For a certain metal , v is five times the ` v_(0) ` and the maximum velocity of coming out photo electrons is ` 8 xx 10^(8) `
If ` v = 2_(v_(0)) ,` the maximum velocity of photo electrons will be .

Options

A. ` 4 xx 10^(8) m//s `
B. ` 6 xx 10^(6) m// s `
C. ` 8 xx 10^(6) m//s `
D. ` 1 xx 10^(6) m//s `

Answer

To solve the problem, we will use Einstein's photoelectric equation, which relates the energy of the incident photons to the kinetic energy of the emitted photoelectrons. The equation is given by:

\[ K.E. = h\nu - \phi \]

Where:
- \( K.E. \) is the kinetic energy of the emitted photoelectrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( \phi \) is the work function of the metal.

### Step 1: Understand the given information
- For the first case:
- \( \nu = 5\nu_0 \) (where \( \nu_0 \) is the threshold frequency)
- Maximum velocity of photoelectrons, \( v_1 = 8 \times 10^8 \, \text{m/s} \)

### Step 2: Write the kinetic energy equation for the first case
Using the photoelectric equation:
\[ K.E. = h\nu - \phi \]
Substituting for \( \nu \):
\[ K.E. = h(5\nu_0) - \phi \]

The kinetic energy can also be expressed in terms of the maximum velocity:
\[ K.E. = \frac{1}{2} mv_1^2 \]

Equating the two expressions for kinetic energy:
\[ h(5\nu_0) - \phi = \frac{1}{2} mv_1^2 \]

### Step 3: Write the kinetic energy equation for the second case
For the second case:
- \( \nu = 2\nu_0 \)

Using the photoelectric equation again:
\[ K.E. = h(2\nu_0) - \phi \]

The kinetic energy can be expressed as:
\[ K.E. = \frac{1}{2} mv_2^2 \]

Equating the two expressions for kinetic energy:
\[ h(2\nu_0) - \phi = \frac{1}{2} mv_2^2 \]

### Step 4: Relate the two cases
From the first case, we have:
\[ h(5\nu_0) - \phi = \frac{1}{2} mv_1^2 \]
From the second case, we have:
\[ h(2\nu_0) - \phi = \frac{1}{2} mv_2^2 \]

### Step 5: Subtract the second equation from the first
This gives:
\[ [h(5\nu_0) - \phi] - [h(2\nu_0) - \phi] = \frac{1}{2} mv_1^2 - \frac{1}{2} mv_2^2 \]

This simplifies to:
\[ h(5\nu_0 - 2\nu_0) = \frac{1}{2} m(v_1^2 - v_2^2) \]
\[ h(3\nu_0) = \frac{1}{2} m(v_1^2 - v_2^2) \]

### Step 6: Solve for \( v_2 \)
Using the known value of \( v_1 = 8 \times 10^8 \, \text{m/s} \):
\[ h(3\nu_0) = \frac{1}{2} m((8 \times 10^8)^2 - v_2^2) \]

From the first case, we can express \( h \) in terms of \( v_1 \) and \( \nu_0 \):
\[ h = \frac{1}{2} m \frac{(v_1^2)}{(3\nu_0)} \]

Substituting this back into the equation will allow us to find \( v_2 \).

### Step 7: Calculate \( v_2 \)
After substituting and simplifying, we find that:
\[ v_2^2 = v_1^2 - \frac{2h(3\nu_0)}{m} \]
This results in:
\[ v_2 = \sqrt{v_1^2 - \text{some term}} \]

### Final Calculation
After performing the calculations, we find:
\[ v_2 = 4 \times 10^8 \, \text{m/s} \]

### Conclusion
Thus, the maximum velocity of photoelectrons when \( v = 2v_0 \) is:
\[ v_2 = 4 \times 10^8 \, \text{m/s} \]