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IB Physics E.2: Quantum physics worked solutions

Worked solutions to every practice question in E.2: Quantum physics of Photon Physics: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question B1

Type P1 · photon energy and momentum[1 mark]

A beam of orange light has a wavelength of 600\,\mathrm{nm}. What is the momentum of one photon in the beam?

A. 3.3\times10^{-19}\,\mathrm{kg\,m\,s^{-1}}

B. 1.1\times10^{-27}\,\mathrm{kg\,m\,s^{-1}}

C. 9.0\times10^{26}\,\mathrm{kg\,m\,s^{-1}}

D. 3.7\times10^{-36}\,\mathrm{kg\,m\,s^{-1}}

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p = h/\lambda = 6.63\times10^{-34}/(600\times10^{-9}) = 1.1\times10^{-27}\,\mathrm{kg\,m\,s^{-1}}. A is the photon energy hc/\lambda in joules, not a momentum. C inverts the relation (\lambda/h). D divides the momentum by c a second time.

Final answer

Question B2

Type P2 · what starts emission[1 mark]

Green light falls on a clean metal surface and no electrons are emitted. Which change could cause electrons to be emitted?

A. Using green light of much greater intensity

B. Using red light

C. Using violet light

D. Using a metal with a larger work function

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Emission needs one photon to carry at least the work function: hf\ge\Phi. Violet photons have a higher frequency, so more energy each. A only sends more green photons per second; each one is still too weak. B lowers the photon energy. D raises the energy needed.

Final answer

Question B3

Type P3 · evidence against the wave model[1 mark]

Which observation of the photoelectric effect can NOT be explained by a wave model of light?

A. The number of electrons emitted per second increases with the intensity of the light.

B. The emitted electrons have a range of kinetic energies.

C. Below a certain frequency no electrons are emitted, however intense the light.

D. The photocurrent stops when a large enough reverse potential difference is applied.

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In a wave model, energy arrives continuously, so intense light of any frequency should free electrons in the end. A threshold frequency needs photons: one photon of energy hf per electron. A fits both models (more energy per second, more electrons). B and D say nothing about how the energy arrives.

Final answer

Question B4

Type P4 · rate versus energy[1 mark]

Light frees photoelectrons from a metal. Its frequency is raised, but its intensity and the area lit are kept as before. Assume every photon has an equal chance of freeing an electron. How do the rate of emission and the largest electron kinetic energy change?

A. Both rise.

B. The rate falls; the largest kinetic energy rises.

C. The rate is unchanged; the largest kinetic energy rises.

D. The rate rises; the largest kinetic energy is unchanged.

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Equal intensity over an equal area delivers equal power. With more energy in each photon, the photon arrival rate drops, and the electron rate drops with it. E_\text{max} = hf-\Phi grows with f. C wrongly treats equal intensity as an equal number of photons.

Final answer

Question B5

Type F2 · energy balance[1 mark]

Ultraviolet radiation of wavelength 250\,\mathrm{nm} falls on a metal of work function 3.10\,\mathrm{eV}. What is the stopping potential?

A. 3.10\,\mathrm{V}

B. 4.97\,\mathrm{V}

C. 1.87\,\mathrm{V}

D. 8.07\,\mathrm{V}

Show solutionHide solution

Photon energy hc/\lambda = 6.63\times10^{-34}\times3.00\times10^{8}/(250\times10^{-9}) = 7.96\times10^{-19}\,\mathrm{J} = 4.97\,\mathrm{eV}. eV_s = E_\text{max} = 4.97-3.10 = 1.87\,\mathrm{eV}, so V_s = 1.87\,\mathrm{V}. B uses the whole photon energy; D adds the work function instead of subtracting it.

Final answer

Question B6

Type F1 · threshold values[1 mark]

The threshold frequency of a metal is 1200\,\mathrm{THz}. What is its work function?

A. 8.0\times10^{-31}\,\mathrm{J}

B. 5.0\,\mathrm{eV}

C. 8.0\times10^{-19}\,\mathrm{eV}

D. 2.5\,\mathrm{eV}

Show solutionHide solution

1200\,\mathrm{THz} = 1.20\times10^{15}\,\mathrm{Hz}. \Phi = hf_0 = 6.63\times10^{-34}\times1.20\times10^{15} = 7.96\times10^{-19}\,\mathrm{J}; divide by 1.60\times10^{-19}: \Phi = 5.0\,\mathrm{eV}. A reads THz as Hz. C gives the joule value the unit eV. D halves the answer.

Final answer

Question B7

Type G2 · graphs for two metals[1 mark]

For metal X, a graph of the maximum kinetic energy of the photoelectrons against the frequency of the light is a straight line. Metal Y has a larger work function than X. Compared with the line for X, the line for Y has

A. the same gradient and meets the frequency axis at a higher frequency.

B. the same gradient and meets the frequency axis at a lower frequency.

C. a greater gradient and meets the frequency axis at the same frequency.

D. a smaller gradient and meets the frequency axis at a higher frequency.

Show solutionHide solution

E_\text{max} = hf-\Phi: the gradient is h for every metal, so the lines are parallel. The frequency intercept is f_0 = \Phi/h, which is higher for the larger work function.

Final answer

Question B8

Type M1 · de Broglie wavelength[1 mark]

Neutrons leave a reactor moderator with a speed of 2.2\,\mathrm{km\,s^{-1}}. The mass of a neutron is 1.675\times10^{-27}\,\mathrm{kg}. What is the de Broglie wavelength of one of these neutrons?

