The threshold frequency for a certain photosensitive metal is `v_0`. When it is illuminated by light of frequency `v=2 v_0`, the stopping potential for photoelectric current is `V_0`. What will be the stopping potential when the same metal is illuminated by light of frequency `v=3v_0`?

Options

A. `1.5V_0`
B. `2 V_0`
C. `2.5 V_0`
D. `3V_0`

Explanation

To solve the problem, we will use the photoelectric effect equation, which relates the stopping potential to the frequency of the incident light and the threshold frequency of the material.

### Step-by-step Solution:

1. **Understanding the Photoelectric Effect**:
The photoelectric effect states that when light of a certain frequency shines on a photosensitive material, it can eject electrons from the surface of the material. The energy of the incident photons must be greater than the work function (related to the threshold frequency) for electrons to be emitted.

2. **Using the Photoelectric Equation**:
The energy of the incident photons can be expressed as:
\[ E = h \cdot v \]
where \( E \) is the energy of the photons, \( h \) is Planck's constant, and \( v \) is the frequency of the light.

3. **Threshold Frequency**:
The threshold frequency \( v_0 \) is the minimum frequency required to eject electrons from the metal. The work function \( \phi \) is given by:
\[ \phi = h \cdot v_0 \]

4. **Stopping Potential**:
The stopping potential \( V \) is related to the maximum kinetic energy of the emitted electrons. The maximum kinetic energy \( K.E. \) of the emitted electrons can be expressed as:
\[ K.E. = h \cdot v - \phi \]
The stopping potential is given by:
\[ eV = K.E. \]
Therefore, we can write:
\[ eV = h \cdot v - h \cdot v_0 \]

5. **First Scenario (Frequency \( v = 2v_0 \))**:
For the first case, where the frequency \( v = 2v_0 \):
\[ eV_0 = h \cdot (2v_0) - h \cdot v_0 \]
Simplifying this gives:
\[ eV_0 = h \cdot v_0 \]

6. **Second Scenario (Frequency \( v = 3v_0 \))**:
Now, we need to find the stopping potential when the frequency is \( v = 3v_0 \):
\[ eV = h \cdot (3v_0) - h \cdot v_0 \]
Simplifying this gives:
\[ eV = h \cdot (3v_0 - v_0) = h \cdot (2v_0) \]

7. **Relating Stopping Potential to Known Values**:
We already know from the first scenario that:
\[ eV_0 = h \cdot v_0 \]
Therefore, we can express \( h \cdot (2v_0) \) in terms of \( V_0 \):
\[ eV = 2 \cdot eV_0 \]
Thus:
\[ V = 2V_0 \]

### Final Answer:
The stopping potential when the same metal is illuminated by light of frequency \( v = 3v_0 \) will be:
\[ V = 2V_0 \]