3.12 Turning points in physics (A-level only) — coverage pack

Answer

• It travels from the cathode towards the anode and consists of electrons.

Method: The cathode is the negative electrode. The ray leaves this electrode and moves towards the positive anode; its field deflections identify its particles as negatively charged electrons.

Answer

• The field ionises the low-pressure gas; positive ions accelerate to the cathode and release electrons, which accelerate towards the anode as a cathode ray.

Method: The large potential difference creates a strong electric field in the rarefied gas. Collisions ionise gas atoms. Positive ions are accelerated towards the negative cathode and their impacts release electrons. These electrons are accelerated away from the cathode towards the anode, producing the cathode ray.

Answer

• Cathode rays are identical negatively charged particles found in every material, so atoms contain smaller constituents and are not indivisible.

Method: Deflection towards the positive plate shows that the beam carries negative charge. Reversal when the magnetic field reverses is consistent with moving charged particles rather than neutral radiation. The same measured behaviour for different gases and cathodes shows that the particles are universal rather than atoms of one material. Thomson identified them as electrons. Because electrons are emitted from atoms of every substance, the evidence overturned the idea that atoms were indivisible.

Answer

• Heating gives some conduction electrons enough energy to overcome the surface barrier and escape.

Method: Thermal energy increases the electrons' energies. Electrons in the high-energy part of the distribution can then overcome the metal's work function and leave the surface.

Answer

• $2.96\times10^7\,\text{m s}^{-1}$

Method: Convert the potential difference: $V=2.50\times10^3\,\text{V}$. Use $eV=\frac{1}{2}m_{\mathrm e}v^2$, so $v=\sqrt{2eV/m_{\mathrm e}}$. Hence $v=\sqrt{2(1.60\times10^{-19})(2.50\times10^3)/(9.11\times10^{-31})}=2.96\times10^7\,\text{m s}^{-1}$.

Answer

• $v=5.13\times10^7\,\text{m s}^{-1}$, $1.88\times10^{16}$ electrons per second, and $22.5\,\text{W}$

Method: For one electron, $eV=\frac{1}{2}m_{\mathrm e}v^2$, so $v=\sqrt{2(1.60\times10^{-19})(7.50\times10^3)/(9.11\times10^{-31})}=5.13\times10^7\,\text{m s}^{-1}$. The emission rate is $N/t=I/e=(3.00\times10^{-3})/(1.60\times10^{-19})=1.88\times10^{16}\,\text{s}^{-1}$. The beam power is $P=IV=(3.00\times10^{-3})(7.50\times10^3)=22.5\,\text{W}$.

Answer

• $3.00\times10^6\,\text{m s}^{-1}$

Method: With no deflection, electric and magnetic forces balance: $eE=evB$. Therefore $v=E/B=(2.40\times10^4)/(8.00\times10^{-3})=3.00\times10^6\,\text{m s}^{-1}$.

Answer

• $1.75\times10^{11}\,\text{C kg}^{-1}$

Method: The magnetic force supplies the centripetal force: $evB=m_{\mathrm e}v^2/r$. Thus $e/m_{\mathrm e}=v/(Br)$. With $r=9.50\times10^{-2}\,\text{m}$, $e/m_{\mathrm e}=(3.00\times10^6)/[(1.80\times10^{-4})(9.50\times10^{-2})]=1.75\times10^{11}\,\text{C kg}^{-1}$.

Answer

• $e/m_{\mathrm e}=1.75\times10^{11}\,\text{C kg}^{-1}$, about $1.83\times10^3$ times the hydrogen-ion value

Method: The plate field is $E=V/d=360/(15.0\times10^{-3})=2.40\times10^4\,\text{V m}^{-1}$. Force balance in the selector gives $v=E/B=(2.40\times10^4)/(8.00\times10^{-3})=3.00\times10^6\,\text{m s}^{-1}$. Magnetic deflection then gives $e/m_{\mathrm e}=v/(Br)=(3.00\times10^6)/[(1.60\times10^{-3})(10.7\times10^{-3})]=1.75\times10^{11}\,\text{C kg}^{-1}$. The ratio is $(1.75\times10^{11})/(9.58\times10^7)=1.83\times10^3$. Since the charge magnitudes are equal, this showed that the electron mass is about $1/1830$ of the hydrogen-ion mass.

