Module 4 · Electrons, Waves and Photons

Photons

Specification: OCR A H556 | Section: 4.5.1 — The Photon Model of Electromagnetic Radiation | Teaching time: 3–4 hours

By the end of this topic you should be able to…

Define a photon as a quantum of electromagnetic energy
Use E = hf and E = hc / λ to calculate photon energy
Convert between joules (J) and electronvolts (eV)
Calculate the number of photons emitted per second by a light source
Describe how Planck's constant h can be determined using LEDs

Part 1

What Is a Photon?

Classical physics treats electromagnetic radiation as a continuous wave. However, experiments such as the photoelectric effect show that EM radiation also behaves as if it comes in discrete packets of energy. Each packet is called a photon.

A photon is a quantum of electromagnetic energy. It has no mass and no charge. It always travels at the speed of light in a vacuum (c = 3.00 × 10⁸ m s⁻¹). The energy of a single photon depends only on its frequency.

Photon energy equation E = hf = hc / λ

Where:

E = photon energy in joules (J)
h = Planck's constant = 6.63 × 10⁻³⁴ J s
f = frequency of the radiation (Hz)
c = speed of light = 3.00 × 10⁸ m s⁻¹
λ = wavelength (m)

⚡ Key Point

Higher frequency means higher energy per photon. Ultraviolet photons each carry more energy than infrared photons — this is why UV can cause sunburn but IR cannot, even if the total power is the same.

The relationship between photon energy and position on the electromagnetic spectrum is summarised below:

γ-rays

X-rays

Visible

Microwave

Radio

← Higher frequency, higher photon energy Lower frequency, lower photon energy →

Part 2

The Electronvolt

Photon energies are extremely small in joules (often ~10⁻¹⁹ J). The electronvolt (eV) is a more convenient unit.

Definition of the electronvolt 1 eV = 1.60 × 10⁻¹⁹ J

One electronvolt is the energy gained by an electron when it is accelerated through a potential difference of 1 volt.

To convert joules → eV: divide by 1.60 × 10⁻¹⁹.
To convert eV → joules: multiply by 1.60 × 10⁻¹⁹.

Worked Example 1 — Converting energy units

A photon has an energy of 4.0 × 10⁻¹⁹ J. Convert this to electronvolts.

E = 4.0 × 10⁻¹⁹ ÷ 1.60 × 10⁻¹⁹ = 2.5 eV

Photon energy = 2.5 eV

Worked Example 2 — Converting eV to joules

An X-ray photon has energy 12.4 keV. Convert to joules.

12.4 keV = 12 400 eV = 12 400 × 1.60 × 10⁻¹⁹ J

E = 1.98 × 10⁻¹⁵ J

Photon energy = 1.98 × 10⁻¹⁵ J

📝 Exam Tip

In the exam, the value e = 1.60 × 10⁻¹⁹ C is given in the data booklet. Remember: 1 eV = e × 1 V = 1.60 × 10⁻¹⁹ J. Always show your conversion step to get the method mark.

Part 3

Photon Energy Calculations

The two key equations are E = hf and E = hc/λ. Use whichever one matches the information given in the question.

Worked Example 3 — Energy from wavelength

A He-Ne laser emits light of wavelength 633 nm. Calculate the energy of one photon in joules and in eV.

λ = 633 nm = 6.33 × 10⁻⁷ m

E = hc / λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (6.33 × 10⁻⁷)

E = 1.989 × 10⁻²⁵ / 6.33 × 10⁻⁷ = 3.14 × 10⁻¹⁹ J

In eV: 3.14 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 1.96 eV

E = 3.14 × 10⁻¹⁹ J ≈ 1.96 eV

Worked Example 4 — Frequency from energy

A UV photon has energy 5.0 eV. Calculate its frequency.

Convert to joules: E = 5.0 × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹⁹ J

Rearrange: f = E / h = 8.0 × 10⁻¹⁹ / 6.63 × 10⁻³⁴

f = 1.21 × 10¹⁵ Hz

Frequency = 1.21 × 10¹⁵ Hz

Worked Example 5 — Wavelength from photon energy

A photon has energy 2.0 eV. Calculate its wavelength.

E = 2.0 × 1.60 × 10⁻¹⁹ = 3.2 × 10⁻¹⁹ J

λ = hc / E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (3.2 × 10⁻¹⁹)

λ = 1.989 × 10⁻²⁵ / 3.2 × 10⁻¹⁹ = 6.2 × 10⁻⁷ m = 620 nm

Wavelength = 620 nm (orange-red light)

Part 4

Number of Photons from a Light Source

If you know the power of a light source and the energy of each photon, you can calculate how many photons are emitted per second.

