Cut off potential for a metal in photoclectric effect for light of wavelength `lambda_(1), lambda_(2)` and `lambda_(3)` is found to be `V_(1),V_(2)` and `V_(3)` volts , If `V_(1),V_(2)` and `V_(3` are in Arithmetic progression then `lambda_(1),lambda_(2)` and `lambda_(3)` will be in

Options

A. Arithmetic progression
B. Geometric progression
C. Harmonic progression
D. None of these

Answer

To solve the problem, we need to analyze the relationship between the cutoff potentials \( V_1, V_2, V_3 \) and the wavelengths \( \lambda_1, \lambda_2, \lambda_3 \) in the context of the photoelectric effect.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The photoelectric effect is described by the equation:
\[ E = h\nu = h\frac{c}{\lambda} = \phi_0 + KE \]
where \( E \) is the energy of the incident photon, \( h \) is Planck's constant, \( c \) is the speed of light, \( \lambda \) is the wavelength, \( \phi_0 \) is the work function, and \( KE \) is the kinetic energy of the emitted electrons.

2. **Relating Potential to Wavelength**:
The cutoff potential \( V \) is related to the energy of the photon and can be expressed as:
\[ eV = h\frac{c}{\lambda} - \phi_0 \]
Rearranging gives:
\[ V = \frac{hc}{e\lambda} - \frac{\phi_0}{e} \]
This shows that the potential \( V \) is inversely proportional to the wavelength \( \lambda \).

3. **Setting Up the Arithmetic Progression**:
Given that \( V_1, V_2, V_3 \) are in arithmetic progression, we have:
\[ 2V_2 = V_1 + V_3 \]

4. **Expressing Wavelengths in Terms of Potentials**:
From the equation for potential, we can express \( V_1, V_2, V_3 \) as:
\[ V_1 = \frac{hc}{e\lambda_1} - \frac{\phi_0}{e}, \quad V_2 = \frac{hc}{e\lambda_2} - \frac{\phi_0}{e}, \quad V_3 = \frac{hc}{e\lambda_3} - \frac{\phi_0}{e} \]

5. **Substituting into the AP Condition**:
Substituting these expressions into the AP condition:
\[ 2\left(\frac{hc}{e\lambda_2} - \frac{\phi_0}{e}\right) = \left(\frac{hc}{e\lambda_1} - \frac{\phi_0}{e}\right) + \left(\frac{hc}{e\lambda_3} - \frac{\phi_0}{e}\right) \]
Simplifying this leads to:
\[ 2\frac{hc}{e\lambda_2} = \frac{hc}{e\lambda_1} + \frac{hc}{e\lambda_3} \]

6. **Cancelling Common Terms**:
We can cancel \( \frac{hc}{e} \) from both sides:
\[ 2\frac{1}{\lambda_2} = \frac{1}{\lambda_1} + \frac{1}{\lambda_3} \]

7. **Identifying the Progression**:
The equation \( 2\frac{1}{\lambda_2} = \frac{1}{\lambda_1} + \frac{1}{\lambda_3} \) indicates that \( \lambda_1, \lambda_2, \lambda_3 \) are in harmonic progression because the reciprocals of the wavelengths are in arithmetic progression.

### Conclusion:
Thus, if \( V_1, V_2, V_3 \) are in arithmetic progression, then \( \lambda_1, \lambda_2, \lambda_3 \) will be in harmonic progression.