A Level Physics 22.1 Energy and Momentum of a Photon MCQs – 30 Practice Questions with Answers (Cambridge 9702)

This page covers Cambridge 9702 A Level Section 22.1, Energy and Momentum of a Photon. It is an A Level (A2) topic: you will practise photon energy in joules and electronvolts, photon momentum, and the photon-rate calculations that 9702 examiners love to set.

Formula recap: E = h f  |  E = h c / λ  |  p = E / c = h / λ  |  number of photons per second = P / E  |  h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, 1 eV = 1.6 × 10−19 J. Full notes: A Level Physics Formula Sheet.

Group 1: Photon Energy

Q1. Light transfers energy to a surface in discrete packets, each carrying energy E = h f. Each packet is called a

  • (A) phonon
  • (B) photon
  • (C) proton
  • (D) neutrino
Show Answer

Answer: B – A photon is the quantum (discrete packet) of electromagnetic radiation. Its energy is E = h f, where h is the Planck constant and f is the frequency of the radiation.

Q2. Green light has frequency 5.7 × 1014 Hz. What is the energy of one photon of this light?

  • (A) 3.8 × 10−20 J
  • (B) 1.2 × 10−28 J
  • (C) 3.8 × 10−19 J
  • (D) 8.6 × 10−19 J
Show Answer

Answer: C – E = h f = 6.63 × 10−34 × 5.7 × 1014 = 3.78 × 10−19 J ≈ 3.8 × 10−19 J. In electronvolts this is 3.78 × 10−19 / 1.6 × 10−19 ≈ 2.4 eV.

Q3. Which type of electromagnetic radiation consists of the most energetic photons?

  • (A) gamma rays
  • (B) X-rays
  • (C) ultraviolet
  • (D) microwaves
Show Answer

Answer: A – Photon energy E = h f, so the radiation with the highest frequency has the most energetic photons. Gamma rays have the highest frequencies (and shortest wavelengths) in the electromagnetic spectrum.

Q4. A photon of red light has wavelength 700 nm. What is its energy in electronvolts?

  • (A) 0.18 eV
  • (B) 0.89 eV
  • (C) 3.6 eV
  • (D) 1.8 eV
Show Answer

Answer: D – E = h c / λ = (6.63 × 10−34 × 3.00 × 108) / (700 × 10−9) = 2.84 × 10−19 J. Converting: 2.84 × 10−19 / 1.6 × 10−19 = 1.78 eV ≈ 1.8 eV.

Q5. The frequency of a beam of light is doubled. The energy of each photon

  • (A) stays the same
  • (B) doubles
  • (C) halves
  • (D) becomes four times larger
Show Answer

Answer: B – E = h f, so photon energy is directly proportional to frequency. Doubling f doubles E. (Equivalently, doubling f halves λ, and E = h c / λ doubles.)

Q6. Ultraviolet radiation has wavelength 250 nm. What is the energy of one photon?

  • (A) 8.0 × 10−19 J
  • (B) 8.0 × 10−20 J
  • (C) 2.7 × 10−19 J
  • (D) 5.0 × 10−19 J
Show Answer

Answer: A – E = h c / λ = (6.63 × 10−34 × 3.00 × 108) / (250 × 10−9) = 7.96 × 10−19 J ≈ 8.0 × 10−19 J, which is about 5.0 eV. Note the distractor in (D): 5.0 is the value in eV, not in joules.

Q7. A photon has energy 3.3 × 10−19 J. What is the frequency of the radiation?

  • (A) 2.0 × 1014 Hz
  • (B) 5.0 × 1013 Hz
  • (C) 2.0 × 1015 Hz
  • (D) 5.0 × 1014 Hz
Show Answer

Answer: D – f = E / h = 3.3 × 10−19 / 6.63 × 10−34 = 4.98 × 1014 Hz ≈ 5.0 × 1014 Hz, which is in the visible range.

Q8. What is the wavelength of a photon whose energy is 2.0 eV?

  • (A) 310 nm
  • (B) 1240 nm
  • (C) 620 nm
  • (D) 155 nm
Show Answer

Answer: C – First convert: E = 2.0 × 1.6 × 10−19 = 3.2 × 10−19 J. Then λ = h c / E = 1.989 × 10−25 / 3.2 × 10−19 = 6.2 × 10−7 m = 620 nm, in the red-orange part of the spectrum.

