When a certain photosensistive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photo current is `-V_(0)//2`. When the surface is illuminated by monochromatic light of frequency `v//2`, the stopping potential is `-V_(0)`. The threshold frequency gor photoelectric emission is :

Options

A. `(5v)/(3)`
B. `(4)/(3)v`
C. `2v`
D. `(3v)/(2)`

Explanation

To solve the problem, we will use the photoelectric effect principles and Einstein's photoelectric equation. We have two scenarios with different frequencies and stopping potentials.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- For frequency \( v \), the stopping potential is \( -\frac{V_0}{2} \).
- For frequency \( \frac{v}{2} \), the stopping potential is \( -V_0 \).

2. **Write the Einstein's Photoelectric Equation:**
The general equation for the photoelectric effect is given by:
\[ eV = h\nu - \phi \]
where:
- \( e \) is the charge of the electron,
- \( V \) is the stopping potential,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( \phi \) is the work function of the material.

3. **Set Up the Equations for Each Case:**
- For the first case (frequency \( v \)):
\[ -\frac{V_0}{2} = \frac{h\nu}{e} - \frac{\phi}{e} \]
Rearranging gives:
\[ e\left(-\frac{V_0}{2}\right) = h\nu - \phi \quad \text{(1)} \]

- For the second case (frequency \( \frac{v}{2} \)):
\[ -V_0 = \frac{h\left(\frac{v}{2}\right)}{e} - \frac{\phi}{e} \]
Rearranging gives:
\[ e(-V_0) = h\left(\frac{v}{2}\right) - \phi \quad \text{(2)} \]

4. **Express the Two Equations:**
- From equation (1):
\[ -\frac{eV_0}{2} = h\nu - \phi \]
- From equation (2):
\[ -eV_0 = \frac{hv}{2} - \phi \]

5. **Eliminate \( \phi \):**
Rearranging both equations to isolate \( \phi \):
- From (1):
\[ \phi = h\nu + \frac{eV_0}{2} \]
- From (2):
\[ \phi = \frac{hv}{2} + eV_0 \]

Setting the two expressions for \( \phi \) equal to each other:
\[ h\nu + \frac{eV_0}{2} = \frac{hv}{2} + eV_0 \]

6. **Simplify the Equation:**
Rearranging gives:
\[ h\nu - \frac{hv}{2} = eV_0 - \frac{eV_0}{2} \]
\[ h\nu - \frac{hv}{2} = \frac{eV_0}{2} \]
Factoring out \( h \):
\[ h\left(\nu - \frac{v}{2}\right) = \frac{eV_0}{2} \]

7. **Find the Threshold Frequency \( \nu_0 \):**
The work function \( \phi \) can also be expressed as:
\[ \phi = h\nu_0 \]
Setting the expression for \( \phi \) equal to \( \frac{3h\nu}{2} \):
\[ h\nu_0 = \frac{3h\nu}{2} \]
Dividing both sides by \( h \):
\[ \nu_0 = \frac{3\nu}{2} \]

### Final Answer:
The threshold frequency \( \nu_0 \) for photoelectric emission is:
\[ \nu_0 = \frac{3\nu}{2} \]