Threshold frequency of a surface is ` v_0`. It is illuminated by 3 `v_0` frequency, then maximum speed of photo electrons is V m/sec. What will be maximum speed if incident frequency is ` 9v _ 0` ?

Answer

To solve the problem, we will use Einstein's photoelectric equation, which relates the kinetic energy of the emitted photoelectrons to the energy of the incident photons and the work function of the material.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a surface with a threshold frequency \( v_0 \).
- When illuminated with a frequency \( 3v_0 \), the maximum speed of the emitted photoelectrons is \( V \) m/s.
- We need to find the maximum speed of the photoelectrons when the frequency is \( 9v_0 \).

2. **Using Einstein's Photoelectric Equation**:
The kinetic energy (KE) of the emitted photoelectrons can be expressed as:
\[ KE = E_{\text{photon}} - \text{Work Function} \]
Where:
- \( E_{\text{photon}} = h \nu \)
- Work Function \( \phi = h v_0 \)

3. **Case 1: Incident Frequency \( 3v_0 \)**:
- The energy of the incident photons when the frequency is \( 3v_0 \) is:
\[ E_{\text{photon}} = h(3v_0) = 3hv_0 \]
- The kinetic energy of the emitted electrons is:
\[ KE_1 = 3hv_0 - hv_0 = 2hv_0 \]
- This kinetic energy can also be expressed in terms of the maximum speed \( V \):
\[ KE_1 = \frac{1}{2} m V^2 \]
- Therefore, we have:
\[ \frac{1}{2} m V^2 = 2hv_0 \quad \text{(1)} \]

4. **Case 2: Incident Frequency \( 9v_0 \)**:
- The energy of the incident photons when the frequency is \( 9v_0 \) is:
\[ E_{\text{photon}} = h(9v_0) = 9hv_0 \]
- The kinetic energy of the emitted electrons is:
\[ KE_2 = 9hv_0 - hv_0 = 8hv_0 \]
- This kinetic energy can also be expressed in terms of the maximum speed \( V_2 \):
\[ KE_2 = \frac{1}{2} m V_2^2 \]
- Therefore, we have:
\[ \frac{1}{2} m V_2^2 = 8hv_0 \quad \text{(2)} \]

5. **Relating the Two Cases**:
- From equation (1):
\[ V^2 = \frac{4hv_0}{m} \]
- From equation (2):
\[ V_2^2 = \frac{16hv_0}{m} \]
- Now, we can relate \( V_2 \) to \( V \):
\[ \frac{V_2^2}{V^2} = \frac{16hv_0/m}{4hv_0/m} = 4 \]
- Taking the square root gives:
\[ \frac{V_2}{V} = 2 \implies V_2 = 2V \]

6. **Final Result**:
- The maximum speed of the photoelectrons when the incident frequency is \( 9v_0 \) is:
\[ V_2 = 2V \text{ m/s} \]