Q9. Blue light has wavelength 450 nm and red light has wavelength 700 nm. The energy of one blue photon is

  • (A) 2.8 × 10−19 J
  • (B) 4.4 × 10−19 J
  • (C) 6.9 × 10−19 J
  • (D) 1.8 × 10−19 J
Show Answer

Answer: B – E = h c / λ = 1.989 × 10−25 / (450 × 10−9) = 4.42 × 10−19 J ≈ 4.4 × 10−19 J. The shorter wavelength gives the larger energy: the blue photon carries 700 / 450 ≈ 1.6 times the energy of the red photon (2.8 × 10−19 J, option A).

Q10. A lamp is made brighter (its intensity is increased) without changing the frequency of the light. The energy of each photon

  • (A) stays the same, but more photons are emitted per second
  • (B) increases
  • (C) decreases
  • (D) becomes zero between the photons
Show Answer

Answer: A – Photon energy depends only on frequency (E = h f). Intensity = (photons per second) × h f, so greater intensity at the same frequency means a greater photon rate, not more energetic photons. This is a key idea behind the photoelectric effect in Section 22.2.

Group 2: Photon Momentum

Q11. The momentum p of a photon of wavelength λ is given by

  • (A) p = h f
  • (B) p = E c
  • (C) p = h / λ
  • (D) p = m c
Show Answer

Answer: C – A photon has momentum p = E / c. Substituting E = h f and f = c / λ gives p = h / λ. Option (B) has the wrong dependence on c, and (D) does not apply because a photon has no rest mass.

Q12. What is the momentum of a photon of light of wavelength 500 nm?

  • (A) 1.3 × 10−27 kg m s−1
  • (B) 1.3 × 10−26 kg m s−1
  • (C) 6.6 × 10−27 kg m s−1
  • (D) 3.3 × 10−34 kg m s−1
Show Answer

Answer: A – p = h / λ = 6.63 × 10−34 / (500 × 10−9) = 1.33 × 10−27 kg m s−1 ≈ 1.3 × 10−27 kg m s−1. Tiny, but not zero.

Q13. A gamma-ray photon has energy 1.0 MeV. What is its momentum?

  • (A) 5.3 × 10−28 kg m s−1
  • (B) 5.3 × 10−19 kg m s−1
  • (C) 1.6 × 10−21 kg m s−1
  • (D) 5.3 × 10−22 kg m s−1
Show Answer

Answer: D – E = 1.0 × 106 × 1.6 × 10−19 = 1.6 × 10−13 J. Then p = E / c = 1.6 × 10−13 / 3.00 × 108 = 5.3 × 10−22 kg m s−1.

Q14. The wavelength of a photon is halved. Its momentum

  • (A) halves
  • (B) doubles
  • (C) stays the same
  • (D) becomes four times larger
Show Answer

Answer: B – p = h / λ, so momentum is inversely proportional to wavelength. Halving λ doubles p (the frequency and the energy also double).

Q15. Which statement about photons is correct?

  • (A) A photon has a small rest mass.
  • (B) A photon has no momentum because it has no mass.
  • (C) A photon has momentum p = E / c even though its rest mass is zero.
  • (D) A photon only has momentum when it is travelling in a vacuum.
Show Answer

Answer: C – A photon has zero rest mass, but it carries energy E = h f, and anything carrying energy E at speed c has momentum p = E / c = h / λ. Light pressure on mirrors and comet tails curving away from the Sun are direct evidence of photon momentum.

Q16. A photon of wavelength 600 nm strikes a mirror at normal incidence and is reflected back along its own path. What is the magnitude of the momentum transferred to the mirror?

  • (A) zero
  • (B) 1.1 × 10−27 kg m s−1
  • (C) 3.3 × 10−27 kg m s−1
  • (D) 2.2 × 10−27 kg m s−1
Show Answer

Answer: D – The photon momentum is p = h / λ = 6.63 × 10−34 / (600 × 10−9) = 1.1 × 10−27 kg m s−1. Absorption would transfer just p, but reflection reverses the photon’s momentum (from +p to −p), so the mirror gains 2p = 2.2 × 10−27 kg m s−1. This is why reflection gives twice the radiation pressure of absorption.

