Option: Turning points in physics – A-level Physics exam questions with mark scheme

1Core5 marks

Cathode rays were first observed in discharge tubes: glass tubes containing gas at low pressure with a large pd between two electrodes. In later experiments, the beam was produced by thermionic emission in an evacuated tube.

(a) Explain why the gas in a discharge tube emits light. [2]

(b) State what is meant by thermionic emission. [1]

(c) In an electron gun the electrons are accelerated from rest through a pd of 2.5 kV. Calculate the speed of the electrons as they leave the anode. Assume the electrons are non-relativistic. [2]

(a) Electrons (cathode rays) / ions accelerated by the pd collide with gas atoms and excite them (allow ionise them)(1)
(a) The atoms de-excite (electrons return to lower energy levels) emitting photons (of light); allow recombination of ions and electrons emits light(1)
(b) Release of electrons from the surface of a (metal) filament/cathode when it is heated (allow: electrons gain enough (thermal) energy to leave the metal surface)(1)
(c) $\tfrac12 m_e v^2 = eV$ with $V = 2500$ V, e.g. $v^2 = \dfrac{2\times1.60\times10^{-19}\times2500}{9.11\times10^{-31}}$(1)
(c) $v = 3.0\times10^{7}$ m s$^{-1}$ (allow $2.96\times10^{7}$; do not allow use of 2.5 V)(1)
Answer: (c) $v \approx 3.0\times10^{7}$ m s$^{-1}$
Examiner tip: In (a) 'electrons hit the gas and it glows' is not enough: you must say the atoms are excited and then emit photons when they de-excite.

2Standard5 marks

A narrow beam of electrons enters the region between two parallel horizontal plates 40 mm apart with a pd of 2400 V between them. A uniform magnetic field of flux density 2.0 mT acts perpendicular to both the beam and the electric field. The fields are adjusted so that the beam passes through undeflected.

(a) Show that the speed of the electrons is about $3\times10^{7}$ m s$^{-1}$. [2]

(b) The electric field is switched off. The electrons now move in a circular arc of radius 85 mm in the magnetic field. Calculate the specific charge $e/m_e$ of the electron. [2]

(c) At the time of this experiment, the largest known specific charge was that of the hydrogen ion, $9.6\times10^{7}$ C kg$^{-1}$. State the significance of the result in (b). [1]

(a) $E = V/d = 2400/0.040 = 6.0\times10^{4}$ V m$^{-1}$(1)
(a) $eE = Bev$ so $v = E/B = 6.0\times10^{4}/2.0\times10^{-3} = 3.0\times10^{7}$ m s$^{-1}$ (must see $v = E/B$ or equivalent force balance)(1)
(b) $Bev = m_e v^2/r$ so $e/m_e = v/(Br)$(1)
(b) $e/m_e = 3.0\times10^{7}/(2.0\times10^{-3}\times0.085) = 1.8\times10^{11}$ C kg$^{-1}$ (allow $1.76\times10^{11}$; ecf from (a))(1)
(c) The specific charge of the electron is (about 1800 times) larger than that of the hydrogen ion, so the electron must have a much smaller mass (and/or larger charge) — it is a subatomic particle / part of the atom (allow: cathode rays are particles, not a form of EM radiation, with the same e/m whatever the cathode metal)(1)
Answer: (a) $3.0\times10^{7}$ m s$^{-1}$; (b) $1.8\times10^{11}$ C kg$^{-1}$
Examiner tip: In 'show that' questions write the force-balance equation $eE = Bev$ first and give the answer to more significant figures than the value quoted.

3Hard6 marks

In a Millikan-type experiment, an oil drop falls between two horizontal plates with no pd applied. It reaches a terminal speed of $1.1\times10^{-4}$ m s$^{-1}$.

Density of oil = 880 kg m$^{-3}$; viscosity of air $\eta = 1.8\times10^{-5}$ N s m$^{-2}$. Ignore the upthrust of the air.

