Question

The maximum kinetic energy of a photoelectron liberated from the surface of lithium with work function $2.35 eV$ 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 $\left(f_{1}=3.6 \times 10^{15} Hz\right.$, 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,
$W_{0}=2.35 \,eV$ and the electric component of electromagnetic radiation $E=a\left[1+\cos \left(2 \pi f_{1} t\right)\right] \cos \left(2 \pi f_{2} t\right)$
$\Rightarrow E=\left[a \cos \left(2 \pi f_{2} t\right)+a \cos \left(2 \pi f_{1} t\right) \cos \left(2 \pi f_{2} t\right)\right]$
$\left(\because \cos \,A \,\cos\, B=\frac{1}{2}[\cos (A+B)-\cos (A-B))\right.$
$\Rightarrow E=a \cos \left(2 \pi f_{2} t\right)+\frac{a}{2} \cos 2 \pi\left(f_{1}+f_{2}\right) t-\frac{a}{2} \cos 2 \pi\left(f_{1}-f_{2}\right) t$
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}} $
$=19.8 \,eV $
Hence, the maximum kinetic energy,
$KE _{\max }=E_{\max }-W_{0}$
$=19.8-2.35=17.45 \,eV$