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 $1.8\times 10^{11} \, Ckg^{- 1}$ , 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, $K_{max}=eV_{0}$ .
Kinetic energy $K_{max}=\frac{1}{2}m v^{2}_{max}$ .
$\frac{1}{2}mv_{max}^{2}=eV_{0}$
$\therefore v_{max}=\sqrt{\frac{2 e V_{0}}{m}}=\sqrt{2 \times 1 . 8 \times 10^{11} \times 9}$
Therefore, maximum velocity of the emitted electron is, $v_{max}=1.8\times 10^{6}ms^{- 1}$ .