Question

The photo-electron liberated from the surface of lithium $(\varphi =2.39eV)$ by electromagnetic radiation whose electric component varies with time as $E=a(1+\cos \omega t)\cos {\omega }_{0}t$ where $a$ = constant, $\omega =6\times {10}^{14}rad/s$ and
${\omega }_0=3.60\times {10}^{15}rad/s$, then the maximum kinetic energy of photo-electron is (in eV)

Answer 1

$E=a\left(1+\cos \omega t\right)\cos {\omega }_{0}t=a\cos {\omega }_{0}t+\frac{1}{2}a\cos \left(\omega +{\omega }_0\right)t+\frac{1}{2}a\cos \left(\omega -{\omega }_0\right)t$
This is a complex vibration consisting of harmonic components of frequencies ${\omega }_0,\left(\omega +{\omega }_0\right)and\left(\omega -{\omega }_0\right).$
Now,
$h\nu =\varphi +k_{max}$
So,
$k_{max}=\frac{h}{2\pi }\left(\omega +{\omega }_0\right)-\varphi =\frac{6.6\times {10}^{-34}}{2\pi }\left(6\times {10}^{14}+3.6\times {10}^{15}\right)-\left(2.39\times 1.6\times {10}^{19}\right)=4.41\times {10}^{-19}-3.82\times {10}^{-19}=0.59\times {10}^{-19}J=0.37eV$