The threshold frequency for a certain metal is `v_(0)`. When a certain radiation of frequency `2v_(0)` is incident on this metal surface the maximum velocity of the photoelectrons emitted is `2xx10^(6)ms^(-1)`. If a radiation of frequency `3v_(0)`, is incident on the same metal surface the maximum velocity of the photoelectrons emitted (in `m.s^(-1)` ) is

Options

A. `2xx10^(6)`
B. `2sqrt2xx10^(6)`
C. `4sqrt2xx10^(6)`
D. `4sqrt3xx10^(6)`

Answer

To solve the problem, we will use the concepts of the photoelectric effect and the relationship between the kinetic energy of emitted photoelectrons and the frequency of the incident radiation.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Threshold frequency of the metal: \( v_0 \)
- Frequency of incident radiation: \( 2v_0 \)
- Maximum velocity of photoelectrons emitted when frequency is \( 2v_0 \): \( v = 2 \times 10^6 \, \text{m/s} \)

2. **Use the Einstein's Photoelectric Equation:**
The kinetic energy (KE) of the emitted photoelectrons can be expressed as:
\[ KE = E_{incident} - W \]
where \( E_{incident} \) is the energy of the incident photons and \( W \) is the work function of the metal.

3. **Calculate the Work Function:**
The energy of the incident radiation at frequency \( 2v_0 \) is:
\[ E_{incident} = h \cdot 2v_0 \]
The work function \( W \) is given by:
\[ W = h \cdot v_0 \]
Therefore, the kinetic energy can be expressed as:
\[ KE = h \cdot 2v_0 - h \cdot v_0 = h \cdot v_0 \]

4. **Relate Kinetic Energy to Velocity:**
The kinetic energy of the emitted photoelectrons is also given by:
\[ KE = \frac{1}{2} m v^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ h \cdot v_0 = \frac{1}{2} m (2 \times 10^6)^2 \]

5. **Calculate the Kinetic Energy for the Second Case:**
Now, consider the second case where the frequency of the incident radiation is \( 3v_0 \):
\[ E_{incident} = h \cdot 3v_0 \]
The kinetic energy in this case will be:
\[ KE' = h \cdot 3v_0 - h \cdot v_0 = h \cdot 2v_0 \]

6. **Relate Kinetic Energy to Velocity for the Second Case:**
Using the kinetic energy formula again:
\[ KE' = \frac{1}{2} m v_2^2 \]
Setting this equal to the expression for kinetic energy gives:
\[ h \cdot 2v_0 = \frac{1}{2} m v_2^2 \]

7. **Relate the Two Cases:**
From the first case, we had:
\[ h \cdot v_0 = \frac{1}{2} m (2 \times 10^6)^2 \]
We can express \( h \cdot v_0 \) in terms of \( v \):
\[ h \cdot v_0 = \frac{1}{2} m (2 \times 10^6)^2 \]
Now substituting this into the second case:
\[ 2 \cdot \frac{1}{2} m (2 \times 10^6)^2 = \frac{1}{2} m v_2^2 \]
This simplifies to:
\[ v_2^2 = 2 \cdot (2 \times 10^6)^2 \]
Taking the square root:
\[ v_2 = \sqrt{2} \cdot 2 \times 10^6 = 2\sqrt{2} \times 10^6 \]

8. **Final Calculation:**
The maximum velocity of the photoelectrons emitted when the frequency is \( 3v_0 \) is:
\[ v_2 = 2\sqrt{2} \times 10^6 \, \text{m/s} \]

### Final Answer:
The maximum velocity of the photoelectrons emitted when a radiation of frequency \( 3v_0 \) is incident on the same metal surface is \( 2\sqrt{2} \times 10^6 \, \text{m/s} \).