Answer

• $QE=mg$, with the electric force upward.

Method: Stationary means zero resultant force. The weight $mg$ acts downward, so the electric force must act upward with equal magnitude: $QE=mg$.

Answer

• $3.20\times10^{-19}\,\text{C}=2.00e$

Method: The field is $E=V/d=600/(3.00\times10^{-3})=2.00\times10^5\,\text{V m}^{-1}$. At equilibrium, $QE=mg$, so $Q=(6.52\times10^{-15})(9.81)/(2.00\times10^5)=3.20\times10^{-19}\,\text{C}$. Dividing by $1.60\times10^{-19}\,\text{C}$ gives $Q/e=2.00$.

Answer

• $Q=3.19\times10^{-19}\,\text{C}\approx2e$, supporting quantisation

Method: At terminal speed, $6\pi\eta rv=mg$ and $m=\frac{4}{3}\pi r^3\rho$. Therefore $6\pi\eta rv=\frac{4}{3}\pi r^3\rho g$, so $r=\sqrt{9\eta v/(2\rho g)}=\sqrt{9(1.80\times10^{-5})(8.00\times10^{-5})/[2(850)(9.81)]}=8.82\times10^{-7}\,\text{m}$. The mass is $m=\frac{4}{3}\pi r^3\rho=2.44\times10^{-15}\,\text{kg}$. When held, $QE=mg$, giving $Q=(2.44\times10^{-15})(9.81)/(7.50\times10^4)=3.19\times10^{-19}\,\text{C}$. This is $Q/e=1.99\approx2$, an integer multiple of the elementary charge within experimental uncertainty.

Answer

• Light appears to travel in straight lines and can form sharp shadows.

Method: Corpuscles moving along straight paths would be blocked geometrically by an obstacle, so the model gave a simple explanation of sharp shadows.

Answer

• Newton's theory predicted a speed increase, whereas wave theory predicted a decrease; the measured decrease contradicted the corpuscular account and supported wave theory.

Method: Newton attributed refraction towards the normal to an attractive force increasing a corpuscle's component of velocity in glass, so his model predicted greater speed. Huygens' construction gives bending towards the normal when wave speed falls. Measurement showed that light travels more slowly in glass, selecting the wave prediction and weakening the corpuscular theory.

Answer

• It explained ray-like behaviour and carried Newton's authority, but diffraction, interference and slower propagation in dense media matched wave theory and contradicted classical corpuscles.

Method: Straight-line corpuscle paths explained sharp shadows, and boundary forces offered accounts of reflection and refraction. Diffraction was not conspicuous in ordinary conditions, while Newton's success and reputation gave his model great weight. Later, fringes containing dark regions showed that light contributions can cancel by superposition, which independent classical particles cannot explain. Diffraction also showed spreading around obstacles, and measured light speeds in dense media agreed with Huygens rather than Newton. The combined evidence therefore required a wave model despite the earlier model's authority.

Answer

• Alternating bright and dark fringes.

Method: The repeated maxima and minima show that two contributions sometimes reinforce and sometimes cancel, which is the signature of interference.

Answer

• Diffracted coherent waves overlap; in-phase waves reinforce to form bright fringes and antiphase waves cancel to form dark fringes.

Method: Both slits are illuminated by the same monochromatic source, so the emerging waves have a constant phase relationship. Each slit diffracts the light and the waves overlap. Where the path difference leaves the waves in phase, their amplitudes add and the intensity is high. Where they arrive in antiphase, their amplitudes cancel and a dark fringe forms.

Answer

• The pattern required diffraction and destructive superposition, behaviours predicted by wave theory but not by independent classical corpuscles, so it drove acceptance of Huygens' model.

Method: A simple corpuscle model would send particles through either opening and produce two accumulated bright regions; adding more corpuscles cannot create regularly spaced zero-intensity bands. Huygens' model predicts that each slit launches spreading wavelets. Their phase difference varies across the screen, so superposition alternates between reinforcement and cancellation. The dark bands therefore provide discriminating evidence for destructive interference, not merely straight-line propagation. Because this directly matched the wave prediction, it overcame the long-standing preference for Newton's theory and established the wave nature of light.

Answer

• The electric and magnetic fields are mutually perpendicular and both are perpendicular to the direction of travel.

