For certain metal incident frequency `nu` is five times threshold frequency `nu_(0)` and the maximum velocity of the photoelectrons is `8xx10^(6)ms^(-1)`. If incident photon frequency is `2nu_(0)`, the maximum velocity of photoelectrons will be

Options

A. `4xx10^(6)ms^(-1)`
B. `6xx10^(6)ms^(-1)`
C. `8xx10^(6)ms^(-1)`
D. `1xx10^(6)ms^(-1)`

Answer

To solve the problem step-by-step, we will use the principles of the photoelectric effect and Einstein's equation.

### Step-by-Step Solution:

1. **Identify Given Information:**
- The incident frequency \( \nu = 5 \nu_0 \)
- The maximum velocity of photoelectrons \( v_1 = 8 \times 10^6 \, \text{m/s} \)
- The new incident frequency \( \nu' = 2 \nu_0 \)

2. **Use Einstein's Photoelectric Equation:**
The kinetic energy (KE) of the emitted photoelectrons can be expressed using Einstein's equation:
\[ KE = h\nu - h\nu_0 \]
where \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \nu_0 \) is the threshold frequency.

3. **Calculate Kinetic Energy for the First Case:**
For the first case where \( \nu = 5 \nu_0 \):
\[ KE_1 = h(5\nu_0) - h\nu_0 = 4h\nu_0 \]

4. **Relate Kinetic Energy to Maximum Velocity:**
The kinetic energy can also be expressed in terms of the maximum velocity of the photoelectrons:
\[ KE = \frac{1}{2} mv^2 \]
Thus, for the first case:
\[ KE_1 = \frac{1}{2} m v_1^2 \]
Setting the two expressions for \( KE_1 \) equal gives:
\[ 4h\nu_0 = \frac{1}{2} m (8 \times 10^6)^2 \]

5. **Calculate Kinetic Energy for the Second Case:**
For the second case where \( \nu' = 2\nu_0 \):
\[ KE_2 = h(2\nu_0) - h\nu_0 = h\nu_0 \]

6. **Relate Kinetic Energy to Maximum Velocity for the Second Case:**
Similarly, for the second case:
\[ KE_2 = \frac{1}{2} m v_2^2 \]
Setting the two expressions for \( KE_2 \) equal gives:
\[ h\nu_0 = \frac{1}{2} m v_2^2 \]

7. **Form a Ratio of Kinetic Energies:**
Now, we can form a ratio of the kinetic energies from both cases:
\[ \frac{KE_1}{KE_2} = \frac{4h\nu_0}{h\nu_0} = 4 \]

8. **Relate Velocities Using Kinetic Energy Ratio:**
Since \( KE \) is proportional to the square of the velocity:
\[ \frac{KE_1}{KE_2} = \frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{v_1^2}{v_2^2} \]
This gives us:
\[ \frac{v_1^2}{v_2^2} = 4 \implies v_1^2 = 4 v_2^2 \implies v_2^2 = \frac{v_1^2}{4} \]

9. **Calculate Maximum Velocity \( v_2 \):**
Taking the square root:
\[ v_2 = \frac{v_1}{2} \]
Substituting \( v_1 = 8 \times 10^6 \, \text{m/s} \):
\[ v_2 = \frac{8 \times 10^6}{2} = 4 \times 10^6 \, \text{m/s} \]

### Final Answer:
The maximum velocity of photoelectrons when the incident photon frequency is \( 2 \nu_0 \) is \( 4 \times 10^6 \, \text{m/s} \).