.
Comparisons should state which quantities, such as radius or distance, are held constant.

Black-body curves showing the higher, shorter-wavelength peak of the hotter star.

Worked example

A star's spectrum peaks at $480\,\text{nm}$ . Estimate its surface temperature.

1.Convert the peak wavelength: $480\,\text{nm}=4.80\times10^{-7}\,\text{m}$ .
2.Rearrange Wien's law to $T=(2.9\times10^{-3})/\lambda_{\max}$ .
3.Substitute to obtain $T=(2.9\times10^{-3})/(4.80\times10^{-7})=6.04\times10^3\,\text{K}$ .

Answer: The estimated surface temperature is $6.0\times10^3\,\text{K}$ .

Common mistakes

Don't substitute a wavelength in nanometres without converting it to metres.
Don't use degrees Celsius in Stefan's law or Wien's law instead of kelvin.
Don't draw the hotter black-body curve with a longer peak wavelength.

Exam tip

On a sketch, show both required changes for higher temperature: the peak moves left and the total area increases.

Tier 1 · Easy

The black-body spectrum of a star peaks at $580\,\text{nm}$ . Estimate its surface temperature using Wien's law.
[2 marks]
Total for this question: 2
State two changes to a star's black-body spectrum when its surface temperature increases.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Star B has twice the radius of star A but $0.800$ times its surface temperature. Using Stefan's law, calculate $P_B/P_A$ .
[3 marks]
Total for this question: 3
The black-body spectra of stars A and B peak at $420\,\text{nm}$ and $630\,\text{nm}$ respectively. Determine $T_A/T_B$ and identify the hotter star.
[2 marks]
Total for this question: 2
Star A has $1.50$ times the radius of star B. Their black-body spectra peak at $500\,\text{nm}$ and $750\,\text{nm}$ respectively. Determine $P_A/P_B$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

A star $50.0\,\text{pc}$ away produces an intensity of $3.20\times10^{-10}\,\text{W m}^{-2}$ at Earth and has peak wavelength $480\,\text{nm}$ . Assuming isotropic emission and black-body behaviour, determine its surface temperature and radius. Use $1\,\text{pc}=3.086\times10^{16}\,\text{m}$ and $\sigma=5.67\times10^{-8}\,\text{W m}^{-2}\text{K}^{-4}$ .
[6 marks]
Total for this question: 6
Stars A and B produce the same intensity at Earth. Star A is $20.0\,\text{pc}$ away at $6.00\times10^3\,\text{K}$ , while star B is $60.0\,\text{pc}$ away at $9.00\times10^3\,\text{K}$ . Determine $R_B/R_A$ to three significant figures, assuming both stars behave as black bodies.
[5 marks]
Total for this question: 5
Star A has angular diameter $0.800$ milliarcseconds (mas) and a black-body peak at $450\,\text{nm}$ . Star B has angular diameter $1.20\,\text{mas}$ and a peak at $600\,\text{nm}$ . Assuming isotropic black-body emission, determine the ratio of the total intensities received at Earth, $I_A/I_B$ .
[5 marks]
Total for this question: 5
Between two observations, the peak wavelength of a star's black-body spectrum changes from $650\,\text{nm}$ to $500\,\text{nm}$ while its total power output becomes $2.20$ times its original value. Determine the ratio of the final to initial surface temperature and the ratio of the final to initial radius. State whether the star has expanded or contracted.
[4 marks]
Total for this question: 4
A spherical star of radius $4.20\times10^9\,\text{m}$ has a black-body spectrum peaking at $510\,\text{nm}$ . A detector can measure the star only if its received intensity is at least $7.20\times10^{-12}\,\text{W m}^{-2}$ . Determine the star's surface temperature, power output and maximum detectable distance in parsecs. Assume isotropic emission and use $\lambda_{\max}T=2.9\times10^{-3}\,\text{m K}$ , $\sigma=5.67\times10^{-8}\,\text{W m}^{-2}\text{K}^{-4}$ and $1\,\text{pc}=3.086\times10^{16}\,\text{m}$ .
[6 marks]
Total for this question: 6

3.9.2.4 · Principles of the use of stellar spectral classes

Explanation

The spectral sequence OBAFGKM runs from hottest to coolest.
O and B stars are blue, A stars blue-white, F white, G yellow-white, K orange and M red.
Prominent lines also change: O spectra show ionised helium, helium and hydrogen; B show helium and hydrogen; A have the strongest hydrogen Balmer lines and ionised metals; F show ionised metals; G show ionised and neutral metals; K show neutral metals; M show neutral atoms and TiO.
Balmer absorption requires hydrogen atoms already in the $n=2$ state.
Cooler stars have too few atoms excited to $n=2$ , while hotter stars have much hydrogen ionised, so Balmer lines are strongest near class A rather than in the hottest stars.

Worked example

A star has a surface temperature of $5800\,\text{K}$ and shows ionised and neutral metal absorption lines. Identify its spectral class and intrinsic colour.

1.A temperature between $5000\,\text{K}$ and $6000\,\text{K}$ corresponds to class G.
2.Ionised and neutral metal lines are consistent with class G.
3.Class G stars have a yellow-white intrinsic colour.

Answer: The star is class G and is intrinsically yellow-white.

Common mistakes

Don't write OBAFGKM from coolest to hottest rather than hottest to coolest.
Don't infer that weak Balmer lines mean little hydrogen is present, ignoring excitation and ionisation.
Don't call a class M star blue even though M is the coolest, red class.

Exam tip

For an identification question, use both the temperature range and the prominent absorption lines before stating the spectral class.

Tier 1 · Easy

Three stars have spectral classes B, G and M. State which has the highest surface temperature.
[1 mark]
Total for this question: 1
State the intrinsic colour of a class K star.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A star's continuous spectrum peaks at $410\,\text{nm}$ . Estimate its temperature and hence identify its most likely spectral class. Use Wien's constant $2.9\times10^{-3}\,\text{m K}$ .
[3 marks]
Total for this question: 3
A star has surface temperature $1.8\times10^4\,\text{K}$ and prominent helium and hydrogen absorption lines. Determine and justify its spectral class using both observations.
[2 marks]
Total for this question: 2
A yellow-white star has prominent absorption lines from both ionised and neutral metals. Determine its spectral class and justify the classification using both observations.
[2 marks]
Total for this question: 2

Tier 3 · Hard

Explain why hydrogen Balmer absorption lines can be weak in both a very hot class O star and a cool class M star but strongest in a class A star.
[5 marks]
Total for this question: 5
Star X shows ionised-helium lines, star Y has the strongest hydrogen Balmer lines, and star Z shows neutral-atom and TiO features. Determine the spectral class of each star and place them in decreasing surface temperature.
[4 marks]
Total for this question: 4
An unresolved source has Balmer absorption lines at their greatest strength and a second set of strong neutral-metal absorption lines associated with an orange component. Deduce why the spectrum is unlikely to come from one normal star, and identify the likely spectral class and temperature range of each component.
[4 marks]
Total for this question: 4
A catalogue contains three entries. P is labelled B and has temperature $1.90\times10^4\,\text{K}$ with helium and hydrogen lines. Q is labelled A and has temperature $6.80\times10^3\,\text{K}$ with ionised-metal lines. R is labelled M and has temperature $4.20\times10^3\,\text{K}$ with neutral-metal lines and an orange colour. Identify every incorrect label, give its corrected spectral class, and place P, Q and R in decreasing surface temperature.
[4 marks]
Total for this question: 4
A model star is heated through three spectral stages. It begins with an orange spectrum dominated by neutral-metal lines, later shows hydrogen Balmer absorption at its greatest strength, and finally shows ionised-helium lines while its Balmer lines weaken. Identify the spectral class at each stage and explain why the Balmer-line strength does not increase continuously with temperature.
[5 marks]
Total for this question: 5

3.9.2.5 · The Hertzsprung-Russell (HR) diagram

Explanation

A Hertzsprung-Russell diagram plots absolute magnitude vertically from about $-10$ at the top to $+15$ at the bottom, against surface temperature decreasing from about $50000\,\text{K}$ on the left to $2500\,\text{K}$ on the right; spectral classes OBAFGKM may replace temperature.
The main sequence runs diagonally from hot, luminous stars at upper left to cool, faint stars at lower right.
Giants occupy the upper-right region, while white dwarfs are hot but faint and lie at lower left.
The Sun is a class G main-sequence star at about $5800\,\text{K}$ and $M\approx+5$ .
A Sun-like star progresses from formation to main sequence, red giant and finally white dwarf.

Schematic HR diagram with the main sequence, giants, white dwarfs and a Sun-like evolution path.

Worked example

A star has surface temperature $12000\,\text{K}$ and absolute magnitude $+12$ . Identify its likely region on an HR diagram.

1. $12000\,\text{K}$ places the star on the hot, left-hand side.
2. $M=+12$ makes it intrinsically faint and therefore low on the diagram.
3.The hot but faint lower-left region contains white dwarfs.

Answer: The star is likely to be a white dwarf.

Common mistakes

Don't draw temperature increasing from left to right instead of decreasing.
Don't place white dwarfs in the cool, faint lower-right region rather than the hot, faint lower-left region.
Don't plot apparent magnitude on the vertical axis instead of absolute magnitude.

Exam tip

Before locating a star, mark that hotter is left and more luminous, more negative absolute magnitude is higher.

Tier 1 · Easy

State the region of an HR diagram occupied by the Sun and give its approximate spectral class.
[2 marks]
Total for this question: 2
State how surface temperature changes from left to right on a standard HR diagram.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A star has surface temperature $12000\,\text{K}$ and absolute magnitude $+12$ . Deduce its region on an HR diagram and justify your answer.
[3 marks]
Total for this question: 3
Stars A and B have the same surface temperature, but star A has a lower absolute magnitude. Deduce which star has the larger radius.
[2 marks]
Total for this question: 2
A star keeps the same radius while its surface temperature falls from $1.00\times10^4\,\text{K}$ to $5.00\times10^3\,\text{K}$ . Determine the factor change in its power output and deduce its direction of movement on an HR diagram.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A star is observed at two stages. Its absolute magnitude changes from $+2.40$ to $+1.20$ while its surface temperature falls from $7.20\times10^3\,\text{K}$ to $6.48\times10^3\,\text{K}$ . Determine the luminosity ratio $L_2/L_1$ . Test whether the observations are consistent with constant radius, then describe the star's motion on an HR diagram.
[5 marks]
Total for this question: 5
Explain the contrasting positions of red giants and white dwarfs on an HR diagram in terms of temperature, absolute magnitude and size.
[5 marks]
Total for this question: 5
Star A has surface temperature $8.00\times10^3\,\text{K}$ and absolute magnitude $+1.00$ . Star B has surface temperature $4.00\times10^3\,\text{K}$ and absolute magnitude $-1.00$ . Determine $R_B/R_A$ , then state and justify the region of an HR diagram occupied by each star.
[6 marks]
Total for this question: 6
A star changes while its total power output remains constant. Its surface temperature falls from $9.00\times10^3\,\text{K}$ to $4.50\times10^3\,\text{K}$ . Determine the factor change in radius and describe its movement on an HR diagram, including whether its absolute magnitude changes.
[4 marks]
Total for this question: 4
Two identical unresolved main-sequence stars each have surface temperature $6.00\times10^3\,\text{K}$ and absolute magnitude $+4.00$ . Determine the absolute magnitude plotted for their combined light. Explain how treating the source as one star would displace it on an HR diagram and could lead to an incorrect conclusion about its size.
[4 marks]
Total for this question: 4

3.9.2.6 · Supernovae, neutron stars and black holes

Explanation

A supernova shows a rapid increase in luminosity and a slower decline. Type Ia supernovae have standardisable peak absolute magnitudes, so comparing their known luminosity with observed brightness gives distance; their light curves helped reveal an accelerating Universe, interpreted using dark energy.
A neutron star is an extremely dense compact remnant composed mainly of neutrons.
Collapse of a supergiant to a neutron star or black hole can produce a gamma-ray burst whose brief energy output is comparable with the Sun's total output over a very long time.
For a black hole, escape velocity exceeds $c$ within the event horizon and $R_s\approx2GM/c^2$ .
Supermassive black holes occur at galactic centres.

Typical type Ia supernova light curve with a rapid rise and slower decline.

Worked example

Calculate the Schwarzschild radius of a black hole of mass $8.0\times10^{30}\,\text{kg}$ . Use $G=6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$ and $c=3.00\times10^8\,\text{m s}^{-1}$ .

1.Use $R_s=2GM/c^2$ .
2.Substitute $R_s=2(6.67\times10^{-11})(8.0\times10^{30})/(3.00\times10^8)^2$ .
3.Evaluate to obtain $R_s=1.19\times10^4\,\text{m}$ .

Answer: The Schwarzschild radius is $1.2\times10^4\,\text{m}$ .

Common mistakes

Don't use $GM/c^2$ and omit the factor of two in the Schwarzschild radius.
Don't describe a neutron star as being composed mainly of protons rather than neutrons.
Don't call a type Ia supernova a standard candle without linking its known peak luminosity to distance.

Exam tip

In an explanation of cosmic acceleration, separate the distance evidence from the dark-energy interpretation.