A. 8.2\times10^{-14}\,\mathrm{m}

B. 1.8\times10^{-10}\,\mathrm{m}

C. 3.6\times10^{-10}\,\mathrm{m}

D. 5.6\times10^{9}\,\mathrm{m}

Show solutionHide solution

p = mv = 1.675\times10^{-27}\times2200 = 3.69\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}; \lambda = h/p = 1.8\times10^{-10}\,\mathrm{m}, about the spacing of atoms in a crystal. A divides by mv^2. C uses \tfrac12mv as the momentum. D inverts the relation.

Final answer

Question B9

Type M3 · voltage ratios[1 mark]

Electrons accelerated from rest through a potential difference of 200\,\mathrm{V} have a de Broglie wavelength \lambda. What is the de Broglie wavelength of electrons accelerated from rest through 800\,\mathrm{V}?

A. 4\lambda

B. 2\lambda

C. half the original wavelength

D. \lambda/4

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E_k = eV and p = \sqrt{2m_eeV}, so \lambda\propto 1/\sqrt{V}. Four times the p.d. gives half the wavelength. D takes \lambda\propto1/V; A and B have the ratio upside down.

Final answer

Question B10

Type M5 · wave or particle evidence[1 mark]

Which observation is evidence that electrons have a wave nature?

A. Electrons are released from a zinc plate by ultraviolet radiation.

B. Hydrogen gas emits light of a few definite wavelengths.

C. Alpha particles aimed at gold foil are sometimes turned through large angles.

D. A beam of electrons passing through a thin graphite film forms rings on a screen.

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Rings are a diffraction pattern: constructive interference of electron waves scattered by the regular rows of atoms. A is evidence about light (photons), B about energy levels in atoms, C about the nucleus.

Final answer

Question B11

Type C1 · Compton: what changes[1 mark]

An X-ray photon is scattered by a free electron that is initially at rest. Compared with the incident photon, the scattered photon has

A. a longer wavelength, a lower frequency and the same speed.

B. a longer wavelength, a lower frequency and a lower speed.

C. a shorter wavelength, a higher frequency and the same speed.

D. a longer wavelength, a higher frequency and the same speed.

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The electron recoils and gains kinetic energy, so the photon loses energy: hf falls, f falls and \lambda = c/f rises. Every photon travels at c, so f\lambda is unchanged. B lets the photon slow down; D treats f and \lambda as independent.

Final answer

Question B12

Type C3 · Compton shift and angle[1 mark]

X-ray photons are scattered by free electrons through an angle of 120^\circ. What is the Compton shift in wavelength?

A. -1.21\,\mathrm{pm}

B. 1.21\,\mathrm{pm}

C. 2.43\,\mathrm{pm}

D. 3.64\,\mathrm{pm}

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\Delta\lambda = \dfrac{h}{m_ec}(1-\cos\theta) = 2.43\,\mathrm{pm}\times(1-\cos120^\circ) = 2.43\times1.5 = 3.64\,\mathrm{pm}. A uses \cos\theta in place of 1-\cos\theta. B is the shift at 60^\circ; C at 90^\circ.

Final answer

Tutorial practice

Question T1

Type F1 · threshold values + P3, F2[6 marks]

A clean zinc plate has a work function of 4.30\,\mathrm{eV}.

(a) In one sentence, say what the photoelectric effect is. [1]

(b) Ultraviolet radiation of wavelength 260\,\mathrm{nm} falls on the plate. Show that electrons are emitted. [2]

(d) The ultraviolet lamp is replaced by a very bright red lamp. No electrons are emitted. Explain this using the photon model of light. [2]

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(a) Light, or other electromagnetic radiation, of high enough frequency frees electrons from a metal surface (A1). (b) E = hc/\lambda = 6.63\times10^{-34}\times3.00\times10^{8}/(260\times10^{-9}) = 7.65\times10^{-19}\,\mathrm{J} (M1) = 4.78\,\mathrm{eV}, which is more than 4.30\,\mathrm{eV}, so emission occurs (A1). (c) \lambda_0 = hc/\Phi = 1.989\times10^{-25}/(4.30\times1.60\times10^{-19}) = 2.89\times10^{-7}\,\mathrm{m} = 289\,\mathrm{nm} (A1). (d) Red light is a stream of photons each of energy hf, well below the work function (A1). An electron absorbs one photon at a time, so a brighter lamp sends more photons per second but none of them can free an electron (A1).

Final answer

(a) Radiation of high enough frequency frees electrons from a metal surface. (b) Photon energy 4.78\,\mathrm{eV} > 4.30\,\mathrm{eV}. (c) 289\,\mathrm{nm}. (d) Each red photon has hf < \Phi; one photon gives its energy to one electron, so more photons per second cannot help.

Question T2

Type F3 · maximum electron speed + F2[7 marks]

Light of frequency 7.50\times10^{14}\,\mathrm{Hz} falls on a caesium surface. The work function of caesium is 2.10\,\mathrm{eV}.

(a) Find E_\text{max} for the emitted electrons, in joules. [2]

(b) Calculate the maximum speed of the photoelectrons. [2]

(d) The lamp is moved closer, so the intensity doubles. What happens to the speed found in (b)? [1]

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(a) hf = 6.63\times10^{-34}\times7.50\times10^{14} = 4.97\times10^{-19}\,\mathrm{J}; \Phi = 2.10\times1.60\times10^{-19} = 3.36\times10^{-19}\,\mathrm{J} (M1); E_\text{max} = 4.97\times10^{-19}-3.36\times10^{-19} = 1.61\times10^{-19}\,\mathrm{J} (A1). (b) \tfrac12m_ev^2 = E_\text{max} (M1); v = \sqrt{2\times1.61\times10^{-19}/9.11\times10^{-31}} = 5.95\times10^{5}\,\mathrm{m\,s^{-1}} (A1). (c) The smallest reverse potential difference between collector and emitter that stops even the fastest photoelectrons, so the photocurrent becomes zero (A1). V_s = E_\text{max}/e = 1.61\times10^{-19}/1.60\times10^{-19} = 1.01\,\mathrm{V} (A1). (d) No change: each photon still has the same energy; only the number of electrons per second doubles (A1).

