Question

Find the maximum kinetic energy (in $eV$ ) of the photoelectron liberated from the surface of lithium (work function $\phi=2.15\,eV$ ) by electromagnetic radiation whose electric component varies with time as $E=a\left(\right.1+cos\omega t\left.\right)cos\left(\omega \right)_{0}t$ , where $a$ is a constant, $\omega =12\,\times 10^{14}rads^{- 1}$ and $\omega _{0}=3.6\,\times 10^{15}rad s^{- 1} h=6.6\times 10^{-34}$ in SI units)

Answer 1

$E=acos\omega _{0}t+acos\omega tcos\omega _{0}t$
$E=acos\left(\omega \right)_{0}t+\frac{a}{2}\left[cos \left(\omega + \left(\omega \right)_{0}\right) t + cos \left(\omega - \left(\omega \right)_{0}\right) t\right]$
$v_{max}=\frac{\omega + \omega _{0}}{2 \pi }=\frac{4 . 8}{2 \pi }\times 10^{15}$
$KE_{max}=hv-\phi$
$=\frac{6 . 6 \times 10^{- 34} \times \frac{4 . 8}{2 \pi } \times 10^{15}}{1 . 6 \times 10^{- 19}}-2.15$
$=\frac{6 . 3}{2}-2.15=1eV$