Method: Electromagnetic waves are transverse. The oscillating electric field is at right angles to the oscillating magnetic field, and each field is at right angles to the propagation direction.

Answer

• $3.00\times10^8\,\text{m s}^{-1}$

Method: Use $c=1/\sqrt{\mu_0\varepsilon_0}$. Thus $c=1/\sqrt{(4\pi\times10^{-7})(8.85\times10^{-12})}=2.999\times10^8\,\text{m s}^{-1}$, which is $3.00\times10^8\,\text{m s}^{-1}$ to three significant figures.

Answer

• $3.01\times10^8\,\text{m s}^{-1}$; the agreement of light and radio-wave speeds with Maxwell's prediction supported their common electromagnetic nature

Method: A gap-to-adjacent-tooth turn is $1/(2N)$ of a revolution, so the round-trip time is $t=1/(2Nf)$. Since the light travels $2D$, $c=2D/t=4DNf$. Therefore $c=4(8.63\times10^3)(720)(12.1)=3.01\times10^8\,\text{m s}^{-1}$. Fizeau showed terrestrially that light has this finite speed. Hertz produced radio waves with wave properties and a speed close to the same value. Their agreement with $1/\sqrt{\mu_0\varepsilon_0}$ supported Maxwell's claim that both light and radio are electromagnetic waves.

Answer

• Below a threshold frequency no electrons are emitted, regardless of intensity.
• Emission starts without a measurable delay.
• Maximum electron kinetic energy depends on frequency rather than intensity.

Method: Any one listed observation conflicts with continuous energy delivery by a classical wave: a threshold, immediate one-event transfer, or frequency-controlled electron energy requires quantised photons.

Answer

• Energy is exchanged only in packets $E=hf$, so high-frequency modes require large quanta and are unlikely to be excited at a fixed temperature, suppressing the classical high-frequency divergence.

Method: Classical equipartition allowed every electromagnetic mode to gain energy continuously, producing an unbounded high-frequency intensity. Planck restricted emission and absorption to quanta with energy $hf$. At high $f$ each quantum is large compared with the available thermal energy, so few such quanta are emitted. The spectrum therefore falls instead of diverging.

Answer

• $6.89\times10^5\,\text{m s}^{-1}$

Method: The photon energy is $E=hc/\lambda=(6.63\times10^{-34})(3.00\times10^8)/(350\times10^{-9})=5.68\times10^{-19}\,\text{J}=3.55\,\text{eV}$. Hence $K_{\max}=3.55-2.20=1.35\,\text{eV}=2.16\times10^{-19}\,\text{J}$. From $K_{\max}=\frac{1}{2}m_{\mathrm e}v^2$, $v=\sqrt{2(2.16\times10^{-19})/(9.11\times10^{-31})}=6.89\times10^5\,\text{m s}^{-1}$. Doubling intensity supplies more photons per second, so more electrons may be emitted, but each photon still has energy $hc/\lambda$ and therefore cannot increase $K_{\max}$.

Answer

• $2.21\times10^{-10}\,\text{m}$

Method: Use $\lambda=h/p=(6.63\times10^{-34})/(3.00\times10^{-24})=2.21\times10^{-10}\,\text{m}$.

Answer

• $1.00\times10^{-10}\,\text{m}$

Method: Use $\lambda=h/\sqrt{2m_{\mathrm e}eV}$. Therefore $\lambda=(6.63\times10^{-34})/\sqrt{2(9.11\times10^{-31})(1.60\times10^{-19})(150)}=1.00\times10^{-10}\,\text{m}$.

Answer

• $6.14\times10^{-11}\,\text{m}$ and $17.8^\circ$

Method: At $100\,\text{V}$, $\lambda_1=h/\sqrt{2m_{\mathrm e}eV}=1.228\times10^{-10}\,\text{m}$. Since $\lambda\propto1/\sqrt{V}$, raising $V$ by a factor of $4$ gives $\lambda_2=\lambda_1/2=6.14\times10^{-11}\,\text{m}$. The same plane spacing gives $\sin\theta_2/\sin\theta_1=\lambda_2/\lambda_1=1/2$, so $\theta_2=\sin^{-1}[0.5\sin(37.8^\circ)]=17.8^\circ$. A diffraction pattern requires coherent wave superposition, so electrons previously treated as particles also have wave behaviour; their localised detection retains the particle aspect.