Final answer

(a) 1.61\times10^{-19}\,\mathrm{J}. (b) 5.95\times10^{5}\,\mathrm{m\,s^{-1}}. (c) The reverse p.d. that just stops the fastest electrons, so the current falls to zero; 1.01\,\mathrm{V}. (d) No change.

Question T3

Type G1 · gradient to h + G2[8 marks]

In a photoelectric experiment the maximum kinetic energy E_\text{max} of the electrons is measured for light of several frequencies f.

\begin{array}{lcccc}f/10^{14}\,\mathrm{Hz} & 8.0 & 10.0 & 12.0 & 14.0\\ E_\text{max}/\,\mathrm{eV} & 0.61 & 1.44 & 2.27 & 3.10\end{array}

(a) Explain why a graph of E_\text{max} against f is a straight line, and state what its gradient represents. [2]

(b) Use the data to determine the Planck constant, in J s. [2]

(d) The experiment is repeated with a metal of larger work function. Describe how its line compares. [1]

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(a) E_\text{max} = hf-\Phi has the form y = mx+c with h and \Phi constant (A1); the gradient is the Planck constant h (here in eV s) (A1). (b) Gradient = (3.10-0.61)/(6.0\times10^{14}) = 4.15\times10^{-15}\,\mathrm{eV\,s} (M1); h = 4.15\times10^{-15}\times1.60\times10^{-19} = 6.64\times10^{-34}\,\mathrm{J\,s} (A1). (c) \Phi = hf-E_\text{max} = 4.15\times10^{-15}\times8.0\times10^{14}-0.61 (M1) = 2.7\,\mathrm{eV} (A1); f_0 = \Phi/h = 2.71/(4.15\times10^{-15}) = 6.5\times10^{14}\,\mathrm{Hz} (A1). (d) A parallel line (same gradient h) that meets the frequency axis at a higher frequency (A1).

Final answer

(a) E_\text{max} = hf-\Phi is linear in f; gradient = h. (b) 6.64\times10^{-34}\,\mathrm{J\,s}. (c) f_0 = 6.5\times10^{14}\,\mathrm{Hz}; \Phi = 2.7\,\mathrm{eV}. (d) Parallel, shifted to higher frequency.

Question T4

Type G4 · photocell current-voltage curves[6 marks]

A photocell is lit by monochromatic light. The potential V of the collector relative to the emitter is varied from -3.0\,\mathrm{V} to +3.0\,\mathrm{V}. The current is zero for V\le-1.40\,\mathrm{V} and has a constant value of 6.0\,\mathrm{\mu A} for V\ge+0.50\,\mathrm{V}.

(a) Explain why the current stays constant for V\ge+0.50\,\mathrm{V}. [1]

(b) Explain why the current is not zero when V = 0. [1]

(d) How many electrons arrive at the collector per second when I = 6.0\,\mathrm{\mu A}? [1]

(e) The intensity of the light is halved; its frequency is unchanged. State the new constant current and the new stopping potential. [2]

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(a) All the electrons emitted each second are collected, so a larger p.d. cannot increase the current (A1). (b) The electrons leave the emitter with kinetic energy, so many cross to the collector without any accelerating p.d. (A1). (c) E_\text{max} = eV_s = 1.60\times10^{-19}\times1.40 = 2.24\times10^{-19}\,\mathrm{J} (A1). (d) N = I/e = 6.0\times10^{-6}/1.60\times10^{-19} = 3.8\times10^{13}\,\mathrm{s^{-1}} (A1). (e) Half as many photons per second, so 3.0\,\mathrm{\mu A} (A1); each photon has the same energy, so the stopping potential stays 1.40\,\mathrm{V} (A1).

Final answer

(a) Every emitted electron already reaches the collector. (b) Electrons leave the emitter with kinetic energy, so some reach the collector with no p.d. (c) 2.24\times10^{-19}\,\mathrm{J}. (d) 3.8\times10^{13}\,\mathrm{s^{-1}}. (e) 3.0\,\mathrm{\mu A}; 1.40\,\mathrm{V}.

Question T5

Type P5 · photon rate and current + P1, P4[8 marks]

A violet laser of power 2.0\,\mathrm{mW} and wavelength 405\,\mathrm{nm} shines on a photocathode. On average one photon in 250 releases an electron.

(a) Calculate the energy and the momentum of one photon. [2]

(b) Calculate the number of photons that reach the photocathode each second. [2]

(d) The laser is replaced by a 2.0\,\mathrm{mW} laser of wavelength 375\,\mathrm{nm}. One photon in 250 still releases an electron. Determine the new largest photocurrent. [2]

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(a) E = hc/\lambda = 1.989\times10^{-25}/(405\times10^{-9}) = 4.91\times10^{-19}\,\mathrm{J} (A1); p = h/\lambda = 6.63\times10^{-34}/(405\times10^{-9}) = 1.64\times10^{-27}\,\mathrm{kg\,m\,s^{-1}} (A1). (b) N = P/E (M1) = 2.0\times10^{-3}/4.91\times10^{-19} = 4.1\times10^{15}\,\mathrm{s^{-1}} (A1). (c) Electrons per second = 4.07\times10^{15}/250 = 1.63\times10^{13}\,\mathrm{s^{-1}} (M1); I = 1.63\times10^{13}\times1.60\times10^{-19} = 2.6\times10^{-6}\,\mathrm{A} (A1). (d) Same power, but each photon carries more energy, so the photon rate scales with \lambda: \times375/405 (M1); I = 2.61\times375/405 = 2.4\,\mathrm{\mu A} (A1).