Tier 1 · Easy

A newly identified black hole has mass $5.20\times10^{30}\,\text{kg}$ . Determine its event-horizon radius using $G=6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$ and $c=3.00\times10^8\,\text{m s}^{-1}$ .
[2 marks]
Total for this question: 2
State what is meant by the event horizon of a black hole and give the escape-velocity condition at this boundary.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A type Ia supernova has absolute magnitude $-19.5$ and peak apparent magnitude $18.5$ . Use the distance modulus to estimate its distance in parsecs.
[4 marks]
Total for this question: 4
Two type Ia supernovae have the same standardised peak absolute magnitude. One appears $2.40$ magnitudes fainter at its peak than the other. Calculate how many times farther away the fainter supernova is.
[3 marks]
Total for this question: 3
A neutron star has mass $2.80\times10^{30}\,\text{kg}$ and radius $1.20\times10^4\,\text{m}$ . Treat the star as a sphere of volume $V=4\pi r^3/3$ . Calculate its mean density and state what it is mainly composed of.
[3 marks]
Total for this question: 3

Tier 3 · Hard

The collapse of a supergiant produces a gamma-ray burst of energy $4.0\times10^{44}\,\text{J}$ and leaves a compact object of mass $6.0\times10^{30}\,\text{kg}$ . Calculate the object's Schwarzschild radius and the time for the Sun, at constant power $3.8\times10^{26}\,\text{W}$ , to emit the burst energy. Give the time in years and use $G=6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ and $1\,\text{year}=3.156\times10^7\,\text{s}$ .
[6 marks]
Total for this question: 6
Distant type Ia supernovae appear fainter than predicted by a model in which the Universe expands at a constant rate. Explain the conclusion drawn from this observation and the proposed cause.
[3 marks]
Total for this question: 3
Model a black hole as mass $M$ within a sphere whose radius is the Schwarzschild radius, $R_s=2GM/c^2$ . Show that its mean density is proportional to $M^{-2}$ . Black holes A and B have masses $8.0\times10^6$ and $4.0\times10^9$ solar masses respectively. Calculate $\rho_B/\rho_A$ and explain why the result does not conflict with B being a black hole.
[5 marks]
Total for this question: 5
A compact remnant has mass $3.20\times10^{30}\,\text{kg}$ and radius $1.15\times10^4\,\text{m}$ . Calculate the Schwarzschild radius for this mass and the ratio of the remnant radius to the Schwarzschild radius. Use the result to decide whether the data describe an object with an event horizon. Take the speed of light as $3.00\times10^8\,\text{m s}^{-1}$ and the gravitational constant as $6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$ .
[4 marks]
Total for this question: 4
A type Ia supernova has standardised peak absolute magnitude $-19.15$ and peak apparent magnitude $+22.05$ . Determine its luminosity distance in parsecs. A constant-expansion model predicts a distance of $1.22\times10^9\,\text{pc}$ at the measured redshift. Calculate the percentage by which the supernova distance exceeds the prediction and explain the cosmological conclusion.
[5 marks]
Total for this question: 5

3.9.3.1 · Doppler effect

Explanation

Relative motion along the line of sight changes observed wavelength and frequency. For speeds much smaller than $c$ , $\Delta f/f=-v/c$ and the redshift $z=\Delta\lambda/\lambda=v/c$ , with positive $v$ for recession.
A receding source has a lower observed frequency and longer wavelength; an approaching source has a higher frequency and shorter wavelength.
The denominator is the unshifted laboratory frequency or wavelength.
In a binary system viewed in the plane of its orbit, each star alternately approaches and recedes, so its spectral lines shift periodically between blue and red.
Calculations must confirm that $v\ll c$ before relying on the non-relativistic approximation.

Periodic blue and red shifts from a binary system viewed in the plane of its orbit.

Worked example

A line of laboratory wavelength $500.00\,\text{nm}$ is observed at $500.40\,\text{nm}$ . Calculate $z$ and the radial speed using $c=3.00\times10^8\,\text{m s}^{-1}$ .

1.Find $\Delta\lambda=500.40-500.00=0.40\,\text{nm}$ .
2.Calculate $z=\Delta\lambda/\lambda=0.40/500.00=8.0\times10^{-4}$ .
3.Use $v=zc=(8.0\times10^{-4})(3.00\times10^8)=2.4\times10^5\,\text{m s}^{-1}$ .

Answer: The redshift is $8.0\times10^{-4}$ and the source is receding at $2.4\times10^5\,\text{m s}^{-1}$ ; $v/c=8.0\times10^{-4}\ll1$ .

Common mistakes

Don't use the observed wavelength rather than the unshifted laboratory wavelength in the denominator.
Don't associate a longer observed wavelength with an approaching source instead of a receding source.
Don't apply $v=zc$ to a large redshift without checking the condition $v\ll c$ .

Exam tip

State both direction and speed: the sign of the shift identifies approach or recession, while its magnitude gives $v$ .

Tier 1 · Easy

A spectral line of laboratory wavelength $500.00\,\text{nm}$ is observed at $500.40\,\text{nm}$ . Calculate the redshift $z$ .
[2 marks]
Total for this question: 2
A stellar absorption line is observed at a shorter wavelength than its laboratory value. State the direction of the star's radial motion.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Radiation emitted at $1.42000\,\text{GHz}$ is received from a gas cloud at $1.41900\,\text{GHz}$ . Determine the cloud's radial speed and state whether it is approaching or receding. Use $c=3.00\times10^8\,\text{m s}^{-1}$ .
[3 marks]
Total for this question: 3
Gas cloud A shifts a $486.100\,\text{nm}$ line to $486.460\,\text{nm}$ , while gas cloud B shifts a $656.300\,\text{nm}$ line to $656.840\,\text{nm}$ . Determine which cloud has the greater radial speed and the ratio of the speeds.
[2 marks]
Total for this question: 2
A gas cloud moves directly away from Earth at $1.20\times10^5\,\text{m s}^{-1}$ . Determine the observed wavelength of a line whose laboratory wavelength is $656.300\,\text{nm}$ . Hence determine the fractional frequency change $\Delta f/f$ , where $\Delta f=f_{\text{observed}}-f_{\text{laboratory}}$ . Use $c=3.00\times10^8\,\text{m s}^{-1}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

In an equal-mass binary system viewed in the plane of its circular orbit, a line of rest wavelength $656.300\,\text{nm}$ alternates between $656.000\,\text{nm}$ and $656.600\,\text{nm}$ for one star. The orbital period is $8.00\,\text{days}$ . Estimate the speed of that star, its orbital radius about the centre of mass, and the separation of the stars. Use $c=3.00\times10^8\,\text{m s}^{-1}$ .
[6 marks]
Total for this question: 6
A binary-star system is viewed in the plane of its circular orbit, and its centre of mass has no radial velocity relative to Earth. At maximum separation of the spectral lines, the same $500.000\,\text{nm}$ absorption line is observed at $499.850\,\text{nm}$ from star A and $500.375\,\text{nm}$ from star B. The total mass is $4.20\times10^{30}\,\text{kg}$ . Calculate the orbital speed and mass of each star. Use $c=3.00\times10^8\,\text{m s}^{-1}$ .
[5 marks]
Total for this question: 5
The hydrogen $21\,\text{cm}$ radio line has laboratory wavelength $0.21106\,\text{m}$ and is observed from a distant galaxy at $0.21169\,\text{m}$ . An optical line from the same galaxy has laboratory wavelength $656.300\,\text{nm}$ and is observed at $658.270\,\text{nm}$ . Estimate the recession speed from each line, decide whether the results are consistent, and determine whether the non-relativistic Doppler approximation is valid. Use $c=3.00\times10^8\,\text{m s}^{-1}$ .
[5 marks]
Total for this question: 5
A binary system is viewed in the plane of its circular orbit. One star moves at $1.85\times10^5\,\text{m s}^{-1}$ in a circle of radius $1.28\times10^{10}\,\text{m}$ about the centre of mass, which has no radial motion relative to Earth. Determine the maximum displacement of a spectral line whose laboratory wavelength is $486.100\,\text{nm}$ and the orbital period. Use $c=3.00\times10^8\,\text{m s}^{-1}$ .
[4 marks]
Total for this question: 4
A binary system is viewed in the plane of its orbit and its centre of mass has no radial motion relative to Earth. A spectral line has laboratory frequency $4.57000\times10^{14}\,\text{Hz}$ . The observed frequency crosses this value while decreasing at day $1.40$ and again at day $8.05$ ; its minimum value is $4.56726\times10^{14}\,\text{Hz}$ . Determine the orbital period and the star's maximum orbital speed, and state the direction of motion at minimum observed frequency. Use $c=3.00\times10^8\,\text{m s}^{-1}$ .
[5 marks]
Total for this question: 5

3.9.3.2 · Hubble's law

Explanation

Hubble's law, $v=Hd$ , states that a distant galaxy's recession speed is proportional to its distance.
A straight-line graph of $v$ against $d$ has gradient $H$ , supporting the interpretation that space is expanding rather than that galaxies left one central point through fixed space.
If $H$ is assumed constant, extrapolation gives an age estimate $t\approx1/H$ ; $H$ must first be converted from $\text{km s}^{-1}\text{Mpc}^{-1}$ to $\text{s}^{-1}$ .
Big Bang evidence also includes cosmological microwave background radiation, interpreted as cooled relic radiation, and the relative abundance of hydrogen and helium produced in the early Universe.
The constant- $H$ age is an estimate, not an exact expansion history.

Hubble plot in which the gradient of recession speed against distance is the Hubble constant.

Worked example

Estimate the age of the Universe for $H=68.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ . Use $1\,\text{Mpc}=3.086\times10^{22}\,\text{m}$ and $1\,\text{year}=3.156\times10^7\,\text{s}$ .

1.Convert $H=(68.0\times10^3)/(3.086\times10^{22})=2.20\times10^{-18}\,\text{s}^{-1}$ .
2.Use $t\approx1/H=4.54\times10^{17}\,\text{s}$ .
3.Convert to years: $t=(4.54\times10^{17})/(3.156\times10^7)=1.44\times10^{10}\,\text{years}$ .

Answer: The estimated age is $1.44\times10^{10}\,\text{years}$ , assuming $H$ has remained constant.

Common mistakes

Don't take the reciprocal of $H$ while it is still in $\text{km s}^{-1}\text{Mpc}^{-1}$ .
Don't call the constant- $H$ result an exact age despite the assumption about expansion history.
Don't describe the cosmological microwave background as radiation emitted by present-day stars.

Exam tip

For an age estimate, show the conversion of $H$ to $\text{s}^{-1}$ before using $t\approx1/H$ .

Tier 1 · Easy

Use $H=70\,\text{km s}^{-1}\text{Mpc}^{-1}$ to calculate the recession speed of a galaxy $120\,\text{Mpc}$ away.
[1 mark]
Total for this question: 1
State what the gradient represents on a graph of galaxy recession speed against distance.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Estimate the age of the Universe for $H=68.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ . Assume $H$ is constant and use $1\,\text{Mpc}=3.086\times10^{22}\,\text{m}$ and $1\,\text{year}=3.156\times10^7\,\text{s}$ .
[4 marks]
Total for this question: 4
A straight line of best fit on a graph of galaxy recession speed against distance passes through $(40.0\,\text{Mpc},\,2.80\times10^3\,\text{km s}^{-1})$ and $(220\,\text{Mpc},\,1.54\times10^4\,\text{km s}^{-1})$ . Determine its gradient and convert it to $\text{s}^{-1}$ . Use $1\,\text{Mpc}=3.086\times10^{22}\,\text{m}$ .
[3 marks]
Total for this question: 3
Galaxy A is $80.0\,\text{Mpc}$ away and recedes at $5.60\times10^3\,\text{km s}^{-1}$ . Galaxy B is $140\,\text{Mpc}$ away and recedes at $1.008\times10^4\,\text{km s}^{-1}$ . Calculate the two values of the Hubble constant and determine the percentage by which the constant- $H$ age estimate from galaxy B is smaller than that from galaxy A.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A galaxy's hydrogen line has laboratory wavelength $486.10\,\text{nm}$ and is observed at $493.40\,\text{nm}$ . Use the non-relativistic Doppler approximation and $H=70.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ to estimate the galaxy's distance. Check whether $v\ll c$ , taking $c=3.00\times10^5\,\text{km s}^{-1}$ .
[6 marks]
Total for this question: 6
Explain how observations of cosmological microwave background radiation and the relative abundance of hydrogen and helium support the Big Bang model.
[5 marks]
Total for this question: 5
A spectrometer adds the same unknown velocity offset $u$ to every galaxy measurement, so the recorded speeds obey $v_{\text{recorded}}=Hd+u$ . The recorded speeds are $4.50\times10^3\,\text{km s}^{-1}$ at $60.0\,\text{Mpc}$ and $1.29\times10^4\,\text{km s}^{-1}$ at $180\,\text{Mpc}$ . Determine $H$ , the offset $u$ , and the corresponding constant- $H$ age of the Universe. Use $1\,\text{Mpc}=3.086\times10^{22}\,\text{m}$ and $1\,\text{year}=3.156\times10^7\,\text{s}$ .
[6 marks]
Total for this question: 6
A galaxy is $144\,\text{Mpc}$ from Earth. A spectral line has laboratory wavelength $517.000\,\text{nm}$ and is observed at $532.882\,\text{nm}$ . Use the non-relativistic Doppler approximation to find the Hubble constant, then estimate $1/H$ in years. Take $1\,\text{year}=3.156\times10^7\,\text{s}$ , $1\,\text{Mpc}=3.086\times10^{22}\,\text{m}$ and $c=3.00\times10^5\,\text{km s}^{-1}$ .
[6 marks]
Total for this question: 6
A calibration error makes every measured galaxy distance $5.00\%$ smaller than its true value, while recession speeds are unaffected. Determine the factor by which the gradient of a $v$ – $d$ graph differs from the true Hubble constant. Hence determine the percentage error in the constant- $H$ age estimate, stating whether the age is overestimated or underestimated.
[4 marks]
Total for this question: 4

3.9.3.3 · Quasars

Explanation

Quasars were first identified as unusually bright radio sources. Their spectra show large optical redshifts, placing them among the most distant measurable objects, yet their received flux can still be substantial.
This combination implies an enormous power output. A quasar is an active galactic nucleus powered by matter accreting onto a supermassive black hole, where gravitational energy is converted into radiation.
For a sufficiently small redshift, $v\approx zc$ and Hubble's law gives $d=v/H$ .
If emission is assumed isotropic, the luminosity follows $P=4\pi d^2F$ .
These estimates require explicit low-redshift, isotropic-emission and inverse-square assumptions; observed flux is not the same quantity as emitted power.

