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 $\omega =10^{15}se{c}^{-1},{\omega }_0=5\times 10^{15}se{c}^{-1}$. This radiation falls on a metal whose stopping potential is $-2ev$. 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 $-2ev$
• D. The maximum kinetic energy of the photo electrons is $1.95{\,} ev.

Answer 1

Answer: (A), (B)

Stopping potential $=-2V$
$\Rightarrow K{E}_{\max }=2ev($ 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 $=hf$
Stopping potential $=V_0=2V=3.95ev$
Now $K{E}_{\max }=e{v}_0=hf-\varphi$
$\Rightarrow 2ev=3.95ev-\varphi$
$\Rightarrow \varphi =1.95ev$
for $w=10^{15}{\sec }^{-1}$
frequency $=\frac{10^{15}}{2\pi }=f_1$
When frequency is $w,f_1=\frac{\omega }{2\pi }=\frac{10^{15}}{2\pi }$
$E=\frac{6.63\times 10^{-34}\times 6\times 10^{15}}{2\pi \times 1.6\times 10^{-19}}ev=3.95ev$
Energy $=h{f}_1=\frac{6.63\times 10^{-34}\times 10^{15}}{2\pi \times 1.6\times 10^{-19}}ev=0.66v<\varphi$ (work function)
Hence photoelectric effect not possible (option A correct)
Also ev ${}_0=hf-\varphi$
(Straight line) (option B correct) $K{E}_{\max }=E-\varphi$
$\Rightarrow \varphi =E-2ev=1.95ev$ (option $c$ is incorrect)