Question

The kinetic energy of the fastest moving photo electron from a metal of work function $2.8eV$ is $2eV$. If the frequency of light is doubled, then find the maximum kinetic energy of photo electron.

Answer 1

Maximum kinetic energy $E=h\nu -\varphi$

Given : $\varphi =2.8eV$ $E=2eV$

So, $2=h\nu -2.8$

$⟹h\nu =4.8eV$

New frequency ${\nu }^{\prime }=2\nu$

So, $E^{\prime }=h{\nu }^{\prime }-\varphi$

Or $E^{\prime }=2h\nu -\varphi =2\times 4.8-2.8=6.8eV$