Q17. A short pulse of light with total energy 3.0 J is completely absorbed by a small target. What total momentum does the target gain?

  • (A) 1.0 × 10−8 kg m s−1
  • (B) 3.0 × 10−8 kg m s−1
  • (C) 9.0 × 108 kg m s−1
  • (D) 1.0 × 10−9 kg m s−1
Show Answer

Answer: A – For the whole pulse, total momentum = total energy / c = 3.0 / 3.00 × 108 = 1.0 × 10−8 kg m s−1. Summing p = E / c over all the photons gives the same result as summing h / λ photon by photon.

Q18. A beam of light of power 100 W is reflected normally by a perfectly reflecting mirror. What force does the light exert on the mirror?

  • (A) 3.3 × 10−7 N
  • (B) 6.7 × 10−7 N
  • (C) 6.7 × 10−9 N
  • (D) 3.3 × 10−5 N
Show Answer

Answer: B – Each second the beam delivers momentum P / c, and reflection doubles the transfer, so F = 2P / c = 2 × 100 / 3.00 × 108 = 6.7 × 10−7 N. (Absorption would give half of this, option A.) This tiny force is called radiation pressure.

Q19. An X-ray photon has wavelength 0.10 nm. What is its momentum?

  • (A) 6.6 × 10−34 kg m s−1
  • (B) 6.6 × 10−25 kg m s−1
  • (C) 6.6 × 10−24 kg m s−1
  • (D) 2.0 × 10−24 kg m s−1
Show Answer

Answer: C – p = h / λ = 6.63 × 10−34 / (0.10 × 10−9) = 6.63 × 10−24 kg m s−1 ≈ 6.6 × 10−24 kg m s−1, about 5000 times the momentum of the visible photon in Q12.

Q20. A radio photon has wavelength 3.0 m and a visible-light photon has wavelength 500 nm. The ratio (momentum of visible photon) / (momentum of radio photon) is

  • (A) 1.7 × 10−7
  • (B) 6.0 × 105
  • (C) 1.5 × 102
  • (D) 6.0 × 106
Show Answer

Answer: D – Since p = h / λ, the ratio of momenta is the inverse ratio of wavelengths: pvisible / pradio = λradio / λvisible = 3.0 / (500 × 10−9) = 6.0 × 106. Short-wavelength radiation carries far more momentum per photon.


Group 3: Photon Calculations

Q21. A laser emits light of wavelength 633 nm with output power 5.0 mW. How many photons does it emit per second?

  • (A) 1.6 × 1015
  • (B) 1.6 × 1016
  • (C) 3.2 × 1016
  • (D) 1.6 × 1018
Show Answer

Answer: B – Energy per photon: E = h c / λ = 1.989 × 10−25 / (633 × 10−9) = 3.14 × 10−19 J. Photon rate = P / E = 5.0 × 10−3 / 3.14 × 10−19 = 1.6 × 1016 photons per second.

Q22. A sodium lamp radiates 10 W of light of wavelength 589 nm. How many photons are emitted per second?

  • (A) 3.0 × 1019
  • (B) 3.0 × 1018
  • (C) 3.4 × 1019
  • (D) 1.7 × 1019
Show Answer

Answer: A – E = h c / λ = 1.989 × 10−25 / (589 × 10−9) = 3.38 × 10−19 J. Rate = P / E = 10 / 3.38 × 10−19 = 2.96 × 1019 ≈ 3.0 × 1019 photons per second.

Q23. Light of wavelength 500 nm falls on a surface with intensity 1.0 × 103 W m−2. How many photons strike each square metre of the surface per second?

  • (A) 2.5 × 1018
  • (B) 4.0 × 1020
  • (C) 2.5 × 1020
  • (D) 2.5 × 1021
Show Answer

Answer: D – E = h c / λ = 1.989 × 10−25 / (500 × 10−9) = 3.98 × 10−19 J. Photon flux = intensity / E = 1.0 × 103 / 3.98 × 10−19 = 2.5 × 1021 photons per second per square metre.

