Question

When radiation is incident on a photoelectron emitter, the stopping potential is found to be $9V$. If $e/m$ for the electron is $1.8\times {10}^{11}Ck{g}^{-1}$. What is the max. speed of the photoelectrons?

• 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 $1.8\times {10}^{6}m{s}^{-1}$

$\frac{1}{2}m{v}_{max}^2=e{V}_0$
$\Rightarrow v_{max}=2\left(\frac{e}{m}\right)V_0=2\times 1.8\times {10}^{11}\times 9$
$=19\times {10}^{15}m/s=1.8\times {10}^{6}m/s$.