How does wavelength affect the photoelectric effect?

Mar 12, 2016

Explanation:

Borrowing some material for this answer:

A single photon, whose wavelength is lower than the threshold wavelength for a specific metal, has the required energy to eject one electron thus creating the observed photoelectric effect.

Einstein used Planck's famous expression $E = h \nu$, to explain other experimental results of photoelectric effect.

$E$ being energy, $\nu$, the frequency of radiated energy and $h$ is Planck's Constant. We know that speed of light in vacuum $c$ is related to its frequency $\nu$ and wavelength $\lambda$

$c = \nu \lambda$

Einstein put forward the following equation:

$\frac{h c}{\lambda} _ \textrm{\in c i \mathrm{de} n t} = \phi + K {E}_{\max}$

where $h$ is Planck's constant, ${\lambda}_{\textrm{\in c i \mathrm{de} n t}}$ wavelength of incident light, $\phi$ work function of the metal and $K {E}_{\max}$ the maximum kinetic energy of the ejected electron(s).

For photoelectric effect to occur, the energy of the photon must be greater than the work function.
$\phi = \frac{h c}{\lambda} _ \textrm{c u t - o f f}$

As the wavelength of the incident light decreases but is lower than the cut-off wavelength, the maximum kinetic energy of the photo electrons increases.