Question

An electromagnetic radiation given by
$E=E_0\left[\sin \right(2\pi {\nu }_{1}t)+\sin (2\pi {\nu }_{2}t\left)\right]$ is incident on a photocell whose work function is $2.0\text{eV}$. Then the maximum kinetic energy (upto $2$ decimal points accuracy) of photoelectrons will be $\text{eV}$

[Take $h=4.14\times {10}^{-15}\text{eVs};{\nu }_1=1.3\times {10}^{15}\text{Hz}$ and ${\nu }_2=1.1\times {10}^{15}\text{Hz}$]

Answer 1

Given, ${\nu }_1=1.3\times {10}^{15}\text{Hz}$ and ${\nu }_2=1.1\times {10}^{15}\text{Hz}$

The light contains two different frequencies. The one with larger frequency will cause photoelectrons with largest kinetic energy.

The maximum kinetic energy of the photoelectrons is,

$K_{max}=h\nu -\varphi$

$=(4.14\times {10}^{-15}\text{eVs})\times (1.3\times {10}^{15}{\text{s}}^{-1})-2.0\text{eV}$

$=5.38-2.0=3.38\text{eV}$