For a certain metal `v = 2 v_(0)` and the electrons come out with a maximum velocity of `4 xx 10^(6)` m/s . If the value of ` v = 5 v_(0)` , then maximum velocity of the photoelectrons will be

Options

A. `2xx10^7" ms"^(-1)`
B. `8xx10^7" ms"^(-1)`
C. `2xx10^6" ms"^(-1)`
D. `8xx10^5" ms"^(-1)`

Answer

To solve the problem, we need to determine the maximum velocity of photoelectrons when the value of \( v \) is increased from \( 2v_0 \) to \( 5v_0 \). We will use the principles of photoelectric effect and kinetic energy.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- For the first case, we have:
- \( v = 2v_0 \)
- Maximum velocity of photoelectrons, \( v_{\text{max1}} = 4 \times 10^6 \, \text{m/s} \)

2. **Using the Photoelectric Equation:**
- The kinetic energy of the emitted electrons can be expressed as:
\[ KE = \frac{1}{2} mv^2 = E - \phi \]
- Where \( E \) is the energy of the incoming photons and \( \phi \) is the work function (threshold energy).
- For the first case:
\[ E = h \nu = h (2 \nu_0) = 2h \nu_0 \]
\[ \phi = h \nu_0 \]
- Therefore, the equation becomes:
\[ \frac{1}{2} m (v_{\text{max1}})^2 = 2h \nu_0 - h \nu_0 = h \nu_0 \]

3. **Substituting Known Values:**
- From the equation:
\[ \frac{1}{2} m (4 \times 10^6)^2 = h \nu_0 \]
- Rearranging gives:
\[ h \nu_0 = \frac{1}{2} m (4 \times 10^6)^2 \]

4. **Considering the Second Case:**
- For the second case where \( v = 5v_0 \):
\[ E = h \nu = h (5 \nu_0) = 5h \nu_0 \]
- The equation becomes:
\[ \frac{1}{2} m (v_{\text{max2}})^2 = 5h \nu_0 - h \nu_0 = 4h \nu_0 \]

5. **Substituting \( h \nu_0 \) from the First Case:**
- We already found that \( h \nu_0 = \frac{1}{2} m (4 \times 10^6)^2 \), so substituting this into the second case gives:
\[ \frac{1}{2} m (v_{\text{max2}})^2 = 4 \left(\frac{1}{2} m (4 \times 10^6)^2\right) \]
- Simplifying:
\[ (v_{\text{max2}})^2 = 4 (4 \times 10^6)^2 \]

6. **Taking the Square Root:**
- Taking the square root of both sides:
\[ v_{\text{max2}} = 2 \times (4 \times 10^6) = 8 \times 10^6 \, \text{m/s} \]

### Final Answer:
The maximum velocity of the photoelectrons when \( v = 5v_0 \) is:
\[ \boxed{8 \times 10^6 \, \text{m/s}} \]