KinexiaLEARN PHYSICS

Photoelectric Effect — Potassium under Violet Light (Worked Example)

Medium
Problem

Light of wavelength 400 nm strikes a potassium surface with work function Φ = 2.29 eV. What is the maximum kinetic energy of the emitted electrons?

Given

  • λ = 400 nm

  • Φ = 2.29 eV

  • hc = 1240 eV·nm

Find

KE_max of photoelectrons.

Principle

Einstein's photoelectric equation: each photon gives its energy hf to one electron; the leftover after escaping the metal is kinetic energy.

Formula

Solution

  1. Photon energy: E=1240/400=3.10 eVE = 1240/400 = 3.10\ \mathrm{eV}

  2. Subtract the work function: KEmax⁡=3.10−2.29KE_{\max} = 3.10 - 2.29

  3. Result: KEmax⁡=0.81 eVKE_{\max} = 0.81\ \mathrm{eV}

0.81 eV

Unit check

eV − eV = eV — energy units throughout.

What this means physically

Each electron escapes with 0.81 eV — slow by particle standards (about 5.3×10⁵ m/s), but the current flows. Brighter light would eject more electrons, not faster ones; only shorter wavelength raises this energy.

Common mistake

Thinking brighter light makes faster electrons. Intensity controls the number of photons (more electrons); frequency/wavelength alone controls each electron's energy.