Question

Electric field component of an EM radiation varies with time as $E=a\left(\cos \omega_{0} t+\sin \omega t \cos \omega_{0} t\right)$, where ' $a$ ' is a constant and $\omega=10^{15} sec ^{-1}, \omega_{0}=5 \times 10^{15} sec ^{-1}$. This radiation falls on a metal whose stopping potential is $-2 \,e v$. Then which of the following statement (s) is/are true $?\left(h=6.62 \times 10^{-34} J-S\right)$

• A. For light of frequency $\omega$, photoelectric effect is not possible
• B. Stopping potential vs. frequency graph will be a straight line
• C. The work function of the metal is $-2 \,ev$
• D. The maximum kinetic energy of the photo electrons is $1.95\, ev.

Answer 1

Answer: (A), (B)

Stopping potential $=-2 V $
$\Rightarrow K E_{\max }=2 e v($ Option $D$ incorrect)
$E=a\left(\cos \omega_{0} t+\sin \omega t \cos \omega_{0} t\right)$
$=a \cos \omega_{0} t+\frac{a}{2}\left\{\sin \left(\omega+\omega_{0}\right) t+\sin \left(\omega-\omega_{0}\right) t\right\}$
Corresponding frequencies are $\frac{\omega_{0}}{\alpha \pi}, \frac{\omega+\omega_{0}}{\alpha \pi} \& \frac{\omega-\omega_{0}}{2 \pi}$
$\left(\frac{\omega+\omega_{0}}{2 \pi}\right)$ determines maximum KEI $\& $ Stopping potential
Now Energy corresponding to $\frac{\omega+\omega_{0}}{2 \pi}$
Corresponding energy $=h f$
Stopping potential $=V_{0}=2 V =3.95 \,e v$
Now $K E_{\max }=e v_{0}=h f-\phi$
$ \Rightarrow 2 \,e v=3.95 \,e v-\phi$
$ \Rightarrow \phi=1.95 \,e v$
for $w=10^{15} \sec ^{-1}$
frequency $=\frac{10^{15}}{2 \pi}=f_{1}$
When frequency is $w, f_{1}=\frac{\omega}{2 \pi}=\frac{10^{15}}{2 \pi}$
$E=\frac{6.63 \times 10^{-34} \times 6 \times 10^{15}}{2 \pi \times 1.6 \times 10^{-19}} e v=3.95\, ev$
Energy $=h f_{1}=\frac{6.63 \times 10^{-34} \times 10^{15}}{2 \pi \times 1.6 \times 10^{-19}} e v=0.66\, v < \phi$ (work function)
Hence photoelectric effect not possible (option A correct)
Also ev $_{0}=h f-\phi$
(Straight line) (option B correct) $K E_{\max }=E-\phi$
$ \Rightarrow \phi=E-2 \,e v=1.95\, ev$ (option $c$ is incorrect)