Worked example

A quasar is $900\,\text{Mpc}$ away and produces flux $2.5\times10^{-15}\,\text{W m}^{-2}$ at Earth. Estimate its power, assuming isotropic emission and $1\,\text{Mpc}=3.086\times10^{22}\,\text{m}$ .

1.Convert the distance: $d=900(3.086\times10^{22})=2.777\times10^{25}\,\text{m}$ .
2.Use the inverse-square relation $P=4\pi d^2F$ .
3.Substitute to obtain $P=4\pi(2.777\times10^{25})^2(2.5\times10^{-15})=2.42\times10^{37}\,\text{W}$ .

Answer: The estimated quasar power is $2.4\times10^{37}\,\text{W}$ .

Common mistakes

Don't equate the measured flux in $\text{W m}^{-2}$ directly with the quasar's power in watts.
Don't describe a quasar as an exceptionally bright ordinary star rather than an active galactic nucleus.
Don't use $v=zc$ for a large optical redshift without qualifying the non-relativistic approximation.

Exam tip

An estimation answer should name the assumptions alongside the result, especially isotropic emission and the validity of the low-redshift approximation.

Tier 1 · Easy

State the central engine believed to power a quasar.
[1 mark]
Total for this question: 1
State the observational property that first led astronomers to identify quasars.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A quasar at distance $900\,\text{Mpc}$ produces a flux of $2.5\times10^{-15}\,\text{W m}^{-2}$ at Earth. Estimate its power output assuming isotropic emission. Use $1\,\text{Mpc}=3.086\times10^{22}\,\text{m}$ .
[3 marks]
Total for this question: 3
Explain why a quasar with large redshift and substantial received flux must have a very large power output.
[3 marks]
Total for this question: 3
A quasar has power output $2.00\times10^{39}\,\text{W}$ and measured flux $4.00\times10^{-14}\,\text{W m}^{-2}$ . Determine its distance in megaparsecs and state the assumption about its emission needed for the calculation. Use $1\,\text{Mpc}=3.086\times10^{22}\,\text{m}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

A low-redshift quasar has $z=0.0200$ and measured flux $1.60\times10^{-13}\,\text{W m}^{-2}$ . Estimate its distance, power output and power in units of the Sun's luminosity. Use $c=3.00\times10^5\,\text{km s}^{-1}$ , $H=70.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ , $1\,\text{Mpc}=3.086\times10^{22}\,\text{m}$ and $L_{\odot}=3.83\times10^{26}\,\text{W}$ . Assume isotropic emission.
[6 marks]
Total for this question: 6
Two low-redshift quasars have the same power output but redshifts $0.0300$ and $0.0600$ . Use Hubble's law and the inverse-square law to determine how many times greater the received flux from the lower-redshift quasar is.
[3 marks]
Total for this question: 3
A low-redshift quasar has redshift $z=0.0400$ . During a $60.0\,\text{s}$ exposure, radiation from the quasar transfers $2.16\times10^{-13}\,\text{J}$ to a detector through an effective collecting area of $0.600\,\text{m}^2$ . Estimate the quasar's distance and power output. Use $v\approx zc$ , $H=70.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ , $c=3.00\times10^5\,\text{km s}^{-1}$ and $1\,\text{Mpc}=3.086\times10^{22}\,\text{m}$ . Assume isotropic emission.
[5 marks]
Total for this question: 5
A quasar has power output $3.60\times10^{39}\,\text{W}$ and is $1.45\times10^3\,\text{Mpc}$ from Earth. Determine the flux received at Earth, assuming the quasar emits isotropically. A detector of collecting area $0.850\,\text{m}^2$ and quantum efficiency $0.250$ records photons of average energy $3.20\times10^{-19}\,\text{J}$ . Calculate the count rate recorded. Use $1\,\text{Mpc}=3.086\times10^{22}\,\text{m}$ .
[5 marks]
Total for this question: 5
A compact source appears star-like in an optical image, is a bright radio source and has optical redshift $z=0.80$ . Its received flux is substantial. A student concludes that it is an unusually bright ordinary star nearby and uses $v=zc$ to calculate its recession speed. Discuss the conclusion and the calculation.
[5 marks]
Total for this question: 5

3.9.3.4 · Detection of exoplanets

Explanation

Direct exoplanet detection is difficult because a planet is much fainter than its host star and the angular separation is very small. The radial-velocity method detects periodic red and blue Doppler shifts in the star's absorption lines as the star orbits the system's centre of mass; it measures only line-of-sight motion.
The transit method detects a repeated fall in brightness when an aligned planet crosses the stellar disc.
A typical light curve has a steady baseline, ingress, a low region and egress, repeated once per orbital period.
Approximately, transit depth satisfies $\Delta I/I=(R_p/R_*)^2$ .
Matching periods from both methods provide stronger evidence, while a missing transit does not rule out a planet because alignment may be unsuitable.

Typical exoplanet transit light curve with ingress, minimum intensity and egress.

Worked example

A transit reduces a star's intensity by $0.810\%$ . Estimate the planet radius if the star radius is $6.90\times10^8\,\text{m}$ .

1.Convert the percentage to a fraction: $0.810\%=0.00810$ .
2.Use $R_p=R_*\sqrt{\Delta I/I}$ .
3.Substitute $R_p=(6.90\times10^8)\sqrt{0.00810}=6.21\times10^7\,\text{m}$ .

Answer: The planet radius is approximately $6.21\times10^7\,\text{m}$ .

Common mistakes

Don't use the percentage value $0.810$ as a fraction instead of converting it to $0.00810$ .
Don't assume that every exoplanet produces a transit even when the orbital plane is not suitably aligned.
Don't attribute the periodic Doppler shift to motion of the planet's spectrum rather than the host star's spectrum.

Exam tip

When interpreting a light curve, identify repeated equal-period dips and distinguish transit depth from transit duration.

Tier 1 · Easy

A star's brightness shows equal, regularly repeated dips. Name the exoplanet-detection method indicated by this observation.
[1 mark]
Total for this question: 1
Give two reasons why direct optical observation of an exoplanet close to its host star is difficult.
[2 marks]
Total for this question: 2

Tier 2 · Standard

During a transit, a star's detected intensity falls by $0.810\%$ . Estimate the radius of the planet if the star's radius is $6.90\times10^8\,\text{m}$ .
[3 marks]
Total for this question: 3
Transit minima occur at times $3.2\,\text{days}$ , $11.7\,\text{days}$ and $20.2\,\text{days}$ . Determine the exoplanet's orbital period and angular speed in $\text{rad s}^{-1}$ .
[2 marks]
Total for this question: 2
A star's spectral line has a maximum redshift on day $0$ , a maximum blueshift on day $6.0$ and its next maximum redshift on day $12.0$ . Transit minima occur on days $3.0$ and $15.0$ . Deduce whether the two signals are consistent with one exoplanet.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A star has radius $8.0\times10^8\,\text{m}$ . Every $5.2\,\text{days}$ its intensity falls by $0.160\%$ , while a $600.0\,\text{nm}$ absorption line shifts with the same period by up to $0.014\,\text{nm}$ to either side of its mean. Calculate the planet's radius and the star's maximum radial speed, then explain why the combined observations are stronger evidence for an exoplanet than either observation alone. Use $c=3.00\times10^8\,\text{m s}^{-1}$ .
[6 marks]
Total for this question: 6
Discuss whether a periodic spectral-line shift is consistent with an exoplanet when no transit is observed. A line with laboratory wavelength $620.00000\,\text{nm}$ shifts by up to $1.0\times10^{-4}\,\text{nm}$ either side of this value, with successive maximum wavelengths $18.0\,\text{days}$ apart. Support the discussion with the maximum radial speed to two significant figures and the orbital period; use $c=3.00\times10^8\,\text{m s}^{-1}$ .
[5 marks]
Total for this question: 5
A small exoplanet makes a central transit across a star of radius $7.50\times10^8\,\text{m}$ . The planet's centre takes $3.60\,\text{h}$ to cross the stellar diameter and the orbital period is $9.00\,\text{days}$ . Assuming a circular edge-on orbit, determine the orbital radius.
[4 marks]
Total for this question: 4
A star of radius $7.25\times10^8\,\text{m}$ has a steady detected intensity of $8.00\times10^{-10}\,\text{W m}^{-2}$ . During each flat-bottomed dip the intensity is $7.936\times10^{-10}\,\text{W m}^{-2}$ ; successive dip centres occur at day $4.30$ and day $17.15$ . Determine the fractional transit depth, the planet radius and the orbital period. Explain what the ingress, flat minimum and egress represent.
[5 marks]
Total for this question: 5
A planet and its host star have predicted angular separation $0.740$ arcseconds. An imaging system can separate sources at angles of at least $0.520$ arcseconds. The planet supplies $2.40\times10^{-9}$ of the star's intensity, while the system reduces the detected starlight at the planet's position by a factor of $8.00\times10^7$ without reducing the planet signal. Decide whether the images are angularly resolved, calculate the residual starlight as a multiple of the planet signal, and determine the minimum starlight-reduction factor for the two signals to be equal. Hence explain why direct detection remains difficult.
[5 marks]
Total for this question: 5

Answer key

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

3.9.1.1 · Astronomical telescope consisting of two converging lenses

Tier 1 · Easy

Mark scheme for 3.9.1.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$40.0$	Convert the eyepiece focal length: $f_e=30.0\,\text{mm}=0.0300\,\text{m}$ . Hence $M=f_o/f_e=1.20/0.0300=40.0$ .	1
02.1	In the focal plane shared by the objective and eyepiece.	Normal adjustment places the objective's real intermediate image at the eyepiece focal plane. The two focal planes therefore coincide.	1

Tier 2 · Standard

Mark scheme for 3.9.1.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.2\times10^{-2}\,\text{rad}$ and $1.525\,\text{m}$	The angular magnification is $M=1.500/0.0250=60.0$ . The angle at the eye is therefore $60.0\times2.0\times10^{-4}=1.2\times10^{-2}\,\text{rad}$ . In normal adjustment the separation is $f_o+f_e=1.500+0.0250=1.525\,\text{m}$ .	3
02.1	The objective forms a real image at its focal plane. This plane is also the focal plane of the eyepiece, so the eyepiece sends parallel rays to the eye and the final virtual image is at infinity. The coincident focal planes make the lens separation $f_o+f_e$ .	Follow the image through the system: a distant object sends parallel rays to the objective, which forms its real image one objective focal length behind that lens. Placing the eyepiece one eyepiece focal length beyond this image makes the emerging rays parallel, so the relaxed eye focuses them without accommodation. Adding the two focal distances gives $L=f_o+f_e$ .	3
03.1	$0.630\,\text{mm}$ ; the intermediate image is inverted.	For a small angle, the separation in the objective focal plane is $h=f_o\alpha=(1.80)(3.50\times10^{-4})=6.30\times10^{-4}\,\text{m}=0.630\,\text{mm}$ . A converging objective forms a real, inverted intermediate image.	2

Tier 3 · Hard

Mark scheme for 3.9.1.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$30.0\,\text{mm}$ and $0.990\,\text{m}$	The required magnification is at least $M=(4.80\times10^{-3})/(1.50\times10^{-4})=32.0$ . Since $M=f_o/f_e$ , the greatest acceptable focal length is $f_e=0.960/32.0=0.0300\,\text{m}=30.0\,\text{mm}$ . At this limiting eyepiece focal length, the normal-adjustment length is $f_o+f_e=0.960+0.0300=0.990\,\text{m}$ ; smaller acceptable eyepiece focal lengths give shorter telescopes. A longer-focal-length eyepiece would give less than the required angular magnification.	5
02.1	Objective: $1.26\,\text{m}$ ; eyepiece: $35.0\,\text{mm}$	Use $f_o/f_e=36.0$ and $f_o+f_e=1.295\,\text{m}$ . Substitution gives $36.0f_e+f_e=1.295$ , hence $f_e=1.295/37.0=0.0350\,\text{m}=35.0\,\text{mm}$ . Therefore $f_o=36.0(0.0350)=1.26\,\text{m}$ to three significant figures.	3
03.1	$3.20\times10^{-3}\,\text{rad}$	The objective focal length is $f_o=h/\alpha=(0.192\times10^{-3})/(1.60\times10^{-4})=1.20\,\text{m}$ . Normal adjustment gives $f_e=1.260-1.20=0.0600\,\text{m}$ . The angular magnification is $M=f_o/f_e=1.20/0.0600=20.0$ , so the angle at the eye is $M\alpha=20.0(1.60\times10^{-4})=3.20\times10^{-3}\,\text{rad}$ .	4
04.1	The objective forms a real, inverted intermediate image at its focal plane. This plane is also the focal plane of the eyepiece, so the lenses are separated by $f_o+f_e$ . The eyepiece is beyond the intermediate image and sends parallel rays to the eye, giving a final virtual image at infinity and relaxed viewing. The magnitude of the angular magnification is $f_o/f_e$ , not $f_o+f_e$ .	Follow the corrected ray path. Parallel rays from the distant object are brought to a real, inverted image by the objective. Normal adjustment places that image in the focal plane of the eyepiece, making the objective-to-eyepiece distance $f_o+f_e$ . The eyepiece then produces parallel emergent rays, so the eye does not need to accommodate and the final image is virtual at infinity. Angular magnification is a dimensionless ratio of angles, $M=f_o/f_e$ in magnitude; $f_o+f_e$ has units of length and cannot be an angular magnification.	5
05.1	Initially, the angular magnification is $37.5$ and the length is $2.002\,\text{m}$ . After interchange, the angular magnification is $0.0267$ and the length remains $2.002\,\text{m}$ . The interchanged arrangement reduces the angle subtended at the eye, so it is not useful as an astronomical magnifier.	Initially, $M=f_o/f_e=1.950/0.0520=37.5$ and $L=f_o+f_e=1.950+0.0520=2.002\,\text{m}$ . After interchange, $M'=0.0520/1.950=0.026666\ldots=0.0267$ , while addition is unchanged, so $L'=0.0520+1.950=2.002\,\text{m}$ . Because $M'<1$ , the final image subtends a smaller angle than the unaided object; merely retaining normal adjustment does not make the reversed arrangement a useful telescope.	4

