KinexiaLEARN PHYSICS

Photoelectric Effect — Cesium under Green Light (Worked Example)

Medium
Problem

Green light of wavelength 500 nm shines on cesium metal with work function 2.1 eV. What is the maximum kinetic energy of the photoelectrons?

Given

  • λ = 500 nm = 5.00 × 10⁻⁷ m

  • Φ = 2.1 eV

Find

The maximum kinetic energy KE_max.

Principle

KE = hc/λ − Φ: cesium's low work function lets even visible light eject electrons.

Formula

Solution

  1. Photon energy: E=1.986×10−25/5.00×10−7=3.97×10−19 JE = 1.986 \times 10^{-25}/5.00 \times 10^{-7} = 3.97 \times 10^{-19}\ \mathrm{J}

  2. In eV: 3.97×10−19/1.602×10−19=2.48 eV3.97 \times 10^{-19}/1.602 \times 10^{-19} = 2.48\ \mathrm{eV}

  3. Subtract: KE=2.48−2.1=0.38 eVKE = 2.48 - 2.1 = 0.38\ \mathrm{eV}

0.38 eV

Unit check

Energy minus energy — correct.

What this means physically

0.38 eV — just barely ejected. Cesium photocathodes exploit this: they respond to visible light, making them useful in photomultiplier tubes and night-vision-adjacent sensors.

Common mistake

Using λ in nm directly in hc/λ. Convert nm → m (×10⁻⁹) first, or the photon energy is off by a billion.