Question

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

Answer 1

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