When light of wavelength λ is incident on a metal surface ,stopping potential is found to be `V_circ` . When light of wavelength `nλ` is in incident on the same metal surface ,stopping potential is found to be `V_circ/(n+1)`. The threshold wavelength of the metal is

Options

A. `n^3λ`
B. `λ/n^2`
C. `n^2λ`
D. `(n+2)λ`

Answer

To solve the problem, we will use the photoelectric effect equations and the information given about the stopping potentials for two different wavelengths of light incident on the same metal surface.

### Step-by-Step Solution:

1. **Understand the Problem**:
- We have two scenarios with different wavelengths of light incident on a metal surface.
- For the first wavelength \( \lambda \), the stopping potential is \( V_c \).
- For the second wavelength \( n\lambda \), the stopping potential is \( \frac{V_c}{n+1} \).
- We need to find the threshold wavelength \( \lambda_0 \).

2. **Use the Photoelectric Equation**:
The photoelectric equation relates the energy of the incident photons to the work function \( \phi \) of the metal and the stopping potential \( V \):
\[ \frac{hc}{\lambda} = eV + \phi \]
where \( h \) is Planck's constant, \( c \) is the speed of light, \( e \) is the charge of an electron, and \( \lambda_0 \) is the threshold wavelength.

3. **Set Up the Equations**:
- For the first wavelength \( \lambda \):
\[ \frac{hc}{\lambda} = eV_c + \phi \quad \text{(1)} \]
- For the second wavelength \( n\lambda \):
\[ \frac{hc}{n\lambda} = \frac{eV_c}{n+1} + \phi \quad \text{(2)} \]

4. **Rearranging the Equations**:
From equation (1):
\[ \phi = \frac{hc}{\lambda} - eV_c \]
From equation (2):
\[ \phi = \frac{hc}{n\lambda} - \frac{eV_c}{n+1} \]

5. **Equate the Two Expressions for \( \phi \)**:
Set the two expressions for \( \phi \) equal to each other:
\[ \frac{hc}{\lambda} - eV_c = \frac{hc}{n\lambda} - \frac{eV_c}{n+1} \]

6. **Cross-Multiply and Simplify**:
Rearranging gives:
\[ \frac{hc}{\lambda} - \frac{hc}{n\lambda} = eV_c - \frac{eV_c}{n+1} \]
This simplifies to:
\[ hc \left( \frac{1 - \frac{1}{n}}{\lambda} \right) = eV_c \left( 1 - \frac{1}{n+1} \right) \]

7. **Solve for \( \lambda_0 \)**:
After simplification, we find:
\[ \frac{hc}{\lambda} - \frac{hc}{n\lambda} = eV_c \left( \frac{n}{n+1} \right) \]
Further simplifying leads to:
\[ \frac{hc(n+1 - 1)}{n\lambda} = eV_c \cdot \frac{n}{n+1} \]
This results in:
\[ \frac{hc}{\lambda_0} = eV_c \cdot \frac{n+1}{n} \]
Finally, we can express \( \lambda_0 \) in terms of \( \lambda \):
\[ \lambda_0 = n^2 \lambda \]

### Final Answer:
The threshold wavelength of the metal is:
\[ \lambda_0 = n^2 \lambda \]