When light of wavelength `lambda` is incident on photosensitive surface, the stopping potential is V. When light of wavelength `3lambda` is incident on same surface, the stopping potential is `V/6` Threshold wavelength for the surface is

Options

A. (a)`2lambda`
B. (b)`3lambda`
C. (c)`4lambda`
D. (d)`5lambda`

Answer

To find the threshold wavelength for the photosensitive surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The stopping potential \( V \) is related to the energy of the incident photons and the work function \( \phi \) of the material. The energy of the photon can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the incident light.

2. **Setting Up the Equations**:
For the first case where the wavelength is \( \lambda \) and the stopping potential is \( V \):
\[ \frac{hc}{\lambda} - \phi = eV \quad \text{(Equation 1)} \]
For the second case where the wavelength is \( 3\lambda \) and the stopping potential is \( \frac{V}{6} \):
\[ \frac{hc}{3\lambda} - \phi = e\left(\frac{V}{6}\right) \quad \text{(Equation 2)} \]

3. **Subtracting the Two Equations**:
Now, we subtract Equation 2 from Equation 1:
\[ \left(\frac{hc}{\lambda} - \phi\right) - \left(\frac{hc}{3\lambda} - \phi\right) = eV - e\left(\frac{V}{6}\right) \]
Simplifying the left side:
\[ \frac{hc}{\lambda} - \frac{hc}{3\lambda} = \frac{3hc - hc}{3\lambda} = \frac{2hc}{3\lambda} \]
And simplifying the right side:
\[ eV - \frac{eV}{6} = \frac{6eV - eV}{6} = \frac{5eV}{6} \]
Thus, we have:
\[ \frac{2hc}{3\lambda} = \frac{5eV}{6} \]

4. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 2hc \cdot 6 = 5eV \cdot 3\lambda \]
Simplifying this:
\[ 12hc = 15eV\lambda \]

5. **Solving for the Work Function**:
Rearranging gives:
\[ \lambda = \frac{12hc}{15eV} = \frac{4hc}{5eV} \]

6. **Finding the Work Function**:
The work function \( \phi \) can also be expressed in terms of the threshold wavelength \( \lambda_0 \):
\[ \phi = \frac{hc}{\lambda_0} \]
Setting the two expressions for the work function equal gives:
\[ \frac{hc}{\lambda_0} = \frac{hc}{5} \]
Thus, we find:
\[ \lambda_0 = 5\lambda \]

### Conclusion:
The threshold wavelength for the surface is \( \lambda_0 = 5\lambda \).