Answer

• The electrons gain momentum, so their de Broglie wavelength decreases and smaller detail can be resolved.

Method: Acceleration through a larger potential difference increases electron momentum. Since $\lambda=h/p$, the wavelength falls; the diffraction limit is then smaller, allowing finer detail to be distinguished.

Answer

• $151\,\text{V}$

Method: Rearrange $\lambda=h/\sqrt{2m_{\mathrm e}eV}$ to $V=h^2/(2m_{\mathrm e}e\lambda^2)$. Thus $V=(6.63\times10^{-34})^2/[2(9.11\times10^{-31})(1.60\times10^{-19})(0.100\times10^{-9})^2]=151\,\text{V}$.

Answer

• $6.03\times10^4\,\text{V}$

Method: Use $V=h^2/(2m_{\mathrm e}e\lambda^2)$. With $\lambda=5.00\times10^{-12}\,\text{m}$, $V=(6.63\times10^{-34})^2/[2(9.11\times10^{-31})(1.60\times10^{-19})(5.00\times10^{-12})^2]=6.03\times10^4\,\text{V}$. In a TEM the high-voltage electrons pass through a thin specimen and electromagnetic lenses form an image from transmitted and scattered electrons; the short wavelength supports high resolution. An STM instead scans a sharp conducting tip across a conducting surface and measures the strongly distance-dependent tunnelling current. Its atomic sensitivity is therefore a tunnelling effect, not a consequence of TEM-style anode-voltage wavelength reduction.

Answer

• A fringe shift when the interferometer was rotated.

Method: An ether wind was expected to change the relative travel times along the perpendicular arms. Rotation would exchange their orientations and change the phase difference, moving the fringes.

Answer

• Rotation exchanged the arm parallel to the proposed ether wind with the perpendicular arm, so an ether should change the phase difference; no shift meant absolute motion was not detected and supported an invariant light speed.

Method: The beam splitter sends light along two perpendicular paths. In the ether model, their round-trip times differ according to orientation. A $90^\circ$ rotation swaps those orientations, so the predicted time difference changes sign and should move the interference pattern. No significant shift was observed. This removed evidence for a stationary ether or detectable absolute motion and was consistent with light having the same speed in every inertial direction.

Answer

• $0.400$ fringes

Method: The initial time difference is $\Delta t=Lv^2/c^3=11.0(3.00\times10^4)^2/(3.00\times10^8)^3=3.67\times10^{-16}\,\text{s}$. Rotation exchanges the arms, so the change is twice the initial path difference and $n=2c\Delta t/\lambda=2(3.00\times10^8)(3.67\times10^{-16})/(550\times10^{-9})=0.400$. A shift of this scale was the ether model's testable prediction. Its absence meant the expected directional difference in light speed was not present, undermining absolute ether motion and supporting invariant $c$.

Answer

• A non-accelerating frame in which an object with no resultant force moves at constant velocity.

Method: An inertial frame obeys Newton's first law. It may move at a constant velocity relative to another inertial frame, but it must not accelerate or rotate.

Answer

• Physical laws have the same form in all inertial frames and light in free space has invariant speed $c$; therefore perpendicular arms cannot reveal an ether wind through different measured light speeds.

Method: First, no inertial frame is privileged because physical laws have the same form in all of them. Second, every inertial observer measures the vacuum speed of light as $c$, regardless of source or observer motion. The two interferometer arms therefore do not acquire the classical directional speed difference required by an ether wind, so the absence of a rotation-dependent fringe shift is expected.

Answer

• The passenger moves towards the front flash and away from the rear flash, so receives the front flash first; because both observers measure each light pulse at $c$, the passenger concludes the front event occurred earlier, showing simultaneity is frame-dependent.

Method: In the platform frame, equal distances from simultaneous events mean the midpoint observer receives the pulses together. The passenger moves towards the pulse from the front and away from the pulse from the rear, so the front pulse reaches the passenger first. Einstein's second postulate forbids explaining this by assigning different light speeds: the passenger measures both pulses at $c$. The passenger must instead infer unequal emission times in the train frame. This preserves the same physical laws in both inertial frames and demonstrates that simultaneity, unlike $c$, is not invariant.

Answer

• The time between two events measured by one clock in the frame where the events occur at the same position.

