Photoelectric Effect — Zinc under Ultraviolet (Worked Example)

Medium

Problem

Ultraviolet light of wavelength 200 nm strikes zinc (Φ = 4.30 eV). Find the maximum kinetic energy of the emitted electrons.

Given

λ = 200 nm
Φ = 4.30 eV
hc = 1240 eV·nm

Find

KE_max.

Principle

KE_max = hf − Φ: photon energy minus the binding energy stays as electron kinetic energy.

Formula

K_{\max} = \frac{hc}{\lambda} - \Phi

KE_max = hc/λ − Φ

Solution

Photon energy: $E = 1240/200 = 6.20\ \mathrm{eV}$
Subtract: $KE_{\max} = 6.20 - 4.30 = 1.90\ \mathrm{eV}$
Interpret: $1.90\ \mathrm{eV}$ of kinetic energy per electron — faster than the potassium case because UV photons carry more energy.

1.90 eV

Unit check

eV − eV = eV — consistent.

What this means physically

UV photons carry enough punch (6.2 eV) to beat zinc's 4.3 eV grip, leaving 1.9 eV of motion. Visible light (~2–3 eV) could never do this job — which is why photocells for visible light use low-work-function metals like cesium.

Common mistake

Using wavelength in meters with hc = 1240 eV·nm. The shortcut 1240/λ only works when λ is in nanometers — mixing units is the classic error here.