Question

The electric field of light wave is given as $\vec{E}=10^{-3}cos(\frac{2\pi x}{5\times 10^{-7}}-2\pi \times 6\times 10^{14}t)\hat{x}\frac{N}{C}.$ This light falls on a metal plate of work function 2eV. The stopping potential of the photo-electrons is : Given, E (in eV) = $\frac{12375}{\lambda (in\dot{A)}}$

• A. 0.48 V
• B. 2.0 V
• C. 2.48 V
• D. 0.72 V

Answer 1

Answer: (A)

$\Omega =6\times 10^{14}\times 2\pi$
$f=6\times 10^{14}$
C = f$\lambda$
$\lambda =\frac{C}{f}=\frac{3\times 10^8}{6\times 10^{14}}=5000\dot{A}$
energy of photon $\Rightarrow \frac{12375}{5000}$
= 2.475 eV
from Einstein's equation
$K{E}_{max}=E-\Phi$
eV${}_s=E-\Phi$
eV${}_s$ = 2.475 - 2
eV${}_s$ = 0.475 - 2
eV${}_s$ = 0.475 eV
V${}_s$ = 0.475 V = 0.48 volt
Option (1)