3.9.1.2 · Reflecting telescopes

Tier 1 · Easy

Mark scheme for 3.9.1.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Chromatic aberration.	A mirror reflects different visible wavelengths without refracting them by different amounts, so it does not form wavelength-dependent focal lengths. The avoided aberration is chromatic aberration.	1
02.1	A concave mirror; more specifically, a parabolic concave mirror.	Credit concave. The more precise description is a parabolic concave primary, which converges incident light without the spherical aberration of a spherical surface.	1

Tier 2 · Standard

Mark scheme for 3.9.1.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The parabolic concave primary reflects the rays towards its focus; before they meet, the convex secondary reflects them back through the central hole in the primary and towards the eyepiece.	Award the chain in order: the concave parabolic primary starts to converge the parallel rays; the convex secondary intercepts them before the primary focus; the secondary reflects them back through the hole in the primary so that they continue to the eyepiece.	3
02.1	The larger secondary blocks a greater fraction of the radiation incident on the primary. Less radiation reaches the focus and detector, so the image signal is weaker and faint objects become harder to detect.	The secondary lies in the incoming beam. Increasing its diameter increases the obstructed area, reducing the effective collecting area of the primary. The detector therefore receives fewer photons from the source, producing a fainter or lower-signal image.	3
03.1	The mirror can be supported over its whole back surface, whereas a lens must be supported around its edge. The distributed support reduces distortion under the element's weight.	A mirror has only one optical face that must remain unobstructed, so supports can act across its rear surface. Light must pass through a lens, leaving edge support as the practical option; a large lens is therefore more liable to sag and distort.	2

Tier 3 · Hard

Mark scheme for 3.9.1.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The reflector has no chromatic aberration, and a parabolic primary avoids the on-axis spherical aberration of a spherical surface. Its mirror can be supported from behind, so a large aperture can be made thinner and with less distortion than a lens supported only at its edge. A refractor can also absorb light in its thick objective, although the reflector's secondary obstruction can reduce the collected intensity and affect the image.	Build the comparison from linked points: reflection does not disperse colours, so there is no chromatic aberration; a parabolic primary focuses on-axis parallel rays together, unlike a spherical surface; a mirror is supported across its back, whereas a lens is supported around its edge; this makes large reflector apertures mechanically practical. A balanced discussion may add the reflector disadvantage that the secondary obstructs part of the incoming beam.	5
02.1	A spherical surface reflects marginal and paraxial rays towards different axial points, producing spherical aberration and a blurred image. A parabolic primary directs on-axis parallel rays to one focus, so the rays form a sharper image. This change does not address chromatic aberration because a mirror already reflects the different visible wavelengths without refraction.	Identify the defect first: rays meeting different zones of a spherical mirror do not share one focus. Their spread at the image plane is spherical aberration. For a parabolic surface, every on-axis incident ray parallel to the axis is reflected through the same focus, reducing the blur. Chromatic aberration is not the reason for the improvement because both versions are mirrors.	3
03.1	$1.07$ ; the reflector gives about $6.7\%$ more signal.	Signal is proportional to effective collecting area. The reflector has area proportional to $4.00^2-0.800^2=15.36$ , while the refractor's transmitted area is proportional to $0.900(4.00^2)=14.40$ . The ratio is $15.36/14.40=1.0667=1.07$ . Therefore, despite its central obstruction, the reflector has the greater detected signal under the stated assumptions and gives about $6.7\%$ more signal.	3
04.1	The incoming rays strike the concave parabolic primary first and are reflected towards its focus. Before meeting, they strike the convex secondary, which reflects them back through the central hole in the primary and towards the eyepiece. A spherical primary brings marginal and paraxial rays to different points, whereas a parabolic primary brings on-axis parallel rays to one focus and reduces spherical aberration.	Order the optical elements from the incoming beam: primary mirror, secondary mirror, central hole, then eyepiece. The hole is an exit route for the folded returning beam, not the entry route for the incident rays. The concave primary begins the convergence; the convex secondary intercepts the beam before the primary focus and returns it through the hole. The parabolic profile is required because it directs on-axis parallel rays from all zones of the mirror to the same focus, avoiding the axial blur produced by a spherical surface.	5
05.1	$D=3.67\,\text{m}$ ; $4.00\%$ of the beam area is obstructed. The primary must be parabolic so that on-axis parallel rays share one focus.	Equating clear areas cancels the common factor $\pi/4$ : $D^2-(0.200D)^2=(3.60)^2$ . Hence $0.960D^2=12.96$ and $D=\sqrt{12.96/0.960}=3.674\ldots\,\text{m}=3.67\,\text{m}$ . The obstructed fraction is $(0.200D)^2/D^2=0.0400$ , or $4.00\%$ . Equal clear area fixes collecting power only; a parabolic primary is also needed to bring on-axis parallel rays to a common focus rather than producing spherical aberration.	4

3.9.1.3 · Single dish radio telescopes, I-R, U-V and X-ray telescopes

Tier 1 · Easy

Mark scheme for 3.9.1.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Earth's atmosphere absorbs astronomical X-rays.	X-rays from space are strongly absorbed by the atmosphere, so a detector must be placed above most or all of the atmosphere.	1
02.1	It collects radio waves and reflects them towards a receiver at the focus.	The curved reflecting dish intercepts the incoming radio wavefront and concentrates it at the receiver near its focus.	1

Tier 2 · Standard

Mark scheme for 3.9.1.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$7.64\times10^{4}\,\text{m}$	Equal resolution requires $\lambda_{\rm optical}/D_{\rm optical}=\lambda_{\rm radio}/D_{\rm radio}$ . Therefore $D_{\rm radio}=0.210\times0.200/(550\times10^{-9})=7.64\times10^4\,\text{m}$ . This very large value follows from the much longer radio wavelength.	3
02.1	Both use a concave primary reflector to focus electromagnetic radiation. The radio telescope directs waves to an aerial or receiver at its focus, whereas the Cassegrain uses a convex secondary to direct light through the primary towards an eyepiece or detector.	Compare like with like. The concave primary is the shared focusing element. The receiving path differs: a radio aerial detects the focused radio signal directly, while the Cassegrain's convex secondary folds the optical path back through the primary.	2
03.1	The radio telescope has $100$ times the collecting power. Its minimum angular resolution is $6.00\times10^{-3}\,\text{rad}$ , compared with $2.40\times10^{-7}\,\text{rad}$ for the optical telescope, so the optical telescope has $2.50\times10^4$ times better angular resolution.	Collecting power is proportional to $D^2$ , so $C_{\text{radio}}/C_{\text{optical}}=(25.0/2.50)^2=100$ . For the radio telescope, $\theta=0.150/25.0=6.00\times10^{-3}\,\text{rad}$ . For the optical telescope, $\theta=(600\times10^{-9})/2.50=2.40\times10^{-7}\,\text{rad}$ . The smaller angle means better resolution, and $(6.00\times10^{-3})/(2.40\times10^{-7})=2.50\times10^4$ , so the optical telescope has much better resolving power despite its smaller collecting area.	4

Tier 3 · Hard

Mark scheme for 3.9.1.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	A radio dish can work on the ground, preferably away from human-made radio interference. An infrared telescope should be high and dry to reduce absorption by atmospheric water vapour. Ultraviolet and X-ray instruments should be above the atmosphere, normally on satellites, because those radiations are strongly absorbed. Radio wavelengths are much longer than visible wavelengths, so a much larger $D$ is needed for comparable resolution through $\theta\approx\lambda/D$ ; the large diameter also gives greater collecting power because this is proportional to $D^2$ .	Link each location to atmospheric transmission: radio reaches the ground, so choose a radio-quiet ground site; infrared is absorbed by water vapour, so use a high, dry site; ultraviolet is largely absorbed, so place its telescope above the atmosphere; X-rays are also absorbed, so use a satellite. Then compare apertures: $\theta\approx\lambda/D$ means the much longer radio wavelength requires a much larger $D$ for similar angular resolution, and collecting power proportional to $D^2$ also helps detect weak radio signals.	6
02.1	Atmospheric water vapour absorbs infrared radiation, so altitude and dry air reduce this loss. Warm telescope components and detectors emit infrared radiation themselves, adding background noise. Cooling the equipment reduces this thermal emission and improves detection of weak astronomical infrared signals.	Link each design choice to a distinct limitation. A high, dry site reduces the column of water vapour that absorbs incoming infrared radiation. Cooling reduces infrared emission from the telescope and detector, lowering the thermal background against which the source is measured.	4
03.1	The radio minimum angular resolution is $5.00\times10^{-4}\,\text{rad}$ , so it cannot resolve the sources. The optical value is $2.50\times10^{-7}\,\text{rad}$ , so it can. The radio dish has $2.50\times10^3$ times the collecting power. It should be at a radio-quiet site, whereas the optical reflector should be at a high, dark site.	For the radio dish, $\theta=0.0500/100=5.00\times10^{-4}\,\text{rad}$ , which is greater than the source separation, so the sources are not resolved. For the optical reflector, $\theta=(500\times10^{-9})/2.00=2.50\times10^{-7}\,\text{rad}$ , which is smaller than the separation, so the sources are resolved. Collecting power is proportional to $D^2$ , giving $C_{\text{radio}}/C_{\text{optical}}=(100/2.00)^2=2.50\times10^3$ . The dish needs a radio-quiet site to reduce human-made interference; the optical reflector benefits from a high, dark site with less atmospheric and light-pollution interference.	6
04.1	The single radio telescope and optical telescope have minimum angular resolutions $2.37\times10^{-3}\,\text{rad}$ and $2.17\times10^{-7}\,\text{rad}$ respectively, so the optical telescope has better angular resolution. The linked radio array has minimum angular resolution $4.50\times10^{-5}\,\text{rad}$ , so it outperforms the single radio telescope because its minimum angular resolution is smaller. It does not outperform the optical telescope.	For one radio dish, $\theta_{\rm r}=0.180/76.0=2.368\times10^{-3}\,\text{rad}=2.37\times10^{-3}\,\text{rad}$ . For the optical reflector, $\theta_{\rm o}=(520\times10^{-9})/2.40=2.167\times10^{-7}\,\text{rad}=2.17\times10^{-7}\,\text{rad}$ . The smaller optical value means better angular resolution. For the interferometer, $\theta_{\rm array}\approx\lambda/b=0.180/(4.00\times10^3)=4.50\times10^{-5}\,\text{rad}$ . This is smaller than $2.37\times10^{-3}\,\text{rad}$ but larger than $2.17\times10^{-7}\,\text{rad}$ , so the linked array outperforms the single radio telescope but not the optical telescope.	5
05.1	A radio dish and receiver are designed to collect radio waves, whereas an ultraviolet telescope needs optical surfaces and a detector suitable for ultraviolet radiation; changing the receiver alone does not provide a complete ultraviolet instrument. Astronomical ultraviolet is strongly absorbed by Earth's atmosphere, so a ground site would receive little of it and the telescope normally needs to be above the atmosphere on a satellite. A shorter wavelength would give a smaller diffraction angle for the same usable aperture through $\theta\approx\lambda/D$ , but that theoretical advantage cannot be obtained if the radiation is absorbed before reaching an unsuitable ground-based system.	Assess each part of the proposal separately. The radio dish focuses radio waves onto an aerial or radio receiver, so detector replacement alone does not establish a suitable ultraviolet optical system. The atmosphere is the decisive siting problem because it absorbs most astronomical ultraviolet; placing the detector at a ground focus cannot recover radiation that never reaches the dish. Finally, $\theta\approx\lambda/D$ does predict better diffraction-limited resolution at shorter wavelength for a given effective diameter, but only for a telescope whose collecting surfaces, detector and location can actually operate in that band.	5

3.9.1.4 · Advantages of large diameter telescopes

Tier 1 · Easy

Mark scheme for 3.9.1.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$4.0$	Collecting power is proportional to $D^2$ , so the factor is $(0.60/0.30)^2=2^2=4.0$ .	1
02.1	The fraction of incident photons that the detector records.	Quantum efficiency compares the number of detected photon events with the number of photons incident on the detector.	1

Tier 2 · Standard

Mark scheme for 3.9.1.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.00\times10^{-7}\,\text{rad}$ ; a larger diameter gives a smaller minimum angle.	Convert $500\,\text{nm}=5.00\times10^{-7}\,\text{m}$ . Then $\theta\approx\lambda/D=(5.00\times10^{-7})/2.50=2.00\times10^{-7}\,\text{rad}$ . Since $\theta$ is inversely proportional to $D$ , increasing $D$ improves resolution by reducing the minimum resolvable angle.	3
02.1	$\theta_{\text{UV}}/\theta_{\text{visible}}=0.600$	Use $\theta\approx\lambda/D$ for each observation. The ratio is $[240/(0.800)]/[600/(1.20)]=300/500=0.600$ , so the ultraviolet observation has the smaller diffraction-limited angle despite its smaller aperture.	2
03.1	$5.88\times10^4$ photons	The aperture area is $A=\pi D^2/4=\pi(0.800)^2/4=0.5027\,\text{m}^2$ . The number incident during the exposure is $(1.50\times10^4)(0.5027)(12.0)=9.05\times10^4$ . The recorded number is $0.650(9.05\times10^4)=5.88\times10^4$ photons.	3

