Question

The maximum velocity of the photoelectron emitted by the metal surface is \( V \). Charge and mass of the photoelectron are denoted by \( e \) and \( m \) respectively. The stopping potential in volt is

• A. \( \sqrt{\frac{2V}{m}} \)
• B. \( \frac{V^2}{2 \left( \frac{e}{m} \right)} \)
• C. \( \frac{V^2}{e} \)
• D. \( \sqrt{\frac{V^2}{2m}} \)

Hint:

The stopping potential is the potential that reduces the kinetic energy of the photoelectron to zero. It is related to the kinetic energy by \( K = e V_{\text{stop}} \).

Answer 1

The Correct Option is B

Step 1: Relation between kinetic energy and stopping potential.
The maximum kinetic energy \( K \) of the photoelectron is given by: \[ K = \frac{1}{2} m V^2 \] The stopping potential \( V_{\text{stop}} \) is the potential that stops the photoelectron and is related to its kinetic energy: \[ K = e V_{\text{stop}} \] By equating the kinetic energy and the work done by the stopping potential: \[ \frac{1}{2} m V^2 = e V_{\text{stop}} \] Solving for the stopping potential: \[ V_{\text{stop}} = \frac{V^2}{2 \left( \frac{e}{m} \right)} \]
Step 2: Conclusion.
Thus, the correct answer is (B) \( \frac{V^2}{2 \left( \frac{e}{m} \right) \)}.