Method: For a proper-time measurement the same clock must be present at both events, so no synchronisation of separated clocks is required.

Answer

• $3.67\,\mu\text{s}$ and $880\,\text{m}$

Method: The Lorentz factor is $\gamma=1/\sqrt{1-0.800^2}=1.667$. Hence $t=\gamma t_0=(1.667)(2.20\,\mu\text{s})=3.67\,\mu\text{s}$. The laboratory distance is $x=vt=(0.800)(3.00\times10^8)(3.67\times10^{-6})=8.80\times10^2\,\text{m}$.

Answer

• $0.219$ survive with time dilation, compared with $2.44\times10^{-7}$ without it

Method: For $v=0.995c$, $\gamma=1/\sqrt{1-0.995^2}=10.01$. The laboratory travel time is $t=10.0\times10^3/[0.995(3.00\times10^8)]=33.50\,\mu\text{s}$. The proper time experienced by the muons is $t_0=t/\gamma=3.346\,\mu\text{s}$. Thus the surviving fraction is $N/N_0=e^{-t_0/\tau_0}=e^{-3.346/2.20}=0.219$. Without time dilation one would use $33.50\,\mu\text{s}$ as the decay time, giving $e^{-33.50/2.20}=2.44\times10^{-7}$. The observed survival is therefore evidence for time dilation.

Answer

• Its length measured in the frame in which it is at rest.

Method: The proper length uses the object's rest frame, where its endpoints have fixed positions and the length is greatest.

Answer

• $96.0\,\text{m}$

Method: The Lorentz factor is $\gamma=1/\sqrt{1-0.600^2}=1.25$. Therefore $l=l_0/\gamma=120/1.25=96.0\,\text{m}$.

Answer

• $597\,\text{m}$ and $2.03\,\mu\text{s}$

Method: The Lorentz factor is $\gamma=1/\sqrt{1-0.980^2}=5.025$. In the particle frame the moving accelerator has length $l=l_0/\gamma=(3.00\times10^3)/5.025=597\,\text{m}$. Its transit time there is $t_0=l/v=597/[0.980(3.00\times10^8)]=2.03\,\mu\text{s}$. In the laboratory, $t=(3.00\times10^3)/[0.980(3.00\times10^8)]=10.2\,\mu\text{s}$. Since $t/\gamma=10.2/5.025=2.03\,\mu\text{s}$, the length-contraction and time-dilation descriptions agree.

Answer

• $2.70\times10^6\,\text{J}$

Method: Use $\Delta E=\Delta mc^2=(3.00\times10^{-11})(3.00\times10^8)^2=2.70\times10^6\,\text{J}$.

Answer

• $E=1.37\times10^{-13}\,\text{J}$ and $K=5.47\times10^{-14}\,\text{J}$

Method: The Lorentz factor is $\gamma=1/\sqrt{1-0.800^2}=1.667$. The rest energy is $m_0c^2=(9.11\times10^{-31})(3.00\times10^8)^2=8.20\times10^{-14}\,\text{J}$. Hence $E=\gamma m_0c^2=(1.667)(8.20\times10^{-14})=1.37\times10^{-13}\,\text{J}$. The kinetic energy is $K=E-m_0c^2=(\gamma-1)m_0c^2=5.47\times10^{-14}\,\text{J}$.

Answer

• $v_{\mathrm{rel}}=0.9936c=2.98\times10^8\,\text{m s}^{-1}$; $v_{\mathrm{class}}=1.19\times10^9\,\text{m s}^{-1}$

Method: Relativistically, $K=(\gamma-1)m_0c^2$, so $\gamma=1+4.00/0.511=8.83$. Therefore $v/c=\sqrt{1-1/\gamma^2}=0.9936$ and $v=2.98\times10^8\,\text{m s}^{-1}$. Classically, $K=4.00\times10^6(1.60\times10^{-19})=6.40\times10^{-13}\,\text{J}$, so $v=\sqrt{2K/m_0}=\sqrt{2(6.40\times10^{-13})/(9.11\times10^{-31})}=1.19\times10^9\,\text{m s}^{-1}$, which exceeds $c$. Bertozzi found that measured speed approached but did not exceed $c$ while kinetic energy continued to rise, matching the relativistic prediction and rejecting the classical one.