KinexiaLEARN PHYSICS

Photoelectric Effect — Zinc under UV (Worked Example)

Medium
Problem

Ultraviolet light of wavelength 250 nm strikes a zinc plate with work function 4.3 eV. What is the maximum kinetic energy of the emitted electrons?

Given

  • λ = 250 nm = 2.50 × 10⁻⁷ m

  • Φ = 4.3 eV

  • hc = 1.986 × 10⁻²⁵ J·m

Find

The maximum kinetic energy KE_max.

Principle

Einstein: photon energy hf = hc/λ; the electron keeps KE = hf − Φ after paying the work-function cost.

Formula

Solution

  1. Photon energy: E=hc/λ=1.986×10−25/2.50×10−7=7.94×10−19 JE = hc/\lambda = 1.986 \times 10^{-25}/2.50 \times 10^{-7} = 7.94 \times 10^{-19}\ \mathrm{J}

  2. Convert: 7.94×10−19/1.602×10−19=4.96 eV7.94 \times 10^{-19}/1.602 \times 10^{-19} = 4.96\ \mathrm{eV}

  3. Subtract work function: KE=4.96−4.3=0.66 eVKE = 4.96 - 4.3 = 0.66\ \mathrm{eV}

0.66 eV

Unit check

J − J = J, reported in eV — energy throughout.

What this means physically

0.66 eV — a slow electron. Raise the UV intensity and more electrons emerge, but each still carries 0.66 eV — intensity changes number, not energy. That was Einstein's Nobel insight.

Common mistake

Subtracting in mixed units (joules minus eV). Convert everything to eV (or all to joules) before subtracting.