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, we can use the principles of the photoelectric effect and the kinetic energy of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The incident frequency \( \nu = 5 \nu_0 \) (where \( \nu_0 \) is the threshold frequency).
- The maximum velocity of the photoelectrons at this frequency is \( v_1 = 8 \times 10^6 \, \text{m/s} \).
- We need to find the maximum velocity \( v_2 \) when the incident frequency is \( \nu = 2 \nu_0 \).

2. **Using the Photoelectric Equation**:
The kinetic energy of the emitted photoelectrons can be expressed using the photoelectric effect equation:
\[ K.E. = h\nu - h\nu_0 \]
where \( K.E. \) is the kinetic energy of the photoelectrons.

3. **First Case**:
For the first case where \( \nu = 5 \nu_0 \):
\[ K.E. = h(5\nu_0) - h\nu_0 = 4h\nu_0 \]
The kinetic energy can also be expressed in terms of the maximum velocity:
\[ K.E. = \frac{1}{2} mv_1^2 \]
Therefore, we can equate the two expressions:
\[ \frac{1}{2} mv_1^2 = 4h\nu_0 \tag{1} \]

4. **Second Case**:
For the second case where \( \nu = 2 \nu_0 \):
\[ K.E. = h(2\nu_0) - h\nu_0 = h\nu_0 \]
Again, using the kinetic energy expression:
\[ K.E. = \frac{1}{2} mv_2^2 \]
Thus, we have:
\[ \frac{1}{2} mv_2^2 = h\nu_0 \tag{2} \]

5. **Relating the Two Cases**:
From equation (1), we can express \( h\nu_0 \) in terms of \( v_1 \):
\[ h\nu_0 = \frac{1}{8} mv_1^2 \]
Substituting this into equation (2):
\[ \frac{1}{2} mv_2^2 = \frac{1}{8} mv_1^2 \]
Dividing both sides by \( \frac{1}{2} m \):
\[ v_2^2 = \frac{1}{4} v_1^2 \]
Taking the square root:
\[ v_2 = \frac{1}{2} v_1 \]

6. **Calculating \( v_2 \)**:
Substituting the value of \( v_1 \):
\[ v_2 = \frac{1}{2} (8 \times 10^6) = 4 \times 10^6 \, \text{m/s} \]

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

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