KinexiaLEARN PHYSICS

Photoelectric Effect — Potassium (Worked Example)

Easy
Problem

Blue light of wavelength 450 nm falls on potassium with work function 2.3 eV. Find the maximum kinetic energy of emitted electrons.

Given

  • λ = 450 nm = 4.50 × 10⁻⁷ m

  • Φ = 2.3 eV

Find

KE_max.

Principle

KE = hc/λ − Φ — same equation, new metal.

Formula

Solution

  1. Photon energy: E=1.986×10−25/4.50×10−7=4.41×10−19 JE = 1.986 \times 10^{-25}/4.50 \times 10^{-7} = 4.41 \times 10^{-19}\ \mathrm{J}

  2. In eV: 4.41×10−19/1.602×10−19=2.76 eV4.41 \times 10^{-19}/1.602 \times 10^{-19} = 2.76\ \mathrm{eV}

  3. Subtract: KE=2.76−2.3=0.46 eVKE = 2.76 - 2.3 = 0.46\ \mathrm{eV}

0.46 eV

Unit check

Energy units throughout — correct.

What this means physically

0.46 eV — the electron crawls out. Note red light (700 nm, 1.77 eV) could never eject from potassium no matter how bright — below Φ, zero emission. Frequency, not intensity, rules.

Common mistake

Thinking brighter light gives faster electrons. Brightness adds photons (more electrons); only shorter wavelength adds energy per electron.