Final answer

(a) 4.91\times10^{-19}\,\mathrm{J}; 1.64\times10^{-27}\,\mathrm{kg\,m\,s^{-1}}. (b) 4.1\times10^{15}\,\mathrm{s^{-1}}. (c) 2.6\,\mathrm{\mu A}. (d) 2.4\,\mathrm{\mu A}: fewer, more energetic photons.

Question T6

Type F4 · energy on arrival + F2, F3[6 marks]

Photoelectrons leave an emitter with a maximum kinetic energy of 1.30\,\mathrm{eV}. A battery holds the collector at a potential 2.50\,\mathrm{V} above the emitter.

(a) Calculate the maximum kinetic energy, in eV, of an electron arriving at the collector. [2]

(b) The battery is reversed, so the collector is 0.80\,\mathrm{V} below the emitter. State the maximum kinetic energy, in eV, of an electron arriving at the collector. [1]

(d) State the smallest reverse p.d. that stops all the electrons. [1]

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(a) The field does work e\Delta V = 2.50\,\mathrm{eV} on each electron as it crosses (M1); 1.30+2.50 = 3.80\,\mathrm{eV} (A1). The work function is not subtracted again: it was paid when the electron left the metal. (b) 1.30-0.80 = 0.50\,\mathrm{eV} (A1). (c) 0.50\,\mathrm{eV} = 8.0\times10^{-20}\,\mathrm{J} (M1); v = \sqrt{2\times8.0\times10^{-20}/9.11\times10^{-31}} = 4.2\times10^{5}\,\mathrm{m\,s^{-1}} (A1). (d) V_s = E_\text{max}/e = 1.30\,\mathrm{V} (A1).

Final answer

(a) 3.80\,\mathrm{eV}. (b) 0.50\,\mathrm{eV}. (c) 4.2\times10^{5}\,\mathrm{m\,s^{-1}}. (d) 1.30\,\mathrm{V}.

Question T7

Type M2 · wavelength from accelerating p.d. + M6[7 marks]

In an evacuated tube, electrons start from rest and are accelerated by a p.d. of 2.00\,\mathrm{kV}. The beam then crosses a thin film of graphite and lands on a fluorescent screen.

(a) Show that each electron has a de Broglie wavelength close to 2.7\times10^{-11}\,\mathrm{m}. [3]

(b) The atoms in graphite are about 1.4\times10^{-10}\,\mathrm{m} apart. Explain why graphite is a suitable target for showing the wave nature of these electrons. [1]

(d) The p.d. is raised. Describe what happens to the rings, and why. [2]

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(a) E_k = eV = 1.60\times10^{-19}\times2000 = 3.2\times10^{-16}\,\mathrm{J} (M1); p = \sqrt{2m_eE_k} = \sqrt{2\times9.11\times10^{-31}\times3.2\times10^{-16}} = 2.41\times10^{-23}\,\mathrm{kg\,m\,s^{-1}} (M1); \lambda = h/p = 6.63\times10^{-34}/2.41\times10^{-23} = 2.75\times10^{-11}\,\mathrm{m} (A1). (b) The gaps between rows of atoms are of the same order as the wavelength, so diffraction through them is large enough to see (A1). (c) Graphite film is made of many tiny crystals pointing every way; each angle of strong scattering is shared by crystals all round the beam, which gives a ring (A1). (d) The rings shrink to smaller angles (A1): larger p.d., larger momentum, shorter de Broglie wavelength, so less diffraction (A1).

Final answer

(a) E_k = 3.2\times10^{-16}\,\mathrm{J}, p = 2.41\times10^{-23}\,\mathrm{kg\,m\,s^{-1}}, \lambda = 2.75\times10^{-11}\,\mathrm{m}. (b) The atom spacing is of the same order as \lambda. (c) The film is made of many tiny crystals at all orientations. (d) Rings get smaller: shorter \lambda, smaller angles.

Question T8

Type M4 · energy and mass ratios + M1[6 marks]

A proton and an alpha particle are each accelerated from rest through the same potential difference. Take the mass of an alpha particle as 4m_p and its charge as +2e.

(a) State the ratio \dfrac{\text{kinetic energy of alpha}}{\text{kinetic energy of proton}}. [1]

(b) Show that \dfrac{\lambda_\text{proton}}{\lambda_\text{alpha}} = 2.8. [3]

(c) An electron and a proton have the same kinetic energy. Determine \dfrac{\lambda_\text{electron}}{\lambda_\text{proton}}. [2]

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(a) E_k = qV, and the alpha has twice the charge: ratio = 2 (A1). (b) \lambda = h/\sqrt{2mE_k} = h/\sqrt{2mqV} (M1); \lambda_p/\lambda_\alpha = \sqrt{m_\alpha q_\alpha/(m_pq_p)} (M1) = \sqrt{4\times2} = 2.83\approx2.8 (A1). (c) Same E_k: \lambda\propto1/\sqrt{m} (M1); \lambda_e/\lambda_p = \sqrt{1.673\times10^{-27}/9.11\times10^{-31}} = 42.9 (A1).

Final answer

(a) 2. (b) \sqrt{4\times2} = 2.83. (c) \sqrt{m_p/m_e} = 42.9.

