In an experiment of photoelectric effect the stopping potential was measured to be `V_(1)` and `V_(2)` with incident light of wavelength `lambda` and `(lambda)/(2)`, respectively. The relation between `V_(1)` and `V_(2)` is:

Options

A. `V_(2) gt 2V_(1)`
B. `V_(2) lt V_(1)`
C. `V_(1) lt V_(2) lt 2V_(1)`
D. `V_(2) = 2V_(1)`

Explanation

To solve the problem, we need to establish the relationship between the stopping potentials \( V_1 \) and \( V_2 \) for the two different wavelengths of incident light in the photoelectric effect.

### Step-by-Step Solution:

1. **Understanding Stopping Potential**:
The stopping potential \( V_0 \) is related to the maximum kinetic energy of the emitted electrons. The equation for stopping potential is given by:
\[ eV_0 = \frac{hc}{\lambda} - \phi \]
where \( \phi \) is the work function of the metal, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the incident light.

2. **Expression for \( V_1 \)**:
For the first wavelength \( \lambda \):
\[ eV_1 = \frac{hc}{\lambda} - \phi \]
Rearranging gives:
\[ V_1 = \frac{hc}{e\lambda} - \frac{\phi}{e} \]

3. **Expression for \( V_2 \)**:
For the second wavelength \( \frac{\lambda}{2} \):
\[ eV_2 = \frac{hc}{\frac{\lambda}{2}} - \phi \]
This simplifies to:
\[ eV_2 = \frac{2hc}{\lambda} - \phi \]
Rearranging gives:
\[ V_2 = \frac{2hc}{e\lambda} - \frac{\phi}{e} \]

4. **Comparing \( V_1 \) and \( V_2 \)**:
Now we have:
\[ V_1 = \frac{hc}{e\lambda} - \frac{\phi}{e} \]
\[ V_2 = \frac{2hc}{e\lambda} - \frac{\phi}{e} \]

5. **Finding the Relation**:
To compare \( V_1 \) and \( V_2 \), we can multiply \( V_1 \) by 2:
\[ 2V_1 = 2\left(\frac{hc}{e\lambda} - \frac{\phi}{e}\right) = \frac{2hc}{e\lambda} - \frac{2\phi}{e} \]

6. **Subtracting \( 2V_1 \) from \( V_2 \)**:
Now, we can find the difference:
\[ V_2 - 2V_1 = \left(\frac{2hc}{e\lambda} - \frac{\phi}{e}\right) - \left(\frac{2hc}{e\lambda} - \frac{2\phi}{e}\right) \]
Simplifying this gives:
\[ V_2 - 2V_1 = \frac{2\phi}{e} - \frac{\phi}{e} = \frac{\phi}{e} \]

7. **Conclusion**:
Since \( \frac{\phi}{e} > 0 \), we conclude that:
\[ V_2 > 2V_1 \]

### Final Result:
Thus, the relation between \( V_1 \) and \( V_2 \) is:
\[ V_2 > 2V_1 \]