When a certain metal surface is illuminated wth light of frequency v, the stopping potential for photoelectric current is `V_(0)`. When the same surface is illumiinated by light of frequency `(v)/(2)`, the stopping potential is `(V_(0))/(4)`. The threshold frequency ofr photoelectric emissiohn id

Options

A. `(v)/(6)`
B. `(v)/(3)`
C. `(2v)/(3)`
D. `(4v)/(3)`

Explanation

To solve the problem, we will use the photoelectric effect equations and the information provided about the stopping potentials and frequencies.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two scenarios:
- When light of frequency \( \nu \) is used, the stopping potential is \( V_0 \).
- When light of frequency \( \frac{\nu}{2} \) is used, the stopping potential is \( \frac{V_0}{4} \).

2. **Using the Photoelectric Equation**:
The photoelectric equation is given by:
\[ E = \phi + K.E_{max} \]
where \( E \) is the energy of the incident photon, \( \phi \) is the work function of the metal, and \( K.E_{max} \) is the maximum kinetic energy of the emitted electrons.

3. **First Scenario**:
For the first scenario with frequency \( \nu \):
\[ h\nu = \phi + eV_0 \]
(Equation 1)

4. **Second Scenario**:
For the second scenario with frequency \( \frac{\nu}{2} \):
\[ h\left(\frac{\nu}{2}\right) = \phi + e\left(\frac{V_0}{4}\right) \]
Simplifying this gives:
\[ \frac{h\nu}{2} = \phi + \frac{eV_0}{4} \]
(Equation 2)

5. **Rearranging Equation 2**:
Rearranging Equation 2, we can express \( \phi \):
\[ \phi = \frac{h\nu}{2} - \frac{eV_0}{4} \]

6. **Substituting \( \phi \) in Equation 1**:
Now, substitute \( \phi \) from Equation 2 into Equation 1:
\[ h\nu = \left(\frac{h\nu}{2} - \frac{eV_0}{4}\right) + eV_0 \]
Simplifying this:
\[ h\nu = \frac{h\nu}{2} + \frac{3eV_0}{4} \]

7. **Isolating \( h\nu \)**:
Rearranging gives:
\[ h\nu - \frac{h\nu}{2} = \frac{3eV_0}{4} \]
\[ \frac{h\nu}{2} = \frac{3eV_0}{4} \]
\[ h\nu = \frac{3eV_0}{2} \]

8. **Finding the Work Function \( \phi \)**:
Now substituting \( h\nu \) back into the expression for \( \phi \):
\[ \phi = \frac{3eV_0}{4} - \frac{eV_0}{4} = \frac{2eV_0}{4} = \frac{eV_0}{2} \]

9. **Finding the Threshold Frequency**:
The threshold frequency \( \nu_0 \) is related to the work function by:
\[ \phi = h\nu_0 \]
Thus,
\[ \nu_0 = \frac{\phi}{h} = \frac{eV_0/2}{h} = \frac{eV_0}{2h} \]

### Final Answer:
The threshold frequency for photoelectric emission is:
\[ \nu_0 = \frac{eV_0}{2h} \]