Tier 3 · Hard

Mark scheme for 3.9.1.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Only the $10.0\,\text{m}$ telescope resolves the pair; it has $6.25$ times the collecting power; a CCD detects a greater proportion of the incident photons and can integrate and store the exposure.	For the $4.00\,\text{m}$ telescope, $\theta=600\times10^{-9}/4.00=1.50\times10^{-7}\,\text{rad}$ , which is greater than the separation, so it cannot resolve the pair. For the $10.0\,\text{m}$ telescope, $\theta=600\times10^{-9}/10.0=6.0\times10^{-8}\,\text{rad}$ , which is smaller than the separation, so it can resolve the pair. The collecting-power ratio is $(10.0/4.00)^2=6.25$ . A CCD converts a larger fraction of incident photons into a signal and can accumulate a long exposure and store it, unlike a momentary visual observation.	6
02.1	$15\,\text{s}$ ; the larger telescope has half the minimum resolvable angle.	The detected-photon rate is proportional to $D^2$ times quantum efficiency. Its factor increase is $(3.00/1.50)^2(0.72/0.18)=4\times4=16$ . Equal photon count therefore needs $240/16=15\,\text{s}$ to two significant figures. At fixed wavelength, $\theta\propto1/D$ , so doubling the diameter halves the minimum resolvable angle.	5
03.1	$5.15\,\text{m}$	Resolution requires $D\geq\lambda/\theta=(550\times10^{-9})/(1.10\times10^{-7})=5.00\,\text{m}$ . The photon requirement gives $A\geq N/(\Phi t\eta)=(5.00\times10^5)/[(2.00\times10^2)(300)(0.400)]=20.83\,\text{m}^2$ . Hence $D\geq\sqrt{4A/\pi}=\sqrt{4(20.83)/\pi}=5.15\,\text{m}$ . The photon-count requirement is the stricter one, so the minimum acceptable diameter is $5.15\,\text{m}$ .	5
04.1	The original pixel angle is $4.50\times10^{-7}\,\text{rad}$ and the diffraction angle is $2.13\times10^{-7}\,\text{rad}$ , so the detector pixels limit the resolution. With $3.00\,\mu\text{m}$ pixels the pixel angle is $1.67\times10^{-7}\,\text{rad}$ , so diffraction limits the resolution to about $2.13\times10^{-7}\,\text{rad}$ .	For the original detector, one pixel subtends $p/f=(8.10\times10^{-6})/18.0=4.50\times10^{-7}\,\text{rad}$ . The aperture gives $\theta=\lambda/D=(510\times10^{-9})/2.40=2.125\times10^{-7}\,\text{rad}=2.13\times10^{-7}\,\text{rad}$ . The larger of these two angular scales is the practical limitation, so the original image is pixel-limited. The replacement pixel subtends $(3.00\times10^{-6})/18.0=1.67\times10^{-7}\,\text{rad}$ , now smaller than the diffraction angle. Smaller pixels cannot overcome the aperture limit, so diffraction becomes limiting.	5
05.1	The quantum efficiency is $0.694$ or $69.4\%$ . The masked exposure is $44.4\,\text{s}$ and the minimum angular resolution becomes $1.57$ times as large.	The original collecting area is $A=\pi(1.10)^2/4=0.950332\,\text{m}^2$ . From $N=\Phi At\eta$ , $\eta=(2.85\times10^5)/[(2.40\times10^4)(0.950332)(18.0)]=0.6942=0.694$ . At fixed flux and efficiency, equal photon count requires $t\propto1/D^2$ , so $t'=18.0(1.10/0.700)^2=44.449\ldots\,\text{s}=44.4\,\text{s}$ . Since $\theta\propto1/D$ , the limiting angle increases by $1.10/0.700=1.571\ldots=1.57$ .	5

3.9.2.1 · Classification by luminosity

Tier 1 · Easy

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01.1	The star with apparent magnitude $+1.0$ .	On the magnitude scale, the smaller apparent magnitude denotes the brighter object, so the $+1.0$ star appears brighter.	1
02.1	$+6$	On the Hipparcos apparent-magnitude scale, the unaided-eye limit under suitable dark conditions is approximately $m=+6$ .	1

Tier 2 · Standard

Mark scheme for 3.9.2.1 Tier 2 · Standard
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01.1	$15.8$	The magnitude difference is $5.00-2.00=3.00$ . P has the lower magnitude and is therefore brighter, so $I_P/I_Q=2.51^3=15.8$ .	3
02.1	$4.45$	Use $60.0=2.51^{\Delta m}$ . Taking logarithms gives $\Delta m=\log(60.0)/\log(2.51)=4.449\ldots=4.45$ to three significant figures.	2
03.1	$3.76$	Relative to the brighter star, the fainter star has intensity $2.51^{-1.50}=0.251$ times as large. The combined intensity is therefore $1.251$ times the brighter star's intensity. Its magnitude is $4.00-\log(1.251)/\log(2.51)=3.76$ .	3

Tier 3 · Hard

Mark scheme for 3.9.2.1 Tier 3 · Hard
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01.1	$5.26$ ; yes, because this is brighter than the magnitude- $6$ limit.	Let the magnitude increase be $\Delta m$ . Since the intensity has fallen by a factor of $20.0$ , $2.51^{\Delta m}=20.0$ . Thus $\Delta m=\log(20.0)/\log(2.51)=3.26$ . The new apparent magnitude is $2.00+3.26=5.26$ . This is below the approximate limiting magnitude $6$ , so the star should still be visible under suitable dark conditions.	5
02.1	$4$ stars	The magnitude difference is $6.50-5.00=1.50$ , with the cluster brighter. Its intensity is therefore $2.51^{1.50}=3.98$ times that of one star. Equal stars contribute equal intensities, so the cluster contains approximately $3.98$ , or $4$ stars.	4
03.1	$x=0.510$ (accept $x=0.509$ ), or about $51\%$	The maximum allowed magnitude increase is $6.00-4.40=1.60$ . The corresponding minimum transmitted intensity fraction is $2.51^{-1.60}=0.229$ . Dust and cloud together transmit $0.450x$ , so the limiting condition is $0.450x=0.229$ . Hence $x=0.229/0.450=0.510$ . A smaller cloud transmission would make the star fainter than magnitude $6.00$ .	4
04.1	Star B has apparent magnitude $4.20$ ; it supplies $39.7\%$ of the out-of-eclipse intensity.	The out-of-eclipse to A-only intensity ratio is $I_{A+B}/I_A=2.51^{3.75-3.20}=1.65890$ . Hence $I_B/I_A=1.65890-1=0.658901$ . Using $I_B/I_A=2.51^{m_A-m_B}$ gives $m_B=3.75-\log(0.658901)/\log(2.51)=4.2033\ldots=4.20$ . The fraction supplied by B is $I_B/I_{A+B}=0.658901/1.65890=0.39719$ , or $39.7\%$ .	5
05.1	The apparent magnitudes are $1.03$ at maximum (accept $1.03$ to $1.04$ ) and $4.53$ later. It is not visible before the outburst but is visible at maximum and at the later time.	The brightening changes the magnitude by $\log(320)/\log(2.51)=6.26799\ldots$ , so $m_{\max}=7.30-6.26799=1.032=1.03$ . A fall in intensity by a factor of $25$ increases the magnitude by $\log(25)/\log(2.51)=3.49770\ldots$ , giving $m_{\text{later}}=1.032+3.49770=4.5297=4.53$ . The unaided-eye limit is approximately $+6$ : $7.30$ is too faint, whereas $1.03$ and $4.53$ are visible under the stated conditions.	4

3.9.2.2 · Absolute magnitude, M

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01.1	$M=4.3$	Absolute magnitude is defined as the apparent magnitude at $10\,\text{pc}$ , so at this distance $M=m=4.3$ .	1
02.1	The star with absolute magnitude $-2.0$ .	A lower, more negative absolute magnitude corresponds to greater intrinsic luminosity.	1

Tier 2 · Standard

Mark scheme for 3.9.2.2 Tier 2 · Standard
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01.1	$100\,\text{pc}$	$m-M=8.4-3.4=5.0$ . Therefore $5.0=5\log(d/10)$ , so $\log(d/10)=1.0$ . Hence $d/10=10$ and $d=100\,\text{pc}$ .	3
02.1	$7.36\,\text{pc}$ Absolute magnitude is the apparent magnitude the star would have at $10\,\text{pc}$ .	The distance is $24.0/3.26=7.3619\ldots\,\text{pc}=7.36\,\text{pc}$ . This is not $10\,\text{pc}$ , the reference distance used to define absolute magnitude, so the apparent and absolute magnitudes differ.	2
03.1	$+3.31$ (accept $+3.30$ )	The distance is $d=65.2/3.26=20.0\,\text{pc}$ . The distance modulus is $m-M=5\log_{10}(20.0/10)=5\log_{10}2=1.505$ . Therefore $m=1.80+1.505=+3.31$ .	3

Tier 3 · Hard

Mark scheme for 3.9.2.2 Tier 3 · Hard
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01.1	$3.98\times10^3\,\text{pc}$ and $1.30\times10^4\,\text{ly}$	$m-M=12.0-(-1.0)=13.0$ . Hence $13.0=5\log(d/10)$ and $\log(d/10)=2.60$ . Therefore $d=10\times10^{2.60}=3.98\times10^3\,\text{pc}$ . In light years this is $3.98\times10^3\times3.26=1.30\times10^4\,\text{ly}$ .	5
02.1	$M_A=+4.99$ and $M_B=-0.01$ . $L_B/L_A=1.00\times10^2$ .	For A, $m-M=5\log_{10}(d/10)$ gives $8.00-M_A=5\log_{10}(40.0/10)=3.0103$ , so $M_A=+4.99$ . For B, $8.00-M_B=5\log_{10}(400/10)=8.0103$ , so $M_B=-0.0103\approx-0.01$ . Equal apparent magnitudes mean equal apparent intensities. From $I=L/(4\pi d^2)$ , equal $I$ gives $L\propto d^2$ , so $L_B/L_A=(400/40.0)^2=1.00\times10^2$ .	4
03.1	$1.26\times10^3\,\text{pc}$ ; the distance would be overestimated by a factor of $2.00$ .	Correcting for the dust gives $m=10.00-1.50=8.50$ . Hence $8.50-(-2.00)=5\log_{10}(d/10)$ , so $d=10\times10^{10.50/5}=1.26\times10^3\,\text{pc}$ . Ignoring the dust would give $d_{\text{wrong}}=10\times10^{[10.00-(-2.00)]/5}=2.51\times10^3\,\text{pc}$ . The overestimate factor is $2.51\times10^3/(1.26\times10^3)=2.00$ .	5
04.1	Without absorption, $m=6.73$ . The absorption adds $1.27$ magnitudes and transmits $0.310$ , or $31.0\%$ , of the intensity (accept $0.309$ to $0.310$ , i.e. $30.9\%$ to $31.0\%$ ).	The dust-free distance modulus is $m-M=5\log_{10}(320/10)=7.52575$ , so $m=-0.80+7.52575=6.72575=6.73$ . The observed excess is $8.00-6.72575=1.27425$ magnitudes. If $T$ is the transmitted intensity fraction, $1/T=2.51^{1.27425}$ , so $T=2.51^{-1.27425}=0.30954=0.310$ , equivalent to $31.0\%$ .	5
05.1	Star B is $10.1\,\text{pc}$ away and $L_A/L_B=63.1$ (accept $62.9$ to $63.1$ ).	For A, $m=-3.20+5\log_{10}(80.0/10)=1.31545$ . For B, $1.31545-1.30=5\log_{10}(d_B/10)$ , so $d_B=10\times10^{0.01545/5}=10.0714\,\text{pc}=10.1\,\text{pc}$ . The absolute-magnitude difference is $1.30-(-3.20)=4.50$ , with A more luminous. Using the exact distance-modulus relation, $L_A/L_B=10^{0.4(4.50)}=63.0957=63.1$ ; equivalently, equal apparent intensities give $(80.0/10.0714)^2=63.1$ . Using the rounded per-magnitude factor $2.51$ gives $2.51^{4.50}=62.9$ , which is also accepted.	5

3.9.2.3 · Classification by temperature, black-body radiation

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01.1	$5.0\times10^3\,\text{K}$	$\lambda_{\max}=580\,\text{nm}=5.80\times10^{-7}\,\text{m}$ . Thus $T=(2.9\times10^{-3})/(5.80\times10^{-7})=5.0\times10^3\,\text{K}$ .	2
02.1	The peak moves to a shorter wavelength. The area under the curve increases.	Wien's law moves the peak towards shorter wavelength as temperature rises. Stefan's law increases the total emitted power, represented by a larger area beneath the spectrum.	2

Tier 2 · Standard

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01.1	$1.64$	For spherical stars, $P\propto R^2T^4$ . Therefore $P_B/P_A=(2.00)^2(0.800)^4=4.00\times0.4096=1.6384\approx1.64$ .	3
02.1	$T_A/T_B=1.50$ ; star A is hotter.	Wien's law gives $T\propto1/\lambda_{\max}$ . Hence $T_A/T_B=\lambda_B/\lambda_A=630/420=1.50$ . Star A has the shorter peak wavelength and is therefore hotter.	2
03.1	$11.4$	Wien's law gives $T_A/T_B=\lambda_B/\lambda_A=750/500=1.50$ . Stefan's law for spherical stars gives $P\propto R^2T^4$ . Therefore $P_A/P_B=(1.50)^2(1.50)^4=(1.50)^6=11.39=11.4$ .	4

