Photoelectric Effect — Threshold Wavelength of Cesium (Worked Example)

Medium

Problem

Cesium has a work function of 2.14 eV. What is the maximum wavelength of light that can cause photoemission?

Given

Φ = 2.14 eV
hc = 1240 eV·nm

Find

The threshold wavelength λ_max.

Principle

At threshold the photon energy exactly equals the work function (KE = 0), so λ_max = hc/Φ.

Formula

\lambda_{\max} = \frac{hc}{\Phi}

λ_max = hc/Φ

Solution

Set photon energy equal to work function: $hc/\lambda_{\max} = \Phi$
Rearrange: $\lambda_{\max} = 1240/2.14$
Compute: $\lambda_{\max} \approx 579\ \mathrm{nm}$

≈ 579 nm (yellow-green light)

Unit check

eV·nm / eV = nm — correct for wavelength.

What this means physically

Any wavelength longer (redder) than 579 nm cannot free electrons from cesium, no matter how intense. This sharp cutoff — not a gradual fade — was the smoking gun for photons.

Common mistake

Calling this the 'minimum' wavelength. Longer wavelength means less energy, so 579 nm is the maximum wavelength (minimum frequency) that still works.