For a certain metal v is the five times of `v_(0)` and the maximum velocity of coming out photons is `8 xx 10^(6)m//s`. If `v = 2v_(0)`, then maximum velocity of photoelectrons will be

Options

A. `4xx10^(6)m" "s^(-1)`
B. `6xx10^(6)m" "s^(-1)`
C. `8xx10^(6)m" "s^(-1)`
D. `1xx10^(6)m" "s^(-1)`

Explanation

To solve the problem step by step, we can follow the reasoning based on the principles of the photoelectric effect and the equations derived from it.

### Step 1: Understand the Initial Conditions
We are given:
- The frequency of excitation \( v = 5v_0 \)
- The maximum velocity of photoelectrons when \( v = 5v_0 \) is \( v_1 = 8 \times 10^6 \, \text{m/s} \)

### Step 2: Apply the Photoelectric Equation
According to Einstein's photoelectric equation:
\[ K.E. = E_{\text{photon}} - \phi \]
Where:
- \( K.E. = \frac{1}{2} mv^2 \) (kinetic energy of the photoelectron)
- \( E_{\text{photon}} = h \cdot v \) (energy of the incident photon)
- \( \phi \) is the work function, which can be expressed as \( h \cdot v_0 \)

For the first condition:
\[ \frac{1}{2} mv_1^2 = h \cdot v - h \cdot v_0 \]
Substituting \( v = 5v_0 \):
\[ \frac{1}{2} mv_1^2 = h(5v_0) - h(v_0) = 4h v_0 \]

### Step 3: Rearranging the Equation
From the equation derived:
\[ \frac{1}{2} mv_1^2 = 4h v_0 \]
This can be rearranged to find \( v_1 \):
\[ v_1^2 = \frac{8h v_0}{m} \]

### Step 4: Analyze the Second Condition
Now, we need to analyze the second condition where \( v = 2v_0 \):
\[ \frac{1}{2} mv_2^2 = h(2v_0) - h(v_0) = h v_0 \]
This simplifies to:
\[ \frac{1}{2} mv_2^2 = h v_0 \]

### Step 5: Rearranging the Second Equation
From the equation for the second condition:
\[ v_2^2 = \frac{2h v_0}{m} \]

### Step 6: Relate the Two Conditions
Now we can relate \( v_2 \) to \( v_1 \) by dividing the two equations:
\[ \frac{v_2^2}{v_1^2} = \frac{\frac{2h v_0}{m}}{\frac{8h v_0}{m}} = \frac{2}{8} = \frac{1}{4} \]
Taking the square root gives:
\[ \frac{v_2}{v_1} = \frac{1}{2} \]

### Step 7: Calculate \( v_2 \)
Now substituting \( v_1 = 8 \times 10^6 \, \text{m/s} \):
\[ v_2 = \frac{v_1}{2} = \frac{8 \times 10^6}{2} = 4 \times 10^6 \, \text{m/s} \]

### Final Answer
The maximum velocity of the photoelectrons when \( v = 2v_0 \) is:
\[ \boxed{4 \times 10^6 \, \text{m/s}} \]