Photoelectric Effect

Einstein's photoelectric equation

When light hits a metal surface, each photon can knock out at most one electron, and only if the photon carries enough energy to first overcome the metal's work function $\phi$, the minimum energy binding an electron to the surface. Whatever energy is left over becomes the electron's kinetic energy:

where $h\nu$ is the photon's energy ($h$ is Planck's constant, $6.626\times10^{-34}$ J·s). Below the threshold frequency $\nu_0 = \phi/h$, no electrons are emitted at all, no matter how intense the light is, a single low-energy photon still can't individually clear the work function.

Worked example

Sodium has a work function of 2.3 eV. Light of wavelength 400 nm shines on it. Find the maximum kinetic energy of the emitted electrons, and the threshold wavelength below which no electrons are emitted.

Solution: Working in electron-volts and nanometres, photon energy is $E = hc/\lambda$, and $hc \approx 1240\text{ eV}\cdot\text{nm}$ is a convenient constant to keep on hand:

Subtracting the work function gives the maximum kinetic energy:

The threshold wavelength is where $KE_{max}=0$: the photon energy exactly equals the work function:

Any wavelength longer than this (lower photon energy) produces no photoelectrons at all from sodium, regardless of the light's intensity.