(a) Show that the radius of the drop is about $1.0\times10^{-6}$ m. [2]

(b) Calculate the mass of the drop. [1]

(c) A pd of 474 V is then applied across the plates, which are 6.0 mm apart, as shown in the diagram. The drop becomes stationary. Calculate the charge on the drop. [2]

(d) State what the result of (c) suggests about electric charge. [1]

(a) At terminal speed weight = viscous drag: $\tfrac43\pi r^3\rho g = 6\pi\eta r v$, giving $r^2 = \dfrac{9\eta v}{2\rho g}$(1)
(a) $r = \sqrt{\dfrac{9\times1.8\times10^{-5}\times1.1\times10^{-4}}{2\times880\times9.81}} = 1.02\times10^{-6}$ m (must be seen to at least 3 sf or to more sf than given)(1)
(b) $m = \tfrac43\pi r^3\rho = 3.9\times10^{-15}$ kg (allow $3.6$–$3.9\times10^{-15}$ kg if $r = 1.0\times10^{-6}$ m used)(1)
(c) $\dfrac{QV}{d} = mg$ so $Q = \dfrac{mgd}{V}$(1)
(c) $Q = \dfrac{3.87\times10^{-15}\times9.81\times6.0\times10^{-3}}{474} = 4.8\times10^{-19}$ C (ecf from (b))(1)
(d) The charge is a whole-number multiple (3×) of $1.6\times10^{-19}$ C, so charge is quantised / comes in multiples of $e$ (allow ecf for a value close to an integer multiple; do not allow a bare 'it is negative')(1)
Answer: (a) $1.02\times10^{-6}$ m; (b) $3.9\times10^{-15}$ kg; (c) $4.8\times10^{-19}$ C $= 3e$
Examiner tip: Use the stationary-drop condition $QV/d = mg$, not $QV = mg$: forgetting to divide by the plate separation is the usual lost mark.

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4Core5 marks

In the 17th century, Newton proposed a corpuscular theory of light and Huygens proposed a wave theory. Both theories could explain reflection and refraction.

(a) Compare the predictions of the two theories for the speed of light in water relative to its speed in air, and explain how Newton's theory accounted for refraction towards the normal. [2]

(b) Give one reason why Newton's theory was preferred for more than a century. [1]

(c) Young's double-slit experiment produced a pattern of bright and dark fringes. Explain how this result supported the wave theory, and suggest why the wave theory was still not widely accepted at once. [2]

(a) Newton: light faster in water; Huygens: light slower in water (both needed)(1)
(a) Corpuscles are attracted towards the water surface, increasing the component of velocity perpendicular to the surface (parallel component unchanged), so they bend towards the normal(1)
(b) Any one: Newton's reputation / scientific authority; the wave theory could not explain sharp shadows / why light did not appear to diffract (bend round obstacles); the wave theory could not explain polarisation; speed of light in water could not be measured at the time to decide between them(1)
(c) Dark fringes require (destructive) superposition/interference of waves arriving in antiphase — particles cannot cancel to give darkness (or bright and dark fringes are an interference effect, which only waves show)(1)
(c) Any one: Newton's reputation / supporters of the corpuscular theory opposed it; the experiment was not easily repeated or Young's results/explanation were not convincing to others at first; there was no evidence yet that light was a transverse wave / no medium identified (allow: it took until Fresnel/Foucault for further evidence)(1)
Answer: Newton: faster in water (attraction to surface); Huygens: slower. Interference fringes need waves; acceptance delayed by Newton's authority and lack of further evidence.
Examiner tip: For (a) give both predictions — stating only Huygens' 'slower in water' does not score the comparison mark.

5Standard5 marks

(a) Maxwell's theory predicted that electromagnetic waves travel with speed $c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$. Use values of $\mu_0$ and $\varepsilon_0$ to calculate this speed. [1]

(b) The diagram shows an apparatus of the type used by Fizeau. Light passes through a gap between two teeth of a rotating toothed wheel, travels 8.0 km to a mirror and returns. The wheel has 600 teeth. As the rotation frequency is increased from zero, the returning light is first completely blocked when the wheel turns at 15.6 revolutions per second.

(i) Explain why the returning light is blocked at this rotation frequency. [1]

(ii) Calculate the speed of light from these data. [2]

(c) Hertz produced and detected radio waves. State one observation that showed these waves were electromagnetic. [1]

(a) $c = 1/\sqrt{4\pi\times10^{-7}\times8.85\times10^{-12}} = 3.00\times10^{8}$ m s$^{-1}$(1)
(b)(i) In the time for light to travel to the mirror and back, the wheel turns by half a tooth spacing (from a gap to the next tooth), so the returning light hits a tooth(1)
(b)(ii) Time = $\dfrac{1}{2\times600\times15.6} = 5.34\times10^{-5}$ s (time for wheel to turn through 1/1200 of a revolution)(1)
(b)(ii) $c = \dfrac{2\times8000}{5.34\times10^{-5}} = 3.0\times10^{8}$ m s$^{-1}$ (allow $2.99$–$3.00\times10^{8}$; ecf for time; a distance of 8000 m instead of 16000 m scores max 1)(1)
(c) Any one: waves were polarised (detected signal varied as detector loop/receiver was rotated); stationary waves formed by reflection from a metal sheet gave a wavelength which with the known frequency gave speed $c$; waves were reflected/refracted like light(1)
Answer: (a) $3.00\times10^{8}$ m s$^{-1}$; (b)(ii) $3.0\times10^{8}$ m s$^{-1}$
Examiner tip: The light travels the 8.0 km twice and the wheel turns only half a tooth spacing — missing either factor of 2 gives an answer out by 2 or 4.

