`V_(1)` and `V_(2)` are the stopping potentials for the incident radiations of wave lengths `lamda_(1)` and `lamda_(2)` respectively are incident on a metallic surface. If `lamda_(1)=3lamda_(2)` then

Options

A. `V_(1)gt3V_(2)`
B. `V_(2)gt3V_(1)`
C. `V_(1)gt(V_(2))/(3)`
D. `V_(1)lt(V_(2))/(3)`

Answer

To solve the problem step by step, we will use the concepts of stopping potential and the photoelectric effect.

### Step-by-Step Solution:

1. **Understanding the Stopping Potential**:
The stopping potential \( V \) for a photon of wavelength \( \lambda \) incident on a metallic surface is given by the equation:
\[ eV = \frac{hc}{\lambda} - \phi \]
where \( e \) is the charge of the electron, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \phi \) is the work function of the metal.

2. **Setting Up the Equations**:
For the two wavelengths \( \lambda_1 \) and \( \lambda_2 \), we can write two equations for the stopping potentials \( V_1 \) and \( V_2 \):
\[ eV_1 = \frac{hc}{\lambda_1} - \phi \quad \text{(1)} \]
\[ eV_2 = \frac{hc}{\lambda_2} - \phi \quad \text{(2)} \]

3. **Substituting the Given Condition**:
We are given that \( \lambda_1 = 3\lambda_2 \). We can substitute this into equation (1):
\[ eV_1 = \frac{hc}{3\lambda_2} - \phi \]

4. **Rearranging the Equations**:
Now we can express both equations in terms of \( V_1 \) and \( V_2 \):
- From equation (1):
\[ eV_1 = \frac{hc}{3\lambda_2} - \phi \]
- From equation (2):
\[ eV_2 = \frac{hc}{\lambda_2} - \phi \]

5. **Eliminating the Work Function**:
We can eliminate \( \phi \) by subtracting equation (2) from equation (1):
\[ eV_1 - eV_2 = \left(\frac{hc}{3\lambda_2} - \phi\right) - \left(\frac{hc}{\lambda_2} - \phi\right) \]
Simplifying this gives:
\[ e(V_1 - V_2) = \frac{hc}{3\lambda_2} - \frac{hc}{\lambda_2} \]
\[ e(V_1 - V_2) = \frac{hc}{3\lambda_2} - \frac{3hc}{3\lambda_2} = -\frac{2hc}{3\lambda_2} \]

6. **Finding the Relation Between \( V_1 \) and \( V_2 \)**:
Dividing both sides by \( e \):
\[ V_1 - V_2 = -\frac{2hc}{3e\lambda_2} \]
Rearranging gives:
\[ V_1 = V_2 - \frac{2hc}{3e\lambda_2} \]

7. **Analyzing the Result**:
Since \( \frac{2hc}{3e\lambda_2} \) is a positive quantity, we can conclude:
\[ V_1 < V_2 \]
Thus, we can say that \( V_1 \) is less than \( V_2 \) by a certain amount.

### Final Answer:
The stopping potential \( V_1 \) is less than \( V_2 \) by a factor related to the wavelengths, specifically:
\[ V_1 < V_2 - \frac{2hc}{3e\lambda_2} \]