Question

The electric field of light wave is given as $\vec{E} \, = \, 10^{-3} cos \bigg( \frac{2\pi x}{5 \times 10^{-7}} \, - 2\pi \times 6 \times 10^{14}t\bigg) \hat{x} \frac{N}{C}.$ 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
$KE_{max} = E - \Phi$
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)