The wavelength of light which is incident on a metal surface charge from `lamda/2 " to " lamda/3` . The change is stopping potential will be

Options

A. `(2hc)/(lamdae)`
B. `(lamdae)/(hc)`
C. `(hc)/(lamdae)`
D. `(lamdah)/c`

Explanation

To solve the problem of finding the change in stopping potential when the wavelength of light incident on a metal surface changes from \( \lambda/2 \) to \( \lambda/3 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Photoelectric Equation**:
The maximum kinetic energy (KE) of the emitted electrons can be expressed using the photoelectric equation:
\[ KE = E - \phi \]
where \( E \) is the energy of the incident photons, and \( \phi \) is the work function of the metal.

2. **Photon Energy Calculation**:
The energy of the incident photons can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the light.

3. **Calculate Stopping Potential for \( \lambda/2 \)**:
For the initial wavelength \( \lambda_1 = \frac{\lambda}{2} \):
\[ E_1 = \frac{hc}{\lambda/2} = \frac{2hc}{\lambda} \]
The stopping potential \( V_{s1} \) can be expressed as:
\[ eV_{s1} = E_1 - \phi \]
Rearranging gives:
\[ V_{s1} = \frac{E_1 - \phi}{e} = \frac{2hc/\lambda - \phi}{e} \]

4. **Calculate Stopping Potential for \( \lambda/3 \)**:
For the new wavelength \( \lambda_2 = \frac{\lambda}{3} \):
\[ E_2 = \frac{hc}{\lambda/3} = \frac{3hc}{\lambda} \]
The stopping potential \( V_{s2} \) can be expressed as:
\[ eV_{s2} = E_2 - \phi \]
Rearranging gives:
\[ V_{s2} = \frac{E_2 - \phi}{e} = \frac{3hc/\lambda - \phi}{e} \]

5. **Find the Change in Stopping Potential**:
The change in stopping potential \( \Delta V_s \) is given by:
\[ \Delta V_s = V_{s2} - V_{s1} \]
Substituting the expressions for \( V_{s1} \) and \( V_{s2} \):
\[ \Delta V_s = \left( \frac{3hc/\lambda - \phi}{e} \right) - \left( \frac{2hc/\lambda - \phi}{e} \right) \]
Simplifying this gives:
\[ \Delta V_s = \frac{3hc/\lambda - 2hc/\lambda}{e} = \frac{hc/\lambda}{e} \]

6. **Final Expression**:
Therefore, the change in stopping potential is:
\[ \Delta V_s = \frac{hc}{e\lambda} \]

### Conclusion:
The change in stopping potential when the wavelength changes from \( \lambda/2 \) to \( \lambda/3 \) is given by:
\[ \Delta V_s = \frac{hc}{e\lambda} \]