Question

A piece of metal ejects electrons with a maximum KE of 2.65 eV. If impacted by light of wavelength 300 nm, determine the work function (in eV) of this material?

Answer 1

Solution:

The work function

$\varphi =\frac{hc}{\lambda }$

$\varphi =\frac{\left ( 4.14\times 10^{-15}eV-s \right )\left ( 3\times 10^{8}m/s \right )}{\left ( 300\times 10^{-9}m \right ) }$

$\varphi =4.14eV$