Question

If $e/m$ of electron is $1.76\times 10^{11}Ck{g}^{-1}$ and the stopping potential is 0.71 V, then the maximum velocity of the photoelectron is

• A. $150km{s}^{-1}$
• B. $200km{s}^{-1}$
• C. $500km{s}^{-1}$
• D. $250km{s}^{-1}$

Answer 1

Answer: (C)

Here, stopping potential, $V_S=0.71V$
For electron, $\frac{e}{m}=1\cdot 76\times 10^{11}Ck{g}^{-1}$
The maximum kinetic energy of the photoelectrons is
$K_{\max }=\frac{1}{2}m{v}_{\text{max }}^2=e{V}_s$
$\therefore v_{\text{max }}^2=2\frac{e}{m}{V}_s$
$v_{\max }=\sqrt{2\frac{e}{m}{V}_s}=\sqrt{2\times 1.76\times 10^{11}\times 0.71}$
$=5\times 10^{5}m{s}^{-1}=500\times 10^{3}m{s}^{-1}$
$=500km{s}^{-1}$