Q24. An LED produces a light output of 2.0 mW at wavelength 530 nm. The number of photons emitted per second is

  • (A) 5.3 × 1014
  • (B) 5.3 × 1016
  • (C) 5.3 × 1015
  • (D) 9.4 × 1015
Show Answer

Answer: C – E = h c / λ = 1.989 × 10−25 / (530 × 10−9) = 3.75 × 10−19 J (about 2.3 eV). Rate = 2.0 × 10−3 / 3.75 × 10−19 = 5.3 × 1015 photons per second.

Q25. An X-ray photon has wavelength 25 pm (25 × 10−12 m). What is its energy?

  • (A) 8.0 × 10−15 J
  • (B) 8.0 × 10−17 J
  • (C) 1.6 × 10−14 J
  • (D) 5.0 × 10−15 J
Show Answer

Answer: A – E = h c / λ = 1.989 × 10−25 / (25 × 10−12) = 7.96 × 10−15 J ≈ 8.0 × 10−15 J. In electronvolts: 7.96 × 10−15 / 1.6 × 10−19 ≈ 5.0 × 104 eV, i.e. about 50 keV – typical of a medical X-ray photon.

Q26. An LED takes in electrical power 0.20 W and is 25% efficient at converting it into light of wavelength 470 nm. How many photons does it emit per second?

  • (A) 1.2 × 1016
  • (B) 1.2 × 1017
  • (C) 4.7 × 1017
  • (D) 2.4 × 1016
Show Answer

Answer: B – Light power = 0.25 × 0.20 = 0.050 W. Photon energy: E = h c / λ = 1.989 × 10−25 / (470 × 10−9) = 4.23 × 10−19 J. Rate = 0.050 / 4.23 × 10−19 = 1.2 × 1017 photons per second. (Forgetting the efficiency gives option C.)

Q27. A laser pulse has total energy 0.30 J and wavelength 1064 nm. How many photons are in the pulse?

  • (A) 1.6 × 1017
  • (B) 1.9 × 1018
  • (C) 3.2 × 1017
  • (D) 1.6 × 1018
Show Answer

Answer: D – E per photon = h c / λ = 1.989 × 10−25 / (1064 × 10−9) = 1.87 × 10−19 J. Number of photons = 0.30 / 1.87 × 10−19 = 1.6 × 1018.

Q28. A microwave oven uses radiation of frequency 2.45 GHz. The energy of one photon is

  • (A) 1.6 × 10−26 J
  • (B) 1.0 × 10−23 J
  • (C) 1.6 × 10−24 J
  • (D) 2.5 × 10−24 J
Show Answer

Answer: C – E = h f = 6.63 × 10−34 × 2.45 × 109 = 1.62 × 10−24 J ≈ 1.6 × 10−24 J, only about 10−5 eV. Microwave photons are far too weak to ionise atoms; the oven heats food because it supplies an enormous number of them per second.

Q29. A radio transmitter radiates power 100 kW at frequency 100 MHz. How many photons does it emit per second?

  • (A) 1.5 × 1028
  • (B) 1.5 × 1030
  • (C) 6.6 × 1029
  • (D) 1.5 × 1032
Show Answer

Answer: B – E = h f = 6.63 × 10−34 × 1.00 × 108 = 6.63 × 10−26 J per photon. Rate = P / E = 1.00 × 105 / 6.63 × 10−26 = 1.5 × 1030 photons per second. Radio photons are individually feeble, so useful power needs astronomical photon rates.

Q30. A source emits 4.0 × 1018 photons per second, each of wavelength 400 nm. What is the power of the emitted light?

  • (A) 2.0 W
  • (B) 0.50 W
  • (C) 20 W
  • (D) 8.0 W
Show Answer

Answer: A – E per photon = h c / λ = 1.989 × 10−25 / (400 × 10−9) = 4.97 × 10−19 J. Power = rate × E = 4.0 × 1018 × 4.97 × 10−19 = 1.99 W ≈ 2.0 W.

Keep practising: return to the Physics MCQs hub for more topic pages, and keep the A Level Physics Formula Sheet open while you revise – every equation used on this page is collected there.

These are original practice questions written for revision. They are not Cambridge past-paper questions, and this page is not affiliated with Cambridge Assessment International Education.