Question

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

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

Answer 1

Answer: (D)

We know that the maximum kinetic energy of the electron is equal to the stopping potential of the electron, $K_{max}=e{V}_0$ .
Kinetic energy $K_{max}=\frac{1}{2}m{v}_{max}^2$ .
$\frac{1}{2}m{v}_{max}^2=e{V}_0$
$\therefore v_{max}=\sqrt{\frac{2e{V}_0}{m}}=\sqrt{2\times 1.8\times 10^{11}\times 9}$
Therefore, maximum velocity of the emitted electron is, $v_{max}=1.8\times 10^{6}m{s}^{-1}$ .