Question

When radiation is incident on a photoelectron emitter, the stopping potential is found to be 9V. If elm for the electron is $1.8\times {10}^{11}Ck{g}^{-1}$, the maximum velocity of the ejected electron is

• A. $6\times {10}^{5}m{s}^{-1}$
• B. $8\times {10}^{5}m{s}^{-1}$
• C. $1.8\times {10}^{6}m{s}^{-1}$
• D. $1.8\times {10}^{5}m{s}^{-1}$

Answer 1

The correct option is C $1.8\times {10}^{6}m{s}^{-1}$

$\frac{1}{2}m{v}_{max}^2=e{V}_0$
$\Rightarrow v_{max}=\sqrt{2\frac{e}{m}{V}_0}=\sqrt{2\times 1.8\times {10}^{11}\times 9}$
$=18\times {10}^{5}m/s$
$=1.8\times {10}^{6}m/s$