Tier 3 · Hard

Mark scheme for 3.9.2.3 Tier 3 · Hard
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01.1	$6.0\times10^3\,\text{K}$ and $3.2\times10^9\,\text{m}$	Wien's law gives the guard value $T=(2.9\times10^{-3})/(480\times10^{-9})=6.04\times10^3\,\text{K}$ . The distance is $d=50.0\times3.086\times10^{16}=1.543\times10^{18}\,\text{m}$ . From the inverse-square law, $P=4\pi d^2I=4\pi(1.543\times10^{18})^2(3.20\times10^{-10})=9.57\times10^{27}\,\text{W}$ . Stefan's law gives $P=4\pi R^2\sigma T^4$ , so the guard value is $R=\sqrt{P/(4\pi\sigma T^4)}=3.18\times10^9\,\text{m}$ . To the two significant figures supported by Wien's constant, the final values are $6.0\times10^3\,\text{K}$ and $3.2\times10^9\,\text{m}$ .	6
02.1	$R_B/R_A=1.33$	For a black-body star, received intensity is proportional to $R^2T^4/d^2$ . Equal intensities give $R_A^2T_A^4/d_A^2=R_B^2T_B^4/d_B^2$ . Hence $R_B/R_A=(d_B/d_A)(T_A/T_B)^2=(60.0/20.0)(6000/9000)^2=3(2/3)^2=4/3=1.33$ .	5
03.1	$1.40$	For a star, $I=\sigma T^4(R/d)^2$ . Angular diameter is proportional to $R/d$ , so no conversion of milliarcseconds is needed in the ratio. Wien's law gives $T_A/T_B=\lambda_B/\lambda_A=600/450=4/3$ . Hence $I_A/I_B=(0.800/1.20)^2(4/3)^4=(2/3)^2(4/3)^4=1.40$ .	5
04.1	$T_2/T_1=1.30$ and $R_2/R_1=0.878$ ; the star has contracted.	Wien's law gives $T_2/T_1=\lambda_1/\lambda_2=650/500=1.30$ . Stefan's law gives $P_2/P_1=(R_2/R_1)^2(T_2/T_1)^4$ . Hence $R_2/R_1=\sqrt{2.20/(1.30)^4}=\sqrt{2.20/2.8561}=0.87766=0.878$ . Because the final radius is smaller, the star has contracted even though its power has increased.	4
05.1	$T=5.69\times10^3\,\text{K}$ , $P=1.31\times10^{28}\,\text{W}$ and $d_{\max}=3.91\times10^2\,\text{pc}$ (accept $3.90\times10^2$ – $3.91\times10^2\,\text{pc}$ ).	Wien's law gives $T=(2.9\times10^{-3})/(510\times10^{-9})=5686.27\,\text{K}$ . The emitting area is $4\pi R^2$ , so $P=4\pi(4.20\times10^9)^2(5.67\times10^{-8})(5686.27)^4=1.3140\times10^{28}\,\text{W}$ . At the limiting intensity, $I=P/(4\pi d^2)$ , so $d=\sqrt{P/(4\pi I)}=1.2051\times10^{19}\,\text{m}$ . Converting gives $d=(1.2051\times10^{19})/(3.086\times10^{16})=390.5\,\text{pc}=3.91\times10^2\,\text{pc}$ .	6

3.9.2.4 · Principles of the use of stellar spectral classes

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01.1	The class B star.	The sequence OBAFGKM runs from highest to lowest temperature. Of B, G and M, B occurs first and is therefore hottest.	1
02.1	Orange.	In the OBAFGKM sequence, class K stars are cooler than class G stars and have an orange intrinsic colour.	1

Tier 2 · Standard

Mark scheme for 3.9.2.4 Tier 2 · Standard
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01.1	$7.1\times10^3\,\text{K}$ ; class F	$T=(2.9\times10^{-3})/(410\times10^{-9})=7.07\times10^3\,\text{K}$ . The class F range is approximately $6000$ - $7500\,\text{K}$ , so the most likely class is F.	3
02.1	Class B: $1.8\times10^4\,\text{K}$ lies in the B-class range. B-star spectra show prominent helium and hydrogen lines.	The temperature lies between $11000\,\text{K}$ and $25000\,\text{K}$ , the B-class range. The stated helium and hydrogen absorption lines independently agree with class B.	2
03.1	Class G. G stars are yellow-white and their spectra contain lines from both ionised and neutral metals.	The colour places the star near class G in the OBAFGKM sequence. The simultaneous presence of ionised-metal and neutral-metal absorption lines is also characteristic of a G-star spectrum, so the two observations agree.	2

Tier 3 · Hard

Mark scheme for 3.9.2.4 Tier 3 · Hard
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01.1	Balmer absorption requires bound hydrogen atoms already in the $n=2$ state. In a cool M star too few collisions excite hydrogen to $n=2$ . In a very hot O star much of the hydrogen is ionised, so fewer bound atoms can absorb Balmer photons. An A-star temperature gives a large population of bound hydrogen atoms in $n=2$ , producing the strongest Balmer lines; the difference is an excitation and ionisation effect, not evidence of different hydrogen abundance.	Start with the transition condition: Balmer absorption begins from $n=2$ . For an M star, thermal excitation is insufficient, so the $n=2$ population is small. For an O star, the high temperature ionises much of the hydrogen, leaving fewer bound atoms. At class A temperatures there is a favourable balance: hydrogen remains bound and many atoms occupy $n=2$ . Therefore line strength depends on temperature-dependent excitation and ionisation rather than simply on hydrogen abundance.	5
02.1	X is class O, Y is class A and Z is class M; the decreasing-temperature order is X, Y, Z.	Ionised helium identifies the very hot class O spectrum. The strongest Balmer absorption identifies class A. Neutral atoms together with TiO identify the cool class M spectrum. Since OBAFGKM is ordered from hottest to coolest, the required order is O, A, M, corresponding to X, Y, Z.	4
03.1	The strongest Balmer lines indicate a class A component at about $7500$ – $11000\,\text{K}$ . The orange colour and neutral-metal lines indicate a class K component at about $3500$ – $5000\,\text{K}$ . One normal stellar atmosphere cannot have both temperature ranges, so the source is likely an unresolved binary or multiple system containing A-type and K-type stars.	Use the two diagnostics separately. Hydrogen Balmer absorption is strongest in class A, whose temperature range is $7500$ – $11000\,\text{K}$ . An orange spectrum with neutral-metal absorption identifies class K, at $3500$ – $5000\,\text{K}$ . These temperature regimes cannot describe one normal stellar surface, but the combined spectrum of an unresolved system can contain both.	4
04.1	P is correctly labelled B. Q should be class F, not A: $6.80\times10^3\,\text{K}$ and ionised-metal lines fit F. R should be class K, not M: $4.20\times10^3\,\text{K}$ , neutral-metal lines and orange colour fit K. The decreasing-temperature order is P, Q, R.	Check each row against both its temperature and spectral features. P lies in the B range and its helium and hydrogen lines agree with that label. Q lies in the approximate $6000$ – $7500\,\text{K}$ F range, where ionised-metal lines are prominent, so A is incorrect. R lies in the approximate $3500$ – $5000\,\text{K}$ K range; neutral-metal lines and orange colour support K rather than the cooler red M class. The numerical temperatures directly give P hotter than Q hotter than R.	4
05.1	The stages are class K, then class A, then class O. In the cool K atmosphere relatively few hydrogen atoms are excited to $n=2$ . At class A temperatures many hydrogen atoms remain bound and occupy $n=2$ , so Balmer absorption is strongest. At the much higher class O temperature, much of the hydrogen is ionised, leaving fewer bound atoms able to absorb Balmer photons; ionised helium is then characteristic. Balmer strength therefore depends on excitation and ionisation as well as temperature.	Use both the line species and the temperature order. Neutral metals with orange colour identify class K. Maximum Balmer absorption identifies class A. Ionised helium identifies the much hotter class O. The Balmer series begins from $n=2$ : insufficient excitation limits the line strength in the cool stage, the bound $n=2$ population is favourable near class A, and ionisation removes bound hydrogen in the hottest stage. The result is a maximum rather than a monotonic rise.	5

3.9.2.5 · The Hertzsprung-Russell (HR) diagram

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01.1	The main sequence; spectral class G.	The Sun is an ordinary hydrogen-burning main-sequence star and its temperature of about $5800\,\text{K}$ places it in spectral class G.	2
02.1	Surface temperature decreases from left to right.	The horizontal axis is reversed relative to many graphs: hot O stars are on the left and cool M stars are on the right.	1

Tier 2 · Standard

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01.1	The white-dwarf region: the star is hot but intrinsically faint.	$12000\,\text{K}$ places the star on the hot, left side of the diagram. $M=+12$ is intrinsically faint and therefore low on the vertical axis. Hot but faint stars occupy the lower-left white-dwarf region.	3
02.1	Star A has the larger radius.	The lower absolute magnitude means star A is more luminous. Since both stars have the same temperature and $P\propto R^2T^4$ , the more luminous star must have the larger radius.	2
03.1	Its power falls to $1/16$ of its original value; it moves to the right and downwards.	At constant radius, Stefan's law gives $P\propto T^4$ , so $P_{\text{new}}/P_{\text{old}}=(0.500)^4=1/16$ . The lower temperature moves the star rightwards. Its lower luminosity means a larger absolute magnitude, so it moves downwards.	3

Tier 3 · Hard

Mark scheme for 3.9.2.5 Tier 3 · Hard
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01.1	$L_2/L_1=3.02$ . At constant radius the temperature change would instead give $L_2/L_1=0.656$ , so the radius did not stay constant and must have increased. The star moves rightwards and upwards on the HR diagram.	The magnitude decreases by $1.20$ , so $L_2/L_1=2.51^{2.40-1.20}=2.51^{1.20}=3.017230\ldots=3.02$ . If the radius were constant, Stefan's law would give $L_2/L_1=(T_2/T_1)^4=(6480/7200)^4=0.656100\ldots=0.656$ . The observed luminosity increases rather than falling to this value, so constant radius is ruled out; the radius must increase. The lower temperature moves the star to the right, while its lower absolute magnitude and greater luminosity move it upwards.	5
02.1	Red giants are cool but very luminous, so they have relatively low or negative absolute magnitudes and lie in the upper-right region. Their large radii give high power despite low surface temperature. White dwarfs are hot but faint, so they have large positive absolute magnitudes and lie in the lower-left region. Their very small radii limit their power despite high temperature.	Use both axes for each class. A red giant's low temperature puts it to the right, while its high luminosity and more negative absolute magnitude put it high; its large surface area explains that combination. A white dwarf's high temperature puts it to the left, while its low luminosity and positive absolute magnitude put it low; its small surface area explains why it remains faint.	5
03.1	$R_B/R_A=10.0$ . Star A is on the main sequence because it is hot and moderately luminous. Star B is in the giant region because it is cool but has high luminosity, shown by its lower absolute magnitude.	The two-magnitude difference means $P_B/P_A=2.51^{1.00-(-1.00)}=2.51^2=6.30$ , because B has the lower absolute magnitude. Stefan's law gives $P_B/P_A=(R_B/R_A)^2(T_B/T_A)^4$ . Therefore $R_B/R_A=\sqrt{6.30}(T_A/T_B)^2=\sqrt{6.30}(8000/4000)^2=10.0$ . On an HR diagram, A's high temperature and moderate luminosity place it on the main sequence. B lies in the giant region: it is cool, but its lower absolute magnitude makes it more luminous and its calculated radius is much larger.	6
04.1	The radius increases by a factor of $4.00$ . The star moves to the right at constant vertical position because it becomes cooler while its luminosity, and therefore its absolute magnitude, remains unchanged.	At constant power, $R_1^2T_1^4=R_2^2T_2^4$ , so $R_2/R_1=(T_1/T_2)^2=(9000/4500)^2=4.00$ . Temperature decreases from left to right on an HR diagram, so the star moves rightwards. The vertical axis is absolute magnitude, which represents luminosity; constant power means constant absolute magnitude and therefore no vertical movement.	4
05.1	The combined absolute magnitude is $+3.25$ . It is plotted at the same temperature but $0.753$ magnitudes brighter, i.e. its absolute magnitude is $0.753$ lower ( $+3.25$ against $+4.00$ ), so it lies above the single-star main sequence. If interpreted as one star at the same temperature, its greater luminosity would be attributed to a larger radius even though the excess light actually comes from two stars.	Two identical stars provide twice the intensity at the reference distance. The magnitude decrease is $\log(2)/\log(2.51)=0.75319$ , giving $M_{\text{combined}}=4.00-0.75319=3.24681=+3.25$ . The shared spectral temperature leaves the horizontal position unchanged, while the lower absolute magnitude moves the point upwards. Since $P\propto R^2T^4$ , a one-star interpretation at fixed $T$ would infer a radius larger by $\sqrt{2}$ ; the HR displacement is instead caused by unresolved multiplicity.	4

3.9.2.6 · Supernovae, neutron stars and black holes

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01.1	$7.71\times10^3\,\text{m}$	$R_s=2GM/c^2=2(6.67\times10^{-11})(5.20\times10^{30})/(3.00\times10^8)^2=7.71\times10^3\,\text{m}$ .	2
02.1	It is the boundary of the region from which nothing, including light, can escape. At the event horizon, the escape velocity is $c$ .	The event horizon is the limiting surface around a black hole. At the boundary the escape velocity equals the speed of light; within it the escape velocity exceeds $c$ , so light cannot escape.	2