Number of photons per second (photon flux) N = P / E_photon = Pλ / (hc)

Where:

N = number of photons emitted per second (s⁻¹)
P = power of the source (W = J s⁻¹)
E_photon = energy of one photon (J)

⚡ Key Point

Power is energy per second. Each second, the source emits N photons, each carrying energy E_photon. So total power P = N × E_photon, and therefore N = P / E_photon.

Worked Example 6 — Photons per second from a laser

A helium-neon laser emits light at 633 nm with a power of 5.0 mW. Calculate the number of photons emitted per second.

P = 5.0 mW = 5.0 × 10⁻³ W

Energy of one photon: E = hc / λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 6.33 × 10⁻⁷ = 3.14 × 10⁻¹⁹ J

Number of photons per second: N = P / E = 5.0 × 10⁻³ / 3.14 × 10⁻¹⁹

N = 1.59 × 10¹⁶ s⁻¹

The laser emits approximately 1.6 × 10¹⁶ photons per second

Worked Example 7 — Power from photon count

A green laser pointer emits light at 532 nm. It produces 3.0 × 10¹⁵ photons per second. Calculate its power output.

E_photon = hc / λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 5.32 × 10⁻⁷ = 3.74 × 10⁻¹⁹ J

P = N × E_photon = 3.0 × 10¹⁵ × 3.74 × 10⁻¹⁹ = 1.12 × 10⁻³ W

Power = 1.1 mW

Worked Example 8 — Comparing lasers

Laser A emits at 400 nm with power 2.0 mW. Laser B emits at 700 nm with the same power. Which emits more photons per second, and by what factor?

N = Pλ / (hc). Since P and hc are the same for both, N is proportional to λ.

Ratio: N_B / N_A = λ_B / λ_A = 700 / 400 = 1.75

Laser B (700 nm) emits 1.75× more photons per second than Laser A. Longer wavelength ⇒ lower energy per photon ⇒ more photons needed for the same power.

⚠ Common Mistake

Students often forget to convert mW to W, or nm to m, before calculating. Always convert to SI units first: mW → 10⁻³ W, nm → 10⁻⁹ m, keV → 10³ eV → × 1.60 × 10⁻¹⁹ J.

Part 5

Determining Planck's Constant Using LEDs

Light-emitting diodes (LEDs) emit photons when electrons cross the band gap of the semiconductor. Each photon has energy equal to the work done by a single electron crossing the potential difference:

LED threshold condition hf = eV₀ → h = eV₀λ / c

Where V₀ is the threshold voltage (the minimum p.d. at which the LED just begins to emit light).

Method:

Use several LEDs of different known wavelengths (colours).
For each LED, increase the p.d. slowly until light is just visible. Record this threshold voltage V₀.
Plot a graph of V₀ (y-axis) against 1/λ (x-axis).
The gradient = hc / e. Since c and e are known, calculate h.

📝 Exam Tip

This experiment is a favourite for 6-mark practical questions. You must be able to describe the method, identify the graph axes, and explain how the gradient gives h. A straight line through the origin confirms the relationship eV₀ = hc / λ.

Part 6

More Worked Examples — Lasers and Photon Flux

Worked Example 9 — Total energy from a pulsed laser

A pulsed laser emits 2.0 × 10¹⁸ photons of wavelength 266 nm in a single pulse lasting 10 ns. Calculate the energy of one photon and the peak power of the pulse.

E_photon = hc / λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 2.66 × 10⁻⁷ = 7.48 × 10⁻¹⁹ J

In eV: 7.48 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 4.67 eV

Total energy of pulse: E_total = 2.0 × 10¹⁸ × 7.48 × 10⁻¹⁹ = 1.50 J (wait — let's recheck)

E_total = 2.0 × 10¹⁸ × 7.48 × 10⁻¹⁹ = 1.496 × 10⁰ J ≈ 1.5 J

Peak power: P = E_total / t = 1.5 / (10 × 10⁻⁹) = 1.5 × 10⁸ W = 150 MW

Photon energy = 7.48 × 10⁻¹⁹ J (4.67 eV); Peak power ≈ 150 MW

Worked Example 10 — Photon flux through an area

A 3.0 mW laser emits light at 450 nm. The beam has a cross-sectional area of 2.0 mm². Calculate the number of photons passing through a cross-section per second per mm².

