Question

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

• A. 2.6eV
• B. 7.55eV
• C. 12.5eV
• D. 17.45eV

Answer 1

The correct option is D 17.45eV

$E=a(1+cos\pi f_{1}t)cos2\pi f_{2}t=acos2\pi f_{2}t+acos2\pi f_{1}tcos2\pi f_{2}t$

$\Rightarrow E=acos2\pi f_{2}t+\frac{1}{2}acos2\pi (f_1+f_2)t+\frac{1}{2}acos2\pi (f_1-f_2)t$
This is a complex vibration consists of harmonic vibrations of frequencies $f_2,(f_1+f_2)$ and $(f_1-f_2)$,

The maximum frequency of the electromagnetic radiation = $f_1+f_2$

So, $hv=\varphi +T_{max}$

$T_{max}=h(f_1+f_2)-\varphi =\frac{6.6\times {10}^{-34}}{1.6\times {10}^{-19}}(3.6\times {10}^{15}+1.2\times {10}^{15})-2.35=17.45eV$.