Question

Maximum velocity of the photoelectron emitted by a metal is $1.8 \times 10^6 ms^{-1}$. Take the value of specific charge of the electron is $1.8 \times10^{11} C \, kg^{-1}$. Then the stopping potential in volt is

• A. 1
• B. 3
• C. 9
• D. 6

Answer 1

Answer: (C)

Given, $v=1.8 \times 10^{6} \,m / s ; \frac{e}{m}=1.8 \times 10^{11}\, C / \,,kg$
We have, $e V_{0}=\frac{1}{2} m v^{2}$
$\Rightarrow V_{0} \frac{e}{m}=\frac{v^{2}}{2}$
$\Rightarrow V_{0} \times 1.8 \times 10^{11}=\frac{1.8 \times 1.8 \times\left(10^{6}\right)^{2}}{2}$
$\Rightarrow V_{0}=9 \,V$