Tier 2 · Standard

Mark scheme for 3.9.2.6 Tier 2 · Standard
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01.1	$3.98\times10^8\,\text{pc}$	$m-M=18.5-(-19.5)=38.0$ . Hence $38.0=5\log(d/10)$ , so $\log(d/10)=7.60$ . Therefore $d=10\times10^{7.60}=3.98\times10^8\,\text{pc}$ . The standard-candle assumption supplies the absolute magnitude.	4
02.1	The fainter supernova is $3.02$ times farther away.	For equal absolute magnitudes, subtracting the two distance-modulus equations gives $2.40=5\log_{10}(d_2/d_1)$ . Hence $d_2/d_1=10^{2.40/5}=3.0199\ldots=3.02$ .	3
03.1	$3.87\times10^{17}\,\text{kg m}^{-3}$ . It is composed mainly of neutrons.	$V=4\pi(1.20\times10^4)^3/3=7.24\times10^{12}\,\text{m}^3$ , so $\rho=M/V=(2.80\times10^{30})/(7.24\times10^{12})=3.87\times10^{17}\,\text{kg m}^{-3}$ . A neutron star consists mainly of neutrons.	3

Tier 3 · Hard

Mark scheme for 3.9.2.6 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$8.9\times10^3\,\text{m}$ and $3.3\times10^{10}\,\text{years}$	$R_s=2GM/c^2=2(6.67\times10^{-11})(6.0\times10^{30})/(3.00\times10^8)^2=8.89\times10^3\,\text{m}$ . The equivalent solar emission time is $t=E/P=(4.0\times10^{44})/(3.8\times10^{26})=1.05\times10^{18}\,\text{s}$ . Converting gives $t=(1.05\times10^{18})/(3.156\times10^7)=3.34\times10^{10}\,\text{years}$ , or $3.3\times10^{10}\,\text{years}$ to two significant figures.	6
02.1	The standardisable peak luminosity makes each type Ia supernova a distance indicator. Being fainter than the constant-rate prediction means the events are farther away than that model implies for their measured redshifts. The expansion of the Universe must therefore have accelerated while their light travelled to Earth. Dark energy is proposed as the cause of this accelerated expansion.	Link the evidence in order: the known peak absolute magnitude supplies a luminosity; the measured flux supplies distance; redshift supplies the recession evidence. A supernova farther away than the constant-expansion relation predicts shows that the expansion rate increased. The accepted interpretation attributes the acceleration to dark energy.	3
03.1	$\rho=3c^6/(32\pi G^3M^2)$ , so $\rho\propto M^{-2}$ . $\rho_B/\rho_A=4.0\times10^{-6}$ . A black hole is defined by escape velocity reaching $c$ at its event horizon, not by a minimum mean density.	Using $\rho=M/(4\pi R_s^3/3)$ and $R_s=2GM/c^2$ gives $\rho=3M/[4\pi(2GM/c^2)^3]=3c^6/(32\pi G^3M^2)$ . Hence $\rho_B/\rho_A=(M_A/M_B)^2=[(8.0\times10^6)/(4.0\times10^9)]^2=(1/500)^2=4.0\times10^{-6}$ . Although B has the lower mean density, its escape velocity is $c$ at the event horizon and exceeds $c$ within it, which is the defining condition.	5
04.1	$R_s=4.74\times10^3\,\text{m}$ and $R/R_s=2.42$ . The surface lies outside the Schwarzschild radius, so this model does not have an event horizon at its surface and is consistent with a neutron star rather than a black hole.	$R_s=2GM/c^2=2(6.67\times10^{-11})(3.20\times10^{30})/(3.00\times10^8)^2=4.7431\times10^3\,\text{m}$ . Hence $R/R_s=(1.15\times10^4)/(4.7431\times10^3)=2.4246=2.42$ . A black-hole event horizon for this mass would occur at $R_s$ ; because the stated material surface is farther out, its escape speed is below $c$ and the object is not a black hole under this model.	4
05.1	$d=1.74\times10^9\,\text{pc}$ , which is $42.4\%$ greater than the constant-expansion prediction. The unexpectedly large distance supports an accelerating expansion of the Universe, attributed to dark energy.	The distance modulus is $m-M=22.05-(-19.15)=41.20$ . Hence $41.20=5\log_{10}(d/10)$ , so $d=10\times10^{41.20/5}=1.7378\times10^9\,\text{pc}=1.74\times10^9\,\text{pc}$ . Relative to the model, the excess is $[(1.7378\times10^9)/(1.22\times10^9)-1]\times100=42.4\%$ . A standardised type Ia peak supplies the distance; being farther away, and therefore fainter, than the constant-expansion model predicts is evidence that cosmic expansion has accelerated. Dark energy is the proposed cause.	5

3.9.3.1 · Doppler effect

Tier 1 · Easy

Mark scheme for 3.9.3.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$8.0\times10^{-4}$	$\Delta\lambda=500.40-500.00=0.40\,\text{nm}$ . Therefore $z=\Delta\lambda/\lambda=0.40/500.00=8.0\times10^{-4}$ .	2
02.1	The star is approaching the observer.	A shift to shorter wavelength is a blueshift, which indicates radial motion towards the observer.	1

Tier 2 · Standard

Mark scheme for 3.9.3.1 Tier 2 · Standard
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01.1	$2.11\times10^5\,\text{m s}^{-1}$ , receding	$\Delta f=1.41900-1.42000=-0.00100\,\text{GHz}$ . Using $\Delta f/f=-v/c$ , $v=-c\Delta f/f=-(3.00\times10^8)(-0.00100/1.42000)=2.11\times10^5\,\text{m s}^{-1}$ . The received frequency is lower, so the cloud is receding. Also $v/c=7.04\times10^{-4}\ll1$ , so the approximation is valid.	3
02.1	Cloud B; $v_B/v_A=1.11$ .	For non-relativistic motion, speed is proportional to fractional wavelength shift. Thus $v_B/v_A=(0.540/656.300)/(0.360/486.100)=1.111\ldots=1.11$ , so cloud B has the greater radial speed.	2
03.1	$656.563\,\text{nm}$ ; $\Delta f/f=-4.00\times10^{-4}$	Use $\Delta\lambda/\lambda=v/c$ . Hence $\Delta\lambda=656.300(1.20\times10^5/3.00\times10^8)=0.2625\,\text{nm}$ . The cloud is receding, so $\lambda_{\text{observed}}=656.300+0.2625=656.563\,\text{nm}$ . For a small Doppler shift, $\Delta f/f=-\Delta\lambda/\lambda=-v/c=-4.00\times10^{-4}$ ; the negative sign shows that the observed frequency is lower.	3

Tier 3 · Hard

Mark scheme for 3.9.3.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.37\times10^5\,\text{m s}^{-1}$ , $1.51\times10^{10}\,\text{m}$ , and $3.02\times10^{10}\,\text{m}$	The maximum shift magnitude is $\lvert\Delta\lambda\rvert=0.300\,\text{nm}$ . Hence $v=c\lvert\Delta\lambda\rvert/\lambda=(3.00\times10^8)(0.300/656.300)=1.37\times10^5\,\text{m s}^{-1}$ ; $v/c=4.57\times10^{-4}$ validates the approximation. The period is $T=8.00\times86400=6.912\times10^5\,\text{s}$ . For circular motion, $v=2\pi r/T$ , so $r=vT/(2\pi)=1.51\times10^{10}\,\text{m}$ . Equal masses orbit at equal radii on opposite sides, so their separation is $2r=3.02\times10^{10}\,\text{m}$ .	6
02.1	$v_A=9.00\times10^4\,\text{m s}^{-1}$ and $m_A=3.00\times10^{30}\,\text{kg}$ ; $v_B=2.25\times10^5\,\text{m s}^{-1}$ and $m_B=1.20\times10^{30}\,\text{kg}$ .	Because the orbit is viewed in its plane, the maximum radial speeds are the orbital speeds. For A, $v_A=c(0.150/500.000)=9.00\times10^4\,\text{m s}^{-1}$ ; for B, $v_B=c(0.375/500.000)=2.25\times10^5\,\text{m s}^{-1}$ . The stars orbit their common centre of mass, so $m_Av_A=m_Bv_B$ and $m_B=0.400m_A$ . Combining this with $m_A+m_B=4.20\times10^{30}\,\text{kg}$ gives $m_A=3.00\times10^{30}\,\text{kg}$ and $m_B=1.20\times10^{30}\,\text{kg}$ .	5
03.1	The radio and optical estimates are $8.95\times10^5\,\text{m s}^{-1}$ and $9.01\times10^5\,\text{m s}^{-1}$ , so they are consistent with a recession speed of about $9.0\times10^5\,\text{m s}^{-1}$ . Since $v/c\approx3.0\times10^{-3}\ll1$ , the approximation is valid.	Both observed wavelengths are longer, indicating recession. For the radio line, $v=c\Delta\lambda/\lambda=(3.00\times10^8)(0.21169-0.21106)/0.21106=8.95\times10^5\,\text{m s}^{-1}$ . For the optical line, $v=(3.00\times10^8)(658.270-656.300)/656.300=9.01\times10^5\,\text{m s}^{-1}$ . The estimates differ by less than $1\%$ , so they are consistent. Their ratio to $c$ is approximately $3.0\times10^{-3}\ll1$ , validating the non-relativistic approximation.	5
04.1	The maximum wavelength displacement is $0.300\,\text{nm}$ and the orbital period is $5.03\,\text{days}$ .	Because the system is viewed in the plane of its orbit, the maximum radial speed equals the orbital speed. Hence $\Delta\lambda_{\max}=\lambda v/c=(486.100)(1.85\times10^5)/(3.00\times10^8)=0.29976\,\text{nm}=0.300\,\text{nm}$ . The star travels circumference $2\pi r$ in one orbit, so $T=2\pi r/v=2\pi(1.28\times10^{10})/(1.85\times10^5)=4.347\times10^5\,\text{s}=5.03\,\text{days}$ . Also $v/c=6.17\times10^{-4}\ll1$ , validating the approximation.	4
05.1	The orbital period is $6.65\,\text{days}$ and the maximum orbital speed is $1.80\times10^5\,\text{m s}^{-1}$ . At minimum observed frequency the star is receding.	Crossings of the laboratory frequency in the same direction occur at the same orbital phase, so $T=8.05-1.40=6.65\,\text{days}$ . The maximum frequency change has magnitude $\lvert\Delta f\rvert=(4.57000-4.56726)\times10^{14}=2.74\times10^{11}\,\text{Hz}$ . Using $\Delta f/f=-v/c$ , the speed is $v=c\lvert\Delta f\rvert/f=(3.00\times10^8)(2.74\times10^{11})/(4.57000\times10^{14})=1.7987\times10^5\,\text{m s}^{-1}=1.80\times10^5\,\text{m s}^{-1}$ . The lower frequency is a redshift, so the star is receding. Here $v/c=6.00\times10^{-4}\ll1$ .	5

3.9.3.2 · Hubble's law

Tier 1 · Easy

Mark scheme for 3.9.3.2 Tier 1 · Easy
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01.1	$8.4\times10^3\,\text{km s}^{-1}$	$v=Hd=70\times120=8400\,\text{km s}^{-1}=8.4\times10^3\,\text{km s}^{-1}$ .	1
02.1	The Hubble constant, $H$ .	Hubble's law has the straight-line form $v=Hd$ , so the gradient of $v$ against $d$ is $H$ .	1

Tier 2 · Standard

Mark scheme for 3.9.3.2 Tier 2 · Standard
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01.1	$1.44\times10^{10}\,\text{years}$	Convert $H$ : $H=(68.0\times10^3)/(3.086\times10^{22})=2.20\times10^{-18}\,\text{s}^{-1}$ . Then $t\approx1/H=4.54\times10^{17}\,\text{s}$ . In years, $t=(4.54\times10^{17})/(3.156\times10^7)=1.44\times10^{10}\,\text{years}$ . This assumes a constant expansion rate.	4
02.1	$70.0\,\text{km s}^{-1}\text{Mpc}^{-1}=2.27\times10^{-18}\,\text{s}^{-1}$	The gradient is $H=(1.54\times10^4-2.80\times10^3)/(220-40.0)=12600/180=70.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ . Converting the numerator to metres per second gives $H=(70.0\times10^3)/(3.086\times10^{22})=2.2683\times10^{-18}\,\text{s}^{-1}=2.27\times10^{-18}\,\text{s}^{-1}$ .	3
03.1	$H_A=70.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ , $H_B=72.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ ; the age estimate from B is $2.78\%$ smaller.	The two estimates are $H_A=5600/80.0=70.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ and $H_B=10080/140=72.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ . Since the constant- $H$ estimate has $t\approx1/H$ , $t_B/t_A=H_A/H_B=70.0/72.0$ . Therefore the age from B is smaller by $[1-(70.0/72.0)]\times100=2.78\%$ .	3