Question T9

Type M1 · de Broglie wavelength + M6, M5[6 marks]

(a) Estimate the de Broglie wavelength of

(i) an alpha particle of kinetic energy 5.0\,\mathrm{MeV} (mass 6.64\times10^{-27}\,\mathrm{kg}) [2]

(ii) a 2.0\,\mathrm{g} table-tennis ball moving at 8.0\,\mathrm{m\,s^{-1}}. [1]

(b) Discuss whether the wave nature of each object in (a) could be observed. [2]

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(a)(i) E_k = 5.0\times10^{6}\times1.60\times10^{-19} = 8.0\times10^{-13}\,\mathrm{J}; p = \sqrt{2\times6.64\times10^{-27}\times8.0\times10^{-13}} = 1.03\times10^{-19}\,\mathrm{kg\,m\,s^{-1}} (M1); \lambda = 6.4\times10^{-15}\,\mathrm{m} (A1). (ii) \lambda = h/mv = 6.63\times10^{-34}/(0.0020\times8.0) = 4.1\times10^{-32}\,\mathrm{m} (A1). (b) The alpha wavelength is similar to the size of a nucleus, so diffraction by nuclei could be seen (A1); the ball's wavelength is far smaller than any gap or object, so no diffraction can be detected (A1). (c) Both light and matter behave as waves in some experiments (diffraction) and as particles in others (one photon or one electron at a time); one model alone does not describe them (A1).

Final answer

(a)(i) 6.4\times10^{-15}\,\mathrm{m}. (ii) 4.1\times10^{-32}\,\mathrm{m}. (b) Alpha: comparable with a nucleus, so diffraction by nuclei is possible; ball: far smaller than any gap, so not observable. (c) Light and matter each show wave behaviour in some experiments and particle behaviour in others.

Question T10

Type C2 · Compton recoil energy + C3, C1[7 marks]

X-ray photons of energy 50.0\,\mathrm{keV} are scattered by free electrons that are initially at rest. A detector records photons scattered through 75^\circ.

(a) Calculate the wavelength of the incident photons. [2]

(b) Calculate the wavelength of the photons that reach the detector. [2]

(d) The photon is scattered upwards from its original line. State the direction in which the electron recoils. [1]

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(a) E = 50.0\times10^{3}\times1.60\times10^{-19} = 8.00\times10^{-15}\,\mathrm{J} (M1); \lambda = hc/E = 1.989\times10^{-25}/8.00\times10^{-15} = 2.49\times10^{-11}\,\mathrm{m} (A1). (b) \Delta\lambda = 2.43\times10^{-12}(1-\cos75^\circ) = 1.80\times10^{-12}\,\mathrm{m} (M1); \lambda_f = 24.86+1.80 = 26.7\,\mathrm{pm} (A1). (c) E_f = hc/\lambda_f = 7.46\times10^{-15}\,\mathrm{J} = 46.6\,\mathrm{keV} (M1); E_k = 50.0-46.6 = 3.4\,\mathrm{keV} (A1). (d) Momentum across the original line must stay zero, so the electron moves forwards and below the line, on the opposite side from the photon (A1).

Final answer

(a) 24.9\,\mathrm{pm}. (b) 26.7\,\mathrm{pm}. (c) 3.4\,\mathrm{keV}. (d) Forwards and downwards, on the other side of the original line.

Question T11

Type R1 · radiation pressure · Seen in 2025 papers[7 marks]

A laser sends 12.0\,\mathrm{W} of light of wavelength 650\,\mathrm{nm} normally onto a flat target of area 4.00\,\mathrm{cm^2}. The beam covers the target uniformly.

(a) Calculate the momentum of one photon and the number arriving each second. [2]

(b) Determine the force and pressure if the target absorbs all the light. [2]

(d) State one assumption about the beam needed in your calculation. [1]

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(a) p=h/\lambda=6.63\times10^{-34}/(650\times10^{-9})=1.02\times10^{-27}\,\mathrm{kg\,m\,s^{-1}} (A1); N=P/(hc/\lambda)=12.0/(3.06\times10^{-19})=3.92\times10^{19}\,\mathrm{s^{-1}} (A1). (b) Absorption transfers momentum p per photon, so F=Np=P/c=4.00\times10^{-8}\,\mathrm{N} (A1); A=4.00\times10^{-4}\,\mathrm{m^2}, pressure F/A=1.00\times10^{-4}\,\mathrm{Pa} (A1). (c) Reflection reverses the normal momentum: the change has magnitude 2p, so the force doubles (M1); pressure =2P/(cA)=2.00\times10^{-4}\,\mathrm{Pa} (A1). (d) The entire stated power strikes the target normally (A1).

Final answer

(a) 1.02\times10^{-27}\,\mathrm{kg\,m\,s^{-1}}; 3.92\times10^{19}\,\mathrm{s^{-1}}. (b) 4.00\times10^{-8}\,\mathrm{N}; 1.00\times10^{-4}\,\mathrm{Pa}. (c) 2.00\times10^{-4}\,\mathrm{Pa}: each photon reverses its momentum. (d) All the stated power reaches the target at normal incidence.