6Hard6 marks

At the end of the 19th century, classical wave theory failed to explain both black-body radiation and the photoelectric effect.

Discuss these two failures of classical theory and explain how the work of Planck and Einstein resolved them. [6]

Black-body: classical theory predicted intensity increasing without limit as wavelength decreases / frequency increases (the ultraviolet catastrophe)(1)
Observed black-body spectrum has a peak and intensity falls to zero at short wavelength(1)
Planck: energy of the oscillators/radiation emitted only in discrete packets (quanta) of energy $E = hf$(1)
High-frequency quanta need large energy, so few are emitted — intensity falls at short wavelength, matching experiment(1)
Photoelectricity: wave theory predicted emission at any frequency if intensity high enough / after a time delay at low intensity / KE of electrons increasing with intensity(1)
Observed: threshold frequency below which no emission; emission instantaneous; maximum KE depends on frequency not intensity; intensity affects only number of electrons per second(1)
Einstein: light consists of photons of energy $hf$; one photon is absorbed by one electron (all-or-nothing)(1)
$hf = \phi + E_{k(\max)}$: emission only if $hf \ge \phi$, so threshold frequency $f_0 = \phi/h$; maximum KE increases linearly with $f$(1)
Greater intensity means more photons per second so more electrons per second, but not more energy per electron; no delay because energy is delivered in a single photon(1)

Level	Marks	What it looks like
Level 3	5–6	Both failures are clearly described with the classical prediction contrasted with observation, and both are resolved: Planck's quantisation explains the fall in intensity at short wavelength, and Einstein's photon model explains threshold frequency, instant emission and the intensity/frequency dependence. A coherent, logically linked answer.
Level 2	3–4	Both failures are identified with some explanation, but the resolution of one is incomplete or weakly linked; or one failure and its resolution are covered thoroughly. Reasoning mostly clear.
Level 1	1–2	Some relevant points, e.g. naming the UV catastrophe or stating $E = hf$ or the photoelectric equation, with little linking to why classical theory failed.
Level 0	0	No relevant content.

Answer: UV catastrophe resolved by Planck's quanta $E = hf$; photoelectric threshold, instant emission and KE–frequency dependence explained by Einstein's photon model $hf = \phi + E_{k(\max)}$.
Examiner tip: Top-band answers link each classical prediction to the specific observation it fails and then to the quantum idea that fixes it — listing facts about photons alone stays in Level 2.

7Standard6 marks

(a) In a transmission electron microscope (TEM), the de Broglie wavelength of the electrons must be no larger than the diameter of an atom, about $1\times10^{-10}$ m, for atomic-scale detail to be possible.

Estimate the minimum anode pd needed. Assume the electrons are non-relativistic. [2]

(b) In a low-energy electron diffraction experiment, a beam of electrons is directed at a thin crystal and a pattern of rings is seen on a screen. Describe and explain how the pattern changes when the speed of the electrons is increased. [2]

(c) Explain how a scanning tunnelling microscope (STM) produces an image of a surface, when operated in constant-current mode. [2]

(a) $\lambda = \dfrac{h}{\sqrt{2m_e eV}}$ rearranged: $V = \dfrac{h^2}{2m_e e\lambda^2}$(1)
(a) $V = \dfrac{(6.63\times10^{-34})^2}{2\times9.11\times10^{-31}\times1.60\times10^{-19}\times(1\times10^{-10})^2} \approx 150$ V (accept 100–200 V for an estimate)(1)
(b) Rings become smaller / closer together (smaller radius)(1)
(b) Greater speed means greater momentum, so smaller de Broglie wavelength ($\lambda = h/mv$), so less diffraction / smaller diffraction angles(1)
(c) A very sharp tip is held very close (≈ 1 nm) to the surface with a small pd between them; electrons tunnel across the gap (quantum tunnelling) giving a small current that depends very strongly on the gap width(1)
(c) The tip scans across the surface and its height is adjusted (by feedback) to keep the current constant; the vertical movement of the tip is recorded to map the surface (atomic) contours(1)
Answer: (a) $\approx 150$ V
Examiner tip: In (b) link speed to wavelength through momentum — 'faster electrons diffract less' without the wavelength step loses the explanation mark.

