## In This post you will get answer of Q.

The maximum kinetic energy of a photoelectron liberated from the surface of lithium with work function $2.35 eV$ by electromagnetic radiation whose electric component varies with time as:

$E=aleft[1+cos left(2 pi f_{1} tright)right] cos 2 pi f_{2} t($ where $a$ is a

constant) is $left(f_{1}=3.6 times 10^{15} Hzright.$,

and $f_{2}=1.2 times 10^{15} Hz$ and Planck’s constant $h=6.6 times 10^{-34} Js$ )

Solution:

## Here, work function,

$W_{0}=2.35 ,eV$ and the electric component of electromagnetic radiation $E=aleft[1+cos left(2 pi f_{1} tright)right] cos left(2 pi f_{2} tright)$

$Rightarrow E=left[a cos left(2 pi f_{2} tright)+a cos left(2 pi f_{1} tright) cos left(2 pi f_{2} tright)right]$

$left(because cos ,A ,cos, B=frac{1}{2}[cos (A+B)-cos (A-B))right.$

$Rightarrow E=a cos left(2 pi f_{2} tright)+frac{a}{2} cos 2 pileft(f_{1}+f_{2}right) t-frac{a}{2} cos 2 pileft(f_{1}-f_{2}right) t$

So, the electric component has 3 sub-components with frequencies are, $f_{2},left(f_{1}+f_{2}right)$ and $left(f_{1}-f_{2}right)$

So, for maximum kinetic energy of photoelectron, we take photon of maximum frequency.

Hence, $ E_{max } =frac{h v_{max }}{e}$

$=frac{6.6 times 10^{-34} timesleft(3.6 times 10^{15}+1.2 times 10^{15}right)}{1.6 times 10^{-19}} $

$=19.8 ,eV $

Hence, the maximum kinetic energy,

$KE _{max }=E_{max }-W_{0}$

$=19.8-2.35=17.45 ,eV$