E_photon = hc / λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 4.50 × 10⁻⁷ = 4.42 × 10⁻¹⁹ J

Total photons/s: N = P / E = 3.0 × 10⁻³ / 4.42 × 10⁻¹⁹ = 6.79 × 10¹⁵ s⁻¹

Photons per mm²: 6.79 × 10¹⁵ / 2.0 = 3.4 × 10¹⁵ mm⁻² s⁻¹

Photon flux = 3.4 × 10¹⁵ photons per mm² per second

Worked Example 11 — Comparing photon output at different wavelengths

A 100 W incandescent bulb emits across all visible wavelengths. Estimate how many 550 nm (green) photons per second would be emitted if 5% of the total power were at this wavelength.

Power at 550 nm: P = 0.05 × 100 = 5.0 W

E_photon = hc / λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 5.50 × 10⁻⁷ = 3.62 × 10⁻¹⁹ J

N = P / E = 5.0 / 3.62 × 10⁻¹⁹ = 1.38 × 10¹⁹ s⁻¹

Approximately 1.4 × 10¹⁹ green photons per second

Check your understanding

Knowledge Check

State what is meant by a photon.

2 marks

A discrete packet / quantum of electromagnetic energy (1 mark)
Whose energy depends on the frequency of the radiation: E = hf (1 mark)

A photon has wavelength 400 nm. Calculate its energy in joules and in eV.

3 marks

E = hc / λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 4.00 × 10⁻⁷ = 4.97 × 10⁻¹⁹ J (2 marks)
In eV: 4.97 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 3.11 eV (1 mark)

Explain why a photon of violet light has more energy than a photon of red light.

2 marks

Violet light has a shorter wavelength / higher frequency than red light (1 mark)
E = hf = hc/λ, so higher f (or shorter λ) gives higher energy per photon (1 mark)

A 2.0 mW laser emits photons of wavelength 500 nm. Calculate the number of photons emitted per second.

3 marks

E_photon = hc / λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 5.00 × 10⁻⁷ = 3.98 × 10⁻¹⁹ J (1 mark)
P = 2.0 × 10⁻³ W (1 mark)
N = P / E = 2.0 × 10⁻³ / 3.98 × 10⁻¹⁹ = 5.03 × 10¹⁵ s⁻¹ (1 mark)

Describe how LEDs can be used to determine a value for the Planck constant h.

4 marks

Use LEDs of different known wavelengths (1 mark)
For each LED, find the threshold voltage V₀ at which light is just emitted (1 mark)
Plot V₀ against 1/λ; the graph should be a straight line through the origin (1 mark)
Gradient = hc/e; calculate h from h = gradient × e / c (1 mark)

Exam Practice

Exam-Style Questions

(a) Calculate the energy, in eV, of a photon of wavelength 250 nm.

(b) This photon is in the ultraviolet region of the EM spectrum. State whether a photon of wavelength 500 nm has more, less, or the same energy. Explain your answer.

4 marks

(a) E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 2.50 × 10⁻⁷ = 7.96 × 10⁻¹⁹ J (1 mark)
7.96 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 4.97 eV ≈ 5.0 eV (1 mark)
(b) Less energy (1 mark). Wavelength is doubled, and since E = hc/λ is inversely proportional to λ, the energy is halved to about 2.5 eV (1 mark).

A laser emits light of wavelength 532 nm. The laser has a power output of 50 mW.

(a) Calculate the energy of a single photon.

(b) Calculate the number of photons emitted by the laser in 1.0 s.

(c) The laser beam is incident on a surface of area 4.0 mm². Calculate the number of photons incident per second per unit area.

5 marks

(a) E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 5.32 × 10⁻⁷ = 3.74 × 10⁻¹⁹ J (1 mark)
(b) N = P/E = 5.0 × 10⁻² / 3.74 × 10⁻¹⁹ = 1.34 × 10¹⁷ s⁻¹ (2 marks)
(c) Area = 4.0 × 10⁻⁶ m². Photons per m² per s = 1.34 × 10¹⁷ / 4.0 × 10⁻⁶ = 3.3 × 10²² m⁻² s⁻¹ (2 marks)

The threshold voltage of a red LED (λ = 630 nm) is 1.95 V.

(a) Use this data to calculate a value for the Planck constant.

(b) Compare your answer with the accepted value of h = 6.63 × 10⁻³⁴ J s and suggest a reason for any difference.

5 marks

(a) eV₀ = hc/λ, so h = eV₀λ/c (1 mark)
h = (1.60 × 10⁻¹⁹ × 1.95 × 6.30 × 10⁻⁷) / 3.00 × 10⁸ (1 mark)
h = 1.966 × 10⁻²⁵ / 3.00 × 10⁸ = 6.55 × 10⁻³⁴ J s (1 mark)
(b) This is close to the accepted value of 6.63 × 10⁻³⁴ J s (about 1.2% difference) (1 mark). Difference could be due to difficulty judging the exact threshold voltage visually, or energy losses in the LED not producing photons (1 mark).

A student investigates photon energy using two lasers:

Laser X: λ = 488 nm, P = 15 mW
Laser Y: λ = 633 nm, P = 15 mW

(a) Show that the energy of a photon from Laser X is about 4.1 × 10⁻¹⁹ J.

(b) Without further calculation, state which laser emits more photons per second. Justify your answer.

5 marks

(a) E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 4.88 × 10⁻⁷ = 4.08 × 10⁻¹⁹ J ≈ 4.1 × 10⁻¹⁹ J ✓ (1 mark)
(b) Laser Y emits more photons per second (1 mark). Both have the same power, but Laser Y has longer wavelength so each photon has less energy (E = hc/λ), therefore more photons are needed per second to give the same power: N = P/E (1 mark).
(c) E_Y = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 6.33 × 10⁻⁷ = 3.14 × 10⁻¹⁹ J (1 mark)
N = P/E = 1.5 × 10⁻² / 3.14 × 10⁻¹⁹ = 4.78 × 10¹⁶ s⁻¹ (1 mark)

(a) Define the electronvolt.

(b) A gamma ray photon has energy 1.02 MeV. Calculate its wavelength.

5 marks

(a) The energy gained by an electron accelerated through a p.d. of 1 V; 1 eV = 1.60 × 10⁻¹⁹ J (1 mark)
(b) E = 1.02 × 10⁶ × 1.60 × 10⁻¹⁹ = 1.63 × 10⁻¹³ J (1 mark)
λ = hc/E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 1.63 × 10⁻¹³ = 1.22 × 10⁻¹² m (1 mark)
(c) Much more energy (1 mark). Gamma rays have much higher frequency (and shorter wavelength) than visible light, and E = hf, so each gamma photon carries significantly more energy (1 mark).

A laser emits 4.0 × 10¹⁶ photons per second at a wavelength of 589 nm.

(a) Calculate the energy of one photon in eV.

(b) Calculate the power output of the laser.

(c) The laser is pointed at a perfectly absorbing surface. Calculate the force exerted on the surface. (Hint: use the momentum of a photon, p = h/λ, and F = Δp/Δt.)

6 marks

(a) E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 5.89 × 10⁻⁷ = 3.38 × 10⁻¹⁹ J = 2.11 eV (2 marks)
(b) P = N × E = 4.0 × 10¹⁶ × 3.38 × 10⁻¹⁹ = 1.35 × 10⁻² W ≈ 13.5 mW (2 marks)
(c) Momentum per photon: p = h/λ = 6.63 × 10⁻³⁴ / 5.89 × 10⁻⁷ = 1.13 × 10⁻²⁷ kg m s⁻¹. Force = N × p = 4.0 × 10¹⁶ × 1.13 × 10⁻²⁷ = 4.5 × 10⁻¹¹ N (2 marks)

Topic Summary

What is a photon?

A discrete quantum of electromagnetic energy with no mass or charge. Energy depends on frequency: E = hf = hc/λ.

The electronvolt

1 eV = 1.60 × 10⁻¹⁹ J. Used because photon energies in joules are impractically small numbers.

Photon flux

Number of photons per second from a source: N = P/E_photon = Pλ/(hc). More power or longer wavelength → more photons.

Measuring h with LEDs

Plot threshold voltage V₀ vs 1/λ. Gradient = hc/e → solve for h. A straight line confirms E = hf.

Key comparison

Same power, different wavelength: longer λ → more photons/s (each photon carries less energy). Same wavelength, different power: more power → more photons/s.

Equations to Know

E = hf

E = hc / λ

1 eV = 1.60 × 10⁻¹⁹ J

N = P / E_photon

eV₀ = hc / λ