A photosensitive metallic surfaces has work function `hv_(0)` . If photon of `2 hv_(0)` fall on the surface ,the electrons come out with a maxiumum velocity of `4 xx 10^(6)m//s` . If energy of photon is increased to `5hv_(0)` , the maximum velocity of photoelectrons will be :

Options

A. `2 xx 10^(7) m//s`
B. `8 xx 10^(6) m//s`
C. `2 xx 10^(6) m//s`
D. `8 xx 10^(5) m//s`

Answer

To solve the problem, we need to understand the relationship between the energy of the incident photons, the work function of the metal, and the kinetic energy of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Understanding Work Function and Photon Energy**:
- The work function of the metallic surface is given as \( hv_0 \).
- When a photon of energy \( 2hv_0 \) strikes the surface, the energy absorbed by the electron is \( 2hv_0 \).

2. **Calculating Kinetic Energy**:
- The kinetic energy (KE) of the emitted electrons can be calculated using the equation:
\[ KE = E_{photon} - \text{Work Function} \]
- For the first case with photon energy \( 2hv_0 \):
\[ KE = 2hv_0 - hv_0 = hv_0 \]

3. **Relating Kinetic Energy to Velocity**:
- The kinetic energy is also given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- We know from the problem that the maximum velocity \( v \) of the electrons is \( 4 \times 10^6 \, m/s \).
- Thus, we can write:
\[ hv_0 = \frac{1}{2} m (4 \times 10^6)^2 \]

4. **Increasing Photon Energy**:
- Now, if the energy of the photon is increased to \( 5hv_0 \), we can calculate the new kinetic energy:
\[ KE' = 5hv_0 - hv_0 = 4hv_0 \]

5. **Calculating New Maximum Velocity**:
- Using the kinetic energy formula again:
\[ 4hv_0 = \frac{1}{2} mv'^2 \]
- We can relate this to the previous kinetic energy:
\[ 4hv_0 = 4 \left(\frac{1}{2} mv^2\right) = 4 \left(\frac{1}{2} m (4 \times 10^6)^2\right) \]
- This implies:
\[ v'^2 = 4 \times (4 \times 10^6)^2 \]
- Taking the square root gives:
\[ v' = 2 \times (4 \times 10^6) = 8 \times 10^6 \, m/s \]

### Final Answer:
The maximum velocity of the photoelectrons when the energy of the photon is increased to