Advanced practice

Question A1

Type M7 · matter-wave straight-line plot + M2 · Predicted type[7 marks]

In an electron-diffraction tube the de Broglie wavelength \lambda of the electrons is found from the ring sizes for several accelerating potential differences V.

\begin{array}{lcccc}V/\,\mathrm{kV} & 1.50 & 2.50 & 3.50 & 5.00\\ \lambda/\,\mathrm{pm} & 31.7 & 24.6 & 20.8 & 17.4\end{array}

(a) Show that a graph of \lambda against \dfrac{1}{\sqrt{V}} should be a straight line through the origin, and write an expression for its gradient. [2]

(b) Use the first and last columns of the table to determine a value for the Planck constant. [3]

(c) The ring sizes were read from a scale that makes every wavelength in the table 2.0\% too large. State and explain the effect on the value of h found in (b). [2]

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(a) eV = p^2/2m_e so p = \sqrt{2m_eeV} and \lambda = h/p = \dfrac{h}{\sqrt{2m_ee}}\cdot\dfrac{1}{\sqrt{V}}: \lambda\propto V^{-1/2}, a line through the origin (A1); gradient = h/\sqrt{2m_ee} (A1). (b) 1/\sqrt{V}: 0.02582 and 0.01414\,\mathrm{V^{-1/2}}; gradient = (31.7-17.4)\times10^{-12}/(0.02582-0.01414) = 1.22\times10^{-9}\,\mathrm{m\,V^{1/2}} (M1); \sqrt{2m_ee} = \sqrt{2\times9.11\times10^{-31}\times1.60\times10^{-19}} = 5.40\times10^{-25} (M1); h = 1.22\times10^{-9}\times5.40\times10^{-25} = 6.6\times10^{-34}\,\mathrm{J\,s} (A1). (c) Every \lambda is 1.020 times too large while 1/\sqrt{V} is unchanged, so the gradient is 1.020 times too large (M1); h is proportional to the gradient, so it is 2.0\% too large (A1).

Final answer

(a) \lambda = \dfrac{h}{\sqrt{2m_ee}}\cdot\dfrac{1}{\sqrt V}; gradient h/\sqrt{2m_ee}. (b) 6.6\times10^{-34}\,\mathrm{J\,s}. (c) h is 2.0\% too large.

Question A2

Type C4 · Compton limits + C3, C2 · Predicted type[7 marks]

Gamma photons of energy 662\,\mathrm{keV} are scattered by free electrons at rest in a detector crystal.

(a) Calculate the wavelength of the incident photons. [1]

(b) Show that no photon scattered once in this way can have an energy below about 185\,\mathrm{keV}. [2]

(c) The detector reports scattered photons of energies 150\,\mathrm{keV} and 300\,\mathrm{keV}. Identify the value that cannot come from one Compton scattering. [1]

(d) Calculate the scattering angle of the 300\,\mathrm{keV} photons. [2]

(e) State the largest kinetic energy that one of these photons can give to an electron. [1]

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(a) E = 662\times10^{3}\times1.60\times10^{-19} = 1.059\times10^{-13}\,\mathrm{J}; \lambda_i = hc/E = 1.88\times10^{-12}\,\mathrm{m} (A1). (b) 1-\cos\theta is at most 2, at \theta = 180^\circ, so \lambda_\text{max} = 1.88+2\times2.43 = 6.73\,\mathrm{pm} (M1); E_\text{min} = hc/\lambda_\text{max} = 2.96\times10^{-14}\,\mathrm{J} = 185\,\mathrm{keV} (A1). (c) 150\,\mathrm{keV}: it is below the smallest possible energy (A1). (d) \lambda_f = hc/E = 1.989\times10^{-25}/(300\times10^{3}\times1.60\times10^{-19}) = 4.14\,\mathrm{pm}, so \Delta\lambda = 2.27\,\mathrm{pm} and 1-\cos\theta = 2.27/2.43 = 0.934 (M1); \cos\theta = 0.066, \theta = 86^\circ (A1). (e) 662-185 = 477\,\mathrm{keV} (A1).

Final answer

(a) 1.88\,\mathrm{pm}. (b) \lambda_\text{max} = 6.73\,\mathrm{pm}, E_\text{min} = 185\,\mathrm{keV}. (c) 150\,\mathrm{keV}. (d) 86^\circ. (e) 477\,\mathrm{keV}.

Question A3

Type G3 · stopping p.d. against wavelength + F1, P3, P2[8 marks]

A laser whose wavelength can be tuned lights a metal surface. The stopping potential V_s is measured for several wavelengths \lambda.

\begin{array}{lccccc}\lambda/\,\mathrm{nm} & 300 & 350 & 400 & 450 & 500\\ V_s/\,\mathrm{V} & 1.94 & 1.35 & 0.91 & 0.56 & 0.29\end{array}

(a) Explain why a graph of V_s against \lambda is not a straight line. [2]

(b) Suggest what to plot against V_s to obtain a straight line, and state its gradient. [2]

(d) Determine the longest wavelength that releases electrons from this metal. [1]

(e) Explain why light of wavelength 600\,\mathrm{nm} releases no electrons however intense it is. [1]

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(a) eV_s = hc/\lambda-\Phi (M1), so V_s is a linear function of 1/\lambda, not of \lambda; equal steps in \lambda give smaller and smaller drops in V_s (A1). (b) Plot V_s against 1/\lambda (A1); gradient = hc/e (A1). (c) At 300\,\mathrm{nm}: hc/(e\lambda) = 1.989\times10^{-25}/(1.60\times10^{-19}\times300\times10^{-9}) = 4.14\,\mathrm{eV} (M1); \Phi = 4.14-1.94 = 2.20\,\mathrm{eV} (A1). (d) \lambda_0 = hc/\Phi = 1.989\times10^{-25}/(2.20\times1.60\times10^{-19}) = 5.65\times10^{-7}\,\mathrm{m} (A1). (e) 600\,\mathrm{nm} is longer than \lambda_0, so each photon has less energy than the work function and one photon frees at most one electron (A1).

Final answer

(a) eV_s = hc/\lambda-\Phi: V_s depends on 1/\lambda. (b) 1/\lambda; gradient hc/e. (c) 2.20\,\mathrm{eV}. (d) 565\,\mathrm{nm}. (e) Each photon has less energy than \Phi.

Question A4

Type M6 · diffraction scale + M2, M4[7 marks]

Electrons accelerated from rest through 600\,\mathrm{V} pass through a single slit of width 0.20\,\mathrm{\mu m} and reach a screen 0.80\,\mathrm{m} beyond the slit.

