Question

The maximum kinetic energy of a photoelectron liberated from the surface of lithium with work function $2.35eV$ by electromagnetic radiation whose electric component varies with time as:
$E=a\left[1+\cos \left(2\pi f_{1}t\right)\right]\cos 2\pi f_{2}t($ where $a$ is a constant) is $(f_1=3.6\times 10^{15}Hz$, and $f_2=1.2\times 10^{15}Hz$ and Planck's constant $h=6.6\times 10^{-34}Js$ )

• A. 2.64 eV
• B. 7.55 eV
• C. 12.52 eV
• D. 17.45 eV

Answer 1

Answer: (D)

Here, work function,
$W_0=2.35eV$ and the electric component of electromagnetic radiation $E=a\left[1+\cos \left(2\pi f_{1}t\right)\right]\cos \left(2\pi f_{2}t\right)$
$\Rightarrow E=\left[a\cos \left(2\pi f_{2}t\right)+a\cos \left(2\pi f_{1}t\right)\cos \left(2\pi f_{2}t\right)\right]$
$(\because \cos A\cos B=\frac{1}{2}[\cos (A+B)-\cos (A-B))$
$\Rightarrow E=a\cos \left(2\pi f_{2}t\right)+\frac{a}{2}\cos 2\pi \left(f_1+f_2\right)t-\frac{a}{2}\cos 2\pi \left(f_1-f_2\right)t$
So, the electric component has 3 sub-components with frequencies are, $f_2,\left(f_1+f_2\right)$ and $\left(f_1-f_2\right)$
So, for maximum kinetic energy of photoelectron, we take photon of maximum frequency.
Hence, $E_{\max }=\frac{h{v}_{\max }}{e}$
$=\frac{6.6\times 10^{-34}\times \left(3.6\times 10^{15}+1.2\times 10^{15}\right)}{1.6\times 10^{-19}}$
$=19.8eV$
Hence, the maximum kinetic energy,
$K{E}_{\max }=E_{\max }-W_0$
$=19.8-2.35=17.45eV$