Question

At a place, a light wave described by the equation $E=\left(\frac{100V}{m}\right)\left[\text{sin}\right(2\pi \times {10}^{15}{s}^{-1})t+\text{sin}(3\pi \times {10}^{15}{s}^{-1}\left)t\right]$, where $t$ is in seconds, falls on a metal surface having work function $2.1\text{eV}$. The maximum kinetic energy of the emitted photoelectrons( in $\text{eV}$) is nearly.

Answer 1

$E=\left(\frac{100V}{m}\right)\left[\text{sin}\right(2\pi \times {10}^{15}{s}^{-1})t+\text{sin}(3\pi \times {10}^{15}{s}^{-1}\left)t\right]$
Here, the light contains two different frequencies.
The one with larger frequency will cause photoelectrons with larger kinetic energy. So, the larger frequency is
$\nu =\frac{\omega }{2\pi }=\frac{3\pi \times {10}^{15}}{2\pi }=1.5\times {10}^{15}\text{Hz}$
The maximum kinetic energy of photoelectrons is
$K_{max}=h\nu -W$
$=(4.13\times {10}^{-15}\text{eVs})\times 1.5\times {10}^{15}-2.1$
$=4.095$
$K_{max}\approx 4\text{eV}$