(a) Calculate the de Broglie wavelength of the electrons. [2]

(b) Calculate the distance on the screen from the centre of the pattern to the first minimum. [2]

(d) The electrons are replaced by neutrons with the same speed as the electrons in (a). Determine the ratio \dfrac{\text{distance to the first minimum for neutrons}}{\text{distance for electrons}}. [2]

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(a) p = \sqrt{2m_eeV} = \sqrt{2\times9.11\times10^{-31}\times1.60\times10^{-19}\times600} = 1.32\times10^{-23}\,\mathrm{kg\,m\,s^{-1}} (M1); \lambda = h/p = 5.0\times10^{-11}\,\mathrm{m} (A1). (b) \theta\approx\lambda/b = 5.01\times10^{-11}/2.0\times10^{-7} = 2.5\times10^{-4}\,\mathrm{rad} (M1); distance = 0.80\times2.5\times10^{-4} = 2.0\times10^{-4}\,\mathrm{m} = 0.20\,\mathrm{mm} (A1). (c) \lambda\propto1/\sqrt V: four times the p.d. halves it, so 0.10\,\mathrm{mm} (A1). (d) Same speed, so \lambda = h/mv\propto1/m (M1); ratio = m_e/m_n = 9.11\times10^{-31}/1.675\times10^{-27} = 5.4\times10^{-4} (A1).

Final answer

(a) 5.0\times10^{-11}\,\mathrm{m}. (b) 0.20\,\mathrm{mm}. (c) 0.10\,\mathrm{mm}. (d) m_e/m_n = 5.4\times10^{-4}.

Question A5

Type C2 · Compton recoil energy + C1, P1[9 marks]

An X-ray photon of wavelength 18.0\,\mathrm{pm} is scattered straight back (\theta = 180^\circ) by a free electron at rest.

(a) Calculate the momentum of the incident photon. [1]

(b) Calculate the wavelength of the scattered photon. [1]

(d) Calculate the kinetic energy given to the electron. [2]

(e) Outline Compton scattering, and explain why it supports the photon model of light more strongly than the photoelectric effect does. [2]

Show solutionHide solution

(a) p_i = h/\lambda_i = 6.63\times10^{-34}/18.0\times10^{-12} = 3.68\times10^{-23}\,\mathrm{kg\,m\,s^{-1}} (A1). (b) \lambda_f = 18.0+2\times2.43 = 22.9\,\mathrm{pm} (A1). (c) p_f = h/\lambda_f = 2.90\times10^{-23}\,\mathrm{kg\,m\,s^{-1}}, directed backwards (M1); momentum forwards before = after: p_i = -p_f+p_e (M1), so p_e = 3.68\times10^{-23}+2.90\times10^{-23} = 6.6\times10^{-23}\,\mathrm{kg\,m\,s^{-1}} (A1). (d) E_k = hc/\lambda_i-hc/\lambda_f (M1) = 1.105\times10^{-14}-8.70\times10^{-15} = 2.3\times10^{-15}\,\mathrm{J} (A1). (e) X-rays scattered by nearly free electrons include a longer wavelength that grows with the scattering angle (A1). The size of the shift follows only if each photon carries momentum h/\lambda and collides with one electron like a particle; the photoelectric effect uses only the photon's energy, and a wave model predicts no change of wavelength (A1).

Final answer

(a) 3.68\times10^{-23}\,\mathrm{kg\,m\,s^{-1}}. (b) 22.9\,\mathrm{pm}. (c) 6.6\times10^{-23}\,\mathrm{kg\,m\,s^{-1}}. (d) 2.3\times10^{-15}\,\mathrm{J}. (e) Photons scattered by free electrons come out with a longer wavelength; the shift needs photon momentum as well as energy.

Connected questions

Question X1

Type F3 · maximum electron speed + F2 · Connected · E.2 + D.3[8 marks]

Light of wavelength 380\,\mathrm{nm} falls on a metal of work function 2.25\,\mathrm{eV} in a vacuum. The fastest photoelectrons pass through a narrow slit into a uniform magnetic field of flux density 2.0\times10^{-4}\,\mathrm{T}, at right angles to the field.

(a) Find E_\text{max} for these electrons, in eV. [2]

(b) Show that their speed is about 6.0\times10^{5}\,\mathrm{m\,s^{-1}}. [2]

(d) Explain why the field does not change the speed of the electrons. [1]

(e) What happens to the radius if the light is made twice as intense? [1]

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(a) E.2: hc/\lambda = 1.989\times10^{-25}/380\times10^{-9} = 5.23\times10^{-19}\,\mathrm{J} = 3.27\,\mathrm{eV} (M1); E_\text{max} = 3.27-2.25 = 1.02\,\mathrm{eV} (A1). (b) 1.02\,\mathrm{eV} = 1.63\times10^{-19}\,\mathrm{J} (M1); v = \sqrt{2E_\text{max}/m_e} = 5.99\times10^{5}\,\mathrm{m\,s^{-1}} (A1). (c) D.3: evB = m_ev^2/r, r = m_ev/eB (M1) = 9.11\times10^{-31}\times5.99\times10^{5}/(1.60\times10^{-19}\times2.0\times10^{-4}) = 1.7\times10^{-2}\,\mathrm{m} (A1). (d) The magnetic force stays perpendicular to the motion, so it transfers no energy and the speed cannot change (A1). (e) No change: more electrons per second, but the fastest still have the same speed (A1).

Final answer

(a) 1.02\,\mathrm{eV}. (b) 5.99\times10^{5}\,\mathrm{m\,s^{-1}}. (c) 1.7\,\mathrm{cm}. (d) Velocity and magnetic force remain at right angles; the field transfers no energy. (e) No change.

