Question

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

• A. \( \frac{v^2}{m e} \)
• B. \( \frac{v^2}{e m} \)
• C. \( \frac{v^2}{2 m e} \)
• D. \( \frac{v^2}{2 e m} \)

Hint:

In the photoelectric effect, the stopping potential is related to the maximum kinetic energy of the photoelectron. Use \( e V_s = \frac{1}{2} m v^2 \) to find the stopping potential.

Answer 1

The Correct Option is D

Step 1: Understanding the photoelectric effect.
According to the photoelectric effect, the maximum kinetic energy \( K.E. \) of the photoelectron is given by \( K.E. = \frac{1}{2} m v^2 \), where \( v \) is the maximum velocity of the photoelectron. The work function \( W \) of the metal is related to the stopping potential \( V_s \) by \( W = e V_s \), where \( e \) is the charge of the electron.

Step 2: Using the energy conservation principle.
The energy imparted to the photoelectron is used in two ways: to overcome the work function and to provide kinetic energy. Thus, \[ e V_s = \frac{1}{2} m v^2 \] Solving for \( V_s \), we get: \[ V_s = \frac{v^2}{2 e m} \]
Step 3: Conclusion.
The stopping potential is \( \frac{v^2}{2 e m} \), which is option (D).