Tier 3 · Hard

Mark scheme for 3.9.3.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$64.4\,\text{Mpc}$ ; $v/c=0.0150$ , so $v\ll c$ is a reasonable approximation.	$\Delta\lambda=493.40-486.10=7.30\,\text{nm}$ , so $z=7.30/486.10=0.0150$ . The recession speed is $v\approx zc=0.0150(3.00\times10^5)=4.51\times10^3\,\text{km s}^{-1}$ . Hubble's law gives $d=v/H=(4.51\times10^3)/70.0=64.4\,\text{Mpc}$ . Since $v/c=0.0150\ll1$ , use of the non-relativistic approximation is reasonable.	6
02.1	The Big Bang model predicts relic radiation from the hot early Universe. Expansion stretches and cools this radiation into an almost uniform microwave background arriving from all directions, which is observed. In the early hot Universe, fusion produced helium from hydrogen; expansion then cooled the Universe until fusion stopped. The predicted and observed mixture is approximately three parts hydrogen to one part helium by mass, with few heavier nuclei, supporting the model.	Treat the observations as two independent predictions. The early Universe should leave thermal radiation; expansion shifts it to microwave wavelengths and observation finds it across the sky. Early fusion should convert some hydrogen to helium before cooling stops the reactions. The resulting prediction of roughly $3{:}1$ hydrogen to helium by mass, with little production of heavier elements, agrees with observed cosmic abundances.	5
03.1	$H=70.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ , $u=+300\,\text{km s}^{-1}$ , and $t=1.40\times10^{10}\,\text{years}$ .	Subtracting the two equations eliminates the offset: $H=(12900-4500)/(180-60.0)=8400/120=70.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ . Substitution into the first measurement gives $u=4500-(70.0)(60.0)=+300\,\text{km s}^{-1}$ . Convert $H$ to SI units: $H=(70.0\times10^3)/(3.086\times10^{22})=2.27\times10^{-18}\,\text{s}^{-1}$ . Therefore $t\approx1/H=4.41\times10^{17}\,\text{s}=1.40\times10^{10}\,\text{years}$ .	6
04.1	$H=64.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ and $t=1.53\times10^{10}\,\text{years}$ .	$z=(532.882-517.000)/517.000=0.0307195$ , so $v\approx zc=9.2159\times10^3\,\text{km s}^{-1}$ . The ratio $v/c=0.0307$ is small enough for the stated estimate. Hubble's law gives $H=v/d=(9.2159\times10^3)/144=63.999\,\text{km s}^{-1}\text{Mpc}^{-1}=64.0\,\text{km s}^{-1}\text{Mpc}^{-1}$ . In SI units, $H=(63.999\times10^3)/(3.086\times10^{22})=2.0739\times10^{-18}\,\text{s}^{-1}$ . Hence $t\approx1/H=4.8229\times10^{17}\,\text{s}=1.5279\times10^{10}\,\text{years}=1.53\times10^{10}\,\text{years}$ .	6
05.1	The measured gradient is $1/0.950=1.0526$ times the true Hubble constant, so $H$ is $5.26\%$ too large. The age estimate is $0.950$ times the true constant- $H$ estimate and is therefore $5.00\%$ too small.	Write $d_{\rm measured}=0.950d_{\rm true}$ . With $v$ unchanged, $H_{\rm measured}=v/d_{\rm measured}=H_{\rm true}/0.950=1.05263H_{\rm true}$ , an overestimate of $5.263\%$ . Since the age estimate is the reciprocal, $t_{\rm measured}=1/H_{\rm measured}=0.950/H_{\rm true}=0.950t_{\rm true}$ . It is therefore underestimated by exactly $5.00\%$ under the constant- $H$ model.	4

3.9.3.3 · Quasars

Tier 1 · Easy

Mark scheme for 3.9.3.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Accretion of matter onto an active supermassive black hole.	A quasar is powered by gravitational energy released as matter accretes onto a supermassive black hole in an active galactic nucleus.	1
02.1	They were bright radio sources.	Quasars were first recognised as bright radio sources and initially appeared star-like in optical images.	1

Tier 2 · Standard

Mark scheme for 3.9.3.3 Tier 2 · Standard
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01.1	$2.4\times10^{37}\,\text{W}$	$d=900\times3.086\times10^{22}=2.777\times10^{25}\,\text{m}$ . For isotropic emission, $P=4\pi d^2F=4\pi(2.777\times10^{25})^2(2.5\times10^{-15})=2.42\times10^{37}\,\text{W}$ .	3
02.1	The large redshift indicates that the quasar is very distant. Flux falls with the square of distance, so receiving a substantial flux from such a distance requires an extremely large luminosity at the source.	Use the redshift as evidence of great distance. Then apply the inverse-square relation $F=P/(4\pi d^2)$ . If $d$ is very large while $F$ remains appreciable, the emitted power $P$ must be enormous.	3
03.1	$2.04\times10^3\,\text{Mpc}$ , assuming that the quasar emits isotropically.	Assuming isotropic emission, $F=P/(4\pi d^2)$ . Hence $d=\sqrt{P/(4\pi F)}=\sqrt{(2.00\times10^{39})/[4\pi(4.00\times10^{-14})]}=6.31\times10^{25}\,\text{m}$ . Converting gives $d=(6.31\times10^{25})/(3.086\times10^{22})=2.04\times10^3\,\text{Mpc}$ .	3

Tier 3 · Hard

Mark scheme for 3.9.3.3 Tier 3 · Hard
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01.1	$85.7\,\text{Mpc}$ , $1.41\times10^{37}\,\text{W}$ , and $3.67\times10^{10}L_{\odot}$	$v\approx zc=0.0200(3.00\times10^5)=6.00\times10^3\,\text{km s}^{-1}$ ; the ratio $v/c=0.0200$ is small enough for this estimate. Hubble's law gives $d=v/H=6000/70.0=85.7\,\text{Mpc}=2.645\times10^{24}\,\text{m}$ . Then $P=4\pi d^2F=4\pi(2.645\times10^{24})^2(1.60\times10^{-13})=1.41\times10^{37}\,\text{W}$ . Using the unrounded power, $P/L_{\odot}=3.67\times10^{10}$ .	6
02.1	The quasar at $z=0.0300$ has $4.00$ times the received flux.	At low redshift, $v\approx zc$ , so the recession speeds are $v_{0.0300}\approx9.00\times10^3\,\text{km s}^{-1}$ and $v_{0.0600}\approx1.80\times10^4\,\text{km s}^{-1}$ . Hubble's law then gives $d\propto z$ . Equal source powers give $F\propto1/d^2\propto1/z^2$ . Therefore $F_{0.0300}/F_{0.0600}=(0.0600/0.0300)^2=4.00$ .	3
03.1	$171\,\text{Mpc}$ and $2.11\times10^{36}\,\text{W}$ (accept $2.10\times10^{36}\,\text{W}$ )	The recession speed is $v\approx zc=0.0400(3.00\times10^5)=1.20\times10^4\,\text{km s}^{-1}$ . Hubble's law gives $d=v/H=(1.20\times10^4)/70.0=171\,\text{Mpc}=5.29\times10^{24}\,\text{m}$ . The measured flux is $F=E/(At)=(2.16\times10^{-13})/[0.600(60.0)]=6.00\times10^{-15}\,\text{W m}^{-2}$ . For isotropic emission, $P=4\pi d^2F=4\pi(5.29\times10^{24})^2(6.00\times10^{-15})=2.11\times10^{36}\,\text{W}$ .	5
04.1	$F=1.43\times10^{-13}\,\text{W m}^{-2}$ and the recorded count rate is $9.50\times10^4\,\text{s}^{-1}$ .	The distance is $d=(1.45\times10^3)(3.086\times10^{22})=4.4747\times10^{25}\,\text{m}$ . For isotropic emission, $F=P/(4\pi d^2)=(3.60\times10^{39})/[4\pi(4.4747\times10^{25})^2]=1.4308\times10^{-13}\,\text{W m}^{-2}$ . The incident power on the detector is $P_{\rm det}=FA=(1.4308\times10^{-13})(0.850)=1.2162\times10^{-13}\,\text{W}$ . The incident photon rate is $P_{\rm det}/E_{\rm photon}=(1.2162\times10^{-13})/(3.20\times10^{-19})=3.80\times10^5\,\text{s}^{-1}$ . Applying the quantum efficiency gives a recorded count rate $0.250(3.80\times10^5)=9.50\times10^4\,\text{s}^{-1}$ .	5
05.1	The observations are characteristic of a quasar, not an ordinary nearby star. A large redshift indicates a very distant source, while substantial flux at that distance implies an enormous power output through the inverse-square law. The source is an active galactic nucleus powered by accretion onto a supermassive black hole; appearing point-like only means it is unresolved. Since $z=0.80$ is not small, the non-relativistic relation $v=zc$ is not valid for a reliable recession speed.	Treat each observation as evidence. Radio brightness and a star-like unresolved optical appearance motivated the name quasar, but the large optical redshift places the source at a cosmological distance. In $F=P/(4\pi d^2)$ , a large $d$ combined with appreciable $F$ requires very large $P$ , inconsistent with an ordinary star and consistent with an active galactic nucleus. Accretion onto a supermassive black hole supplies the power. The approximation $v\approx zc$ requires $z\ll1$ ; $0.80$ fails that condition, so no reliable speed should be quoted from that expression.	5

3.9.3.4 · Detection of exoplanets

Tier 1 · Easy

Mark scheme for 3.9.3.4 Tier 1 · Easy
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01.1	The transit method.	A periodic dip occurs when a planet repeatedly passes across the stellar disc, so the observation indicates the transit method.	1
02.1	The planet is much fainter than the star. Their angular separation is very small.	The host star's light overwhelms the weak reflected or emitted light from the planet, and the small apparent separation makes their images hard to resolve.	2

Tier 2 · Standard

Mark scheme for 3.9.3.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$6.21\times10^7\,\text{m}$	The fractional depth is $0.810\%=0.00810$ . Using $\Delta I/I=(R_p/R_)^2$ , $R_p=R_\sqrt{0.00810}=(6.90\times10^8)(0.0900)=6.21\times10^7\,\text{m}$ .	3
02.1	$8.5\,\text{days}$ and $8.6\times10^{-6}\,\text{rad s}^{-1}$	Successive intervals are $11.7-3.2=8.5\,\text{days}$ and $20.2-11.7=8.5\,\text{days}$ , so $T=8.5\,\text{days}=7.344\times10^5\,\text{s}$ . Hence $\omega=2\pi/T=8.5555\times10^{-6}\,\text{rad s}^{-1}=8.6\times10^{-6}\,\text{rad s}^{-1}$ to two significant figures.	2
03.1	Yes. The radial-velocity cycle has period $12.0\,\text{days}$ and the transit period is also $15.0-3.0=12.0\,\text{days}$ . Each transit occurs one-quarter of a cycle after maximum redshift, when the star's radial velocity passes through zero, as required for the same edge-on circular orbit.	A maximum redshift and the next maximum redshift mark one complete radial-velocity cycle, giving $12.0\,\text{days}$ ; the intervening maximum blueshift is half a cycle later. Successive transits are also $12.0\,\text{days}$ apart. In a circular edge-on orbit, conjunction and transit occur when the star's motion is perpendicular to the line of sight, so its radial velocity is zero. Days $3.0$ and $15.0$ are one-quarter of a cycle after the redshift maxima, so the periods and phases are consistent with one planet.	3

Tier 3 · Hard

Mark scheme for 3.9.3.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.2\times10^7\,\text{m}$ and $7.0\times10^3\,\text{m s}^{-1}$ ; matching periodic transit and Doppler signals are independent evidence for the same orbiting companion.	The transit depth is $0.160\%=0.00160$ , so $R_p=R_*\sqrt{0.00160}=(8.0\times10^8)(0.0400)=3.2\times10^7\,\text{m}$ . The maximum radial speed is $v\approx c\Delta\lambda/\lambda=(3.00\times10^8)(0.014/600.0)=7.0\times10^3\,\text{m s}^{-1}$ , with $v/c=2.33\times10^{-5}$ . Repeated transits show an orbiting body crossing the stellar disc, while the alternating Doppler shift independently shows the host star moving towards and away from Earth. If both variations share the $5.2$ -day period, one orbiting companion explains both signals and reduces the chance that brightness variability alone caused the dips.	6
02.1	$48\,\text{m s}^{-1}$ and $18.0\,\text{days}$ ; the orbital plane may not cross the stellar disc from Earth.	The maximum shift is $1.0\times10^{-4}\,\text{nm}$ . Hence $v=c\Delta\lambda/\lambda=(3.00\times10^8)(1.0\times10^{-4}/620.00000)=48\,\text{m s}^{-1}$ to two significant figures. One complete radial-velocity cycle runs between successive maxima, so the orbital period is $18.0\,\text{days}$ . The periodic stellar motion is consistent with an orbiting companion. A transit requires a nearly edge-on alignment; at another inclination the planet does not pass across the stellar disc, so the missing transit does not rule it out.	5
03.1	$1.43\times10^{10}\,\text{m}$	For the stated central transit, the orbital speed is $v=2R_*/t=2(7.50\times10^8)/(3.60\times3600)=1.157\times10^5\,\text{m s}^{-1}$ . The period is $T=9.00(86400)=7.776\times10^5\,\text{s}$ . Circular motion gives $r=vT/(2\pi)=(1.157\times10^5)(7.776\times10^5)/(2\pi)=1.43\times10^{10}\,\text{m}$ .	4
04.1	The transit depth is $0.00800$ or $0.800\%$ , the planet radius is $6.48\times10^7\,\text{m}$ and the orbital period is $12.85\,\text{days}$ . Ingress is the planet moving onto the stellar disc, the flat minimum is its passage across the disc, and egress is its movement off the disc.	The fractional depth is $\Delta I/I=(8.00-7.936)/8.00=0.00800$ . Hence $R_p=R_*\sqrt{\Delta I/I}=(7.25\times10^8)\sqrt{0.00800}=6.4846\times10^7\,\text{m}=6.48\times10^7\,\text{m}$ . The repeat interval gives $T=17.15-4.30=12.85\,\text{days}$ . Intensity falls during ingress as progressively more of the stellar disc is obscured, remains low while the planet is fully in front of the disc, and rises during egress as the planet uncovers it.	5
05.1	The images are angularly resolved because $0.740>0.520$ arcseconds. Residual starlight is $5.21$ times the planet signal, and the minimum reduction factor for equal signals is $4.17\times10^8$ . The planet can therefore remain hidden by stellar glare even when its image is spatially separated.	The predicted separation exceeds the system's minimum separable angle, so angular resolution alone is adequate. After suppression, the residual stellar intensity is $I_/(8.00\times10^7)=1.25\times10^{-8}I_$ . Relative to the planet, this is $(1.25\times10^{-8})/(2.40\times10^{-9})=5.208=5.21$ . Equality requires $I_/S=(2.40\times10^{-9})I_$ , so $S=1/(2.40\times10^{-9})=4.167\times10^8=4.17\times10^8$ . Since the achieved factor is smaller, residual stellar light is still brighter than the planet; spatial separation and contrast are independent obstacles.	5