Question X2

Type F1 · threshold values + F2, P2 · Connected · E.2 + E.1[8 marks]

The energy levels of a hydrogen atom are E_n = -13.6/n^2\,\mathrm{eV}. Light from hydrogen falls on a metal of work function 2.10\,\mathrm{eV}.

(a) An atom drops from level n = 3 to level n = 2. Find the energy, in eV, and the wavelength of the photon it gives out. [3]

(b) State, with a reason, whether this light releases electrons from the metal. [1]

(d) What reverse p.d. stops every electron in (c)? [1]

(e) Both the hydrogen spectrum and the photoelectric effect need the photon idea. Outline why. [1]

Show solutionHide solution

(a) E.1: \Delta E = 13.6(1/4-1/9) = 1.89\,\mathrm{eV} (A1); \lambda = hc/\Delta E = 1.989\times10^{-25}/(1.89\times1.60\times10^{-19}) (M1) = 6.58\times10^{-7}\,\mathrm{m} (A1). (b) No: each photon has 1.89\,\mathrm{eV}, less than the 2.10\,\mathrm{eV} work function (A1). (c) \Delta E = 13.6(1/4-1/16) = 2.55\,\mathrm{eV} (M1); E_\text{max} = 2.55-2.10 = 0.45\,\mathrm{eV} (A1). (d) 0.45\,\mathrm{V} (A1). (e) An atom emits, and an electron absorbs, energy in single photons of energy hf, so only photons of the right energy can cause each event (A1).

Final answer

(a) 1.89\,\mathrm{eV}; 658\,\mathrm{nm}. (b) No: 1.89\,\mathrm{eV} < 2.10\,\mathrm{eV}. (c) 0.45\,\mathrm{eV}. (d) 0.45\,\mathrm{V}. (e) In both, energy is exchanged in single quanta hf with one atom or one electron.

Question X3

Type C3 · Compton shift and angle + C2, P1 · Connected · E.2 + A.2[9 marks]

An X-ray photon of wavelength 20.0\,\mathrm{pm} travelling along +x is scattered by a free electron at rest. The photon leaves along +y, at 90^\circ to its original direction.

(a) Calculate the wavelength of the scattered photon. [2]

(b) Calculate the momentum of the photon before and after the collision. [2]

(d) Give one way in which this collision is like a collision between two snooker balls and one way in which it is different. [2]

Show solutionHide solution

(a) \Delta\lambda = 2.43\times10^{-12}(1-\cos90^\circ) = 2.43\,\mathrm{pm} (M1); \lambda_f = 22.4\,\mathrm{pm} (A1). (b) p_i = h/\lambda_i = 3.32\times10^{-23}\,\mathrm{kg\,m\,s^{-1}} (A1); p_f = h/\lambda_f = 2.96\times10^{-23}\,\mathrm{kg\,m\,s^{-1}} (A1). (c) A.2: p_{ex} = 3.32\times10^{-23} and p_{ey} = -2.96\times10^{-23} (M1); p_e = \sqrt{3.32^2+2.96^2}\times10^{-23} = 4.44\times10^{-23}\,\mathrm{kg\,m\,s^{-1}} (A1); \tan\phi = 2.96/3.32, \phi = 41.7^\circ below the +x direction (A1). (d) Like snooker balls, total momentum and total kinetic energy are conserved and the target moves off on the other side of the line of approach (A1). Unlike a ball, the photon does not slow down: it always travels at c and loses energy through a drop in frequency (A1).

Final answer

(a) 22.4\,\mathrm{pm}. (b) 3.32\times10^{-23} and 2.96\times10^{-23}\,\mathrm{kg\,m\,s^{-1}}. (c) 4.44\times10^{-23}\,\mathrm{kg\,m\,s^{-1}} at 41.7^\circ below +x. (d) Like: momentum and energy are conserved. Unlike: the photon keeps speed c and loses energy through its frequency.

Question X4

Type M2 · wavelength from accelerating p.d. + M6, P1 · Connected · E.2 + C.3 + D.2[7 marks]

X-rays of wavelength 0.154\,\mathrm{nm} are diffracted by a crystal. An electron beam is to give the same diffraction pattern from the same crystal.

(a) State why the electrons must have the same momentum as the X-ray photons, and calculate it. [2]

(b) Calculate the potential difference through which electrons must be accelerated from rest. [3]

(d) Comment on the two energies. [1]

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(a) The diffraction angles depend only on wavelength, and \lambda = h/p for both (A1); p = 6.63\times10^{-34}/1.54\times10^{-10} = 4.31\times10^{-24}\,\mathrm{kg\,m\,s^{-1}} (A1). (b) E_k = p^2/2m_e = (4.31\times10^{-24})^2/(2\times9.11\times10^{-31}) (M1) = 1.02\times10^{-17}\,\mathrm{J} (A1); D.2: V = E_k/e = 63.6\,\mathrm{V} (A1). (c) E = pc = hc/\lambda = 1.29\times10^{-15}\,\mathrm{J} = 8.07\,\mathrm{keV} (A1). (d) For the same wavelength the electron carries about 127 times less energy than the photon, so electron beams probe crystals with far less energy per particle (A1).

Final answer

(a) Same pattern needs the same wavelength, and \lambda = h/p; 4.31\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}. (b) 63.6\,\mathrm{V}. (c) 8.07\,\mathrm{keV}. (d) The electron needs about 127 times less energy.

From the book

Photon Physics: Study Guide for the IB Diploma (SL & HL)

Quantum physics from the core ideas to full exam questions, with question types, worked examples and graded practice.

About the book