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A2 · Practice questions · Energy and momentum of a photon

Count the light in photons.

Six original Cambridge-style questions on the photon as a quantum of energy, E = hf = hc/λ, the electronvolt, photon momentum p = h/λ, and the number of photons a beam delivers each second.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Frequency fixes the energy, wavelength fixes the momentum.

01
Calculation
[3 marks]

The infrared LED in a television remote control emits light of wavelength 940 nm. Calculate the frequency of this light and the energy of one photon, in joules and in electronvolts.

  • f = c / λ = 3.00 × 10⁸ / (940 × 10⁻⁹) = 3.19 × 10¹⁴ Hz. ✓
  • E = hf = 6.63 × 10⁻³⁴ × 3.19 × 10¹⁴ = 2.12 × 10⁻¹⁹ J. ✓
  • E = 2.12 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 1.32 eV. ✓
02
Analysis
[2 marks]

State what is meant by a photon. A monochromatic beam of light is made twice as intense while its wavelength is unchanged. State what changes about the photons in the beam and what does not.

  • A photon is a quantum (discrete packet) of electromagnetic energy, of energy E = hf. ✓
  • Twice as many photons arrive per second, but each photon still carries the same energy hf, because that is fixed by the frequency and not by the intensity. ✓
03
Calculation
[3 marks]

An X-ray photon has an energy of 25 keV. Calculate its energy in joules, its wavelength and its momentum.

  • E = 25 × 10³ × 1.60 × 10⁻¹⁹ = 4.0 × 10⁻¹⁵ J. ✓
  • λ = hc / E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (4.0 × 10⁻¹⁵) = 5.0 × 10⁻¹¹ m. ✓
  • p = h / λ = 6.63 × 10⁻³⁴ / (4.97 × 10⁻¹¹) = 1.3 × 10⁻²³ N s (or p = E / c = 4.0 × 10⁻¹⁵ / 3.00 × 10⁸). ✓
04
Analysis
[2 marks]

A photon has no mass. Explain why it nevertheless has momentum, and state how the momentum of a photon changes when its wavelength is halved.

  • Momentum for a photon is not mv but p = E / c = h / λ: the photon carries energy, and that energy moving at c has a momentum, which is why radiation falling on a surface exerts a force on it. ✓
  • Since p is inversely proportional to λ, halving the wavelength doubles the momentum. ✓
05
Calculation
[3 marks]

A radio transmitter broadcasts at a frequency of 98.0 MHz with an output power of 10 kW. Calculate the energy of one photon and the number of photons emitted per second, and comment on why the radio signal shows no sign of being made of discrete quanta.

  • E = hf = 6.63 × 10⁻³⁴ × 98.0 × 10⁶ = 6.50 × 10⁻²⁶ J. ✓
  • Photons per second = P / E = 1.0 × 10⁴ / 6.50 × 10⁻²⁶ = 1.5 × 10²⁹ s⁻¹. ✓
  • With so enormous a number of photons each carrying so little energy, the energy arrives in steps far too small and too frequent to detect, so the signal behaves as a continuous wave. ✓
06
Calculation
[4 marks]

A green laser pointer emits light of wavelength 532 nm at a power of 5.0 mW. The beam falls on a surface that absorbs it completely.

(a) Calculate the energy of one photon. (b) Calculate the number of photons striking the surface each second. (c) Calculate the momentum of one photon and hence the force the beam exerts on the surface.

  • (a) E = hc / λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (532 × 10⁻⁹) = 3.74 × 10⁻¹⁹ J. ✓
  • (b) N = P / E = 5.0 × 10⁻³ / 3.74 × 10⁻¹⁹ = 1.3 × 10¹⁶ photons per second. ✓
  • (c) p = h / λ = 6.63 × 10⁻³⁴ / (532 × 10⁻⁹) = 1.25 × 10⁻²⁷ N s per photon. ✓
  • Force = rate of change of momentum = Np = 1.34 × 10¹⁶ × 1.25 × 10⁻²⁷ = 1.7 × 10⁻¹¹ N, which equals P / c = 5.0 × 10⁻³ / 3.00 × 10⁸. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.