8Hard5 marks

Muons are produced in the upper atmosphere at a height of 3.0 km above a detector on the ground. They travel vertically downwards at $0.995c$. The half-life of muons measured in their own rest frame is 1.5 μs.

A detector at the top of the 3.0 km path counts $2.0\times10^{4}$ muons per minute.

(a) Calculate the number of these muons expected to reach the ground per minute if time dilation is ignored. [2]

(b) Calculate the number expected to reach the ground per minute when time dilation is taken into account. [2]

(a) Time in Earth frame $t = 3000/(0.995\times3.00\times10^{8}) = 1.01\times10^{-5}$ s, which is 6.7 half-lives(1)
(a) $N = 2.0\times10^{4}\times(\tfrac12)^{6.7} \approx 190$ (allow 180–200)(1)
(b) Dilated half-life $t = \dfrac{t_0}{\sqrt{1-v^2/c^2}} = \dfrac{1.5}{\sqrt{1-0.995^2}} = 15$ μs (allow $\gamma = 10.0$ seen); or proper time of journey $= 1.0$ μs(1)
(b) Number of half-lives $= 10.05/15.0 = 0.67$, $N = 2.0\times10^{4}\times(\tfrac12)^{0.67} \approx 1.3\times10^{4}$ (allow $1.25$–$1.26\times10^{4}$; ecf)(1)
(c) $l = l_0\sqrt{1-v^2/c^2} = 3000\times0.0999 = 300$ m (allow 0.30 km)(1)
Answer: (a) ≈ 190; (b) ≈ $1.3\times10^{4}$; (c) 300 m
Examiner tip: The 1.5 μs is the proper time (measured in the muon frame) — it is the Earth-frame half-life that gets longer, not shorter.

9Stretch6 marks

In an experiment similar to Bertozzi's, pulses of electrons were accelerated to known kinetic energies and timed over a flight path of 8.40 m. Some results are shown.

kinetic energy / MeV	0.50	1.0	1.5	4.5
time of flight / ns	32.5	29.8	29.0	28.1

(a) Calculate the speed predicted by Newtonian mechanics for an electron with kinetic energy 1.5 MeV. [2]

(b) Calculate the measured speed of the 1.5 MeV electrons. [1]

(c) Use $E = \dfrac{m_0c^2}{\sqrt{1-v^2/c^2}}$ to calculate the speed predicted by special relativity for an electron of kinetic energy 1.5 MeV, and comment on how it compares with your answer to (b). [2]

(d) State what the data in the table show about how the kinetic energy of an electron varies with its speed. [1]

(a) $E_k = 1.5\times10^{6}\times1.60\times10^{-19} = 2.4\times10^{-13}$ J and $v = \sqrt{2E_k/m_e}$(1)
(a) $v = 7.3\times10^{8}$ m s$^{-1}$ (allow $7.26\times10^{8}$)(1)
(b) $v = 8.40/29.0\times10^{-9} = 2.90\times10^{8}$ m s$^{-1}$(1)
(c) Total energy $E = E_k + m_0c^2 = 2.4\times10^{-13} + 8.20\times10^{-14} = 3.22\times10^{-13}$ J, so $\sqrt{1-v^2/c^2} = 8.20\times10^{-14}/3.22\times10^{-13} = 0.255$ (allow $\gamma = 3.93$)(1)
(c) $v = 0.967c = 2.90\times10^{8}$ m s$^{-1}$, which agrees with the measured value (b) (Newtonian value is more than twice $c$ and is wrong)(1)
(d) Kinetic energy increases (without limit) while the speed increases only slightly, approaching but never reaching $c$ (allow: $c$ is a limiting speed; KE tends to infinity as $v \to c$)(1)
Answer: (a) $7.3\times10^{8}$ m s$^{-1}$; (b) $2.90\times10^{8}$ m s$^{-1}$; (c) $0.967c = 2.90\times10^{8}$ m s$^{-1}$, agrees with (b)
Examiner tip: In (c) the kinetic energy is $E - m_0c^2$, not $E$: use the total energy (KE plus rest energy) in the relativistic equation.

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