Question

Find the maximum kinetic energy (in $eV$ ) of the photoelectron liberated from the surface of lithium (work function $\varphi =2.15eV$ ) by electromagnetic radiation whose electric component varies with time as $E=a(1+cos\omega t)cos{\left(\omega \right)}_{0}t$ , where $a$ is a constant, $\omega =12\times 10^{14}rad{s}^{-1}$ and ${\omega }_0=3.6\times 10^{15}rad{s}^{-1}h=6.6\times 10^{-34}$ in SI units)

Answer 1

Answer: 1

$E=acos{\omega }_{0}t+acos\omega tcos{\omega }_{0}t$
$E=acos{\left(\omega \right)}_{0}t+\frac{a}{2}\left[cos\left(\omega +{\left(\omega \right)}_0\right)t+cos\left(\omega -{\left(\omega \right)}_0\right)t\right]$
$v_{max}=\frac{\omega +{\omega }_0}{2\pi }=\frac{4.8}{2\pi }\times 10^{15}$
$K{E}_{max}=hv-\varphi$
$=\frac{6.6\times 10^{-34}\times \frac{4.8}{2\pi }\times 10^{15}}{1.6\times 10^{-19}}-2.15$
$=\frac{6.3}{2}-2.15=1eV$