The maximum kinetic energy of a photoelectron liberated from the surface of lithium with work function $2.35…
The maximum kinetic energy of a photoelectron liberated from the surface of lithium with work function $2.35 \mathrm{eV}$ 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 $\left(f_1=3.6 \times 10^{15} \mathrm{~Hz}\right.$,
and $f_2=1.2 \times 10^{15} \mathrm{~Hz}$ and Planck's constant $\left.h=6.6 \times 10^{-34} \mathrm{Js}\right)$
2.64 eV
7.55 eV
12.52 eV
17.45 eV
Solution
Here, work function, $W_0=2.35 \mathrm{eV}$ and the electric component of electromagnetic radiation
$
\begin{aligned}
& 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
\end{aligned}
$
So, the electric component has 3 sub-components with frequencies are,
$
f_2,\left(f_1+f_2\right) \text { and }\left(f_1-f_2\right)
$
So, for maximum kinetic energy of photoelectron, we take photon of maximum frequency. Hence,
$
\begin{aligned}
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 \mathrm{eV}
\end{aligned}
$
Hence, the maximum kinetic energy,
$
\mathrm{KE}_{\text {max }}=E_{\text {max }}-W_0=19.8-2.35=17.45 \mathrm{eV}
$
Hence, the correct option is (d)