Photons and the photoelectric effect

MDCAT Physics · Physics · Dawn of modern physics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Light arrives in packets of energy called photons. The energy of one photon depends only on its frequency: higher frequency (shorter wavelength) means more energy. Brighter light means more photons each second, not bigger photons.

In the photoelectric effect one photon gives all its energy to one electron in a metal. The electron escapes only if the photon energy is at least the work function, so there is a threshold frequency. Above it, the maximum kinetic energy rises with frequency, and the number of electrons rises with intensity.

In the Compton effect an X-ray photon hits a free electron and loses some energy, so its wavelength increases. In pair production a γ-ray photon near a heavy nucleus changes into an electron and a positron.

Rules to remember

photon energy = Planck's constant × frequency (E = hf = hc/λ), h = 6.6 × 10⁻³⁴ J s
Einstein's equation: hf = φ + KE_max, with work function φ = hf₀ = hc/λ₀
stopping potential: eV₀ = KE_max; 1 eV = 1.6 × 10⁻¹⁹ J
Graph of KE_max against f: straight line, slope h, intercept on the frequency axis f₀
Compton shift: Δλ = (h/m₀c)(1 − cos θ); zero at θ = 0°, greatest at θ = 180°
Pair production needs a photon of at least 1.02 MeV (2 × 0.51 MeV); any extra becomes kinetic energy of the pair
Annihilation of an electron and a positron at rest gives two 0.51 MeV photons moving in opposite directions

Common mistake

Students think brighter light gives faster electrons. Intensity changes only the number of electrons; the maximum kinetic energy depends on frequency. Also, KE_max is not proportional to f, because the work function is subtracted first.

Worked example

Light of wavelength 330 nm falls on a metal of work function 3.6 × 10⁻¹⁹ J. Find the stopping potential. (h = 6.6 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m s⁻¹, e = 1.6 × 10⁻¹⁹ C)

Photon energy = hc ÷ λ = (6.6 × 10⁻³⁴ × 3.0 × 10⁸) ÷ (3.3 × 10⁻⁷) = 6.0 × 10⁻¹⁹ J.
KE_max = 6.0 × 10⁻¹⁹ − 3.6 × 10⁻¹⁹ = 2.4 × 10⁻¹⁹ J.
V₀ = KE_max ÷ e = (2.4 × 10⁻¹⁹) ÷ (1.6 × 10⁻¹⁹) = 1.5 V.

Answer: Stopping potential = 1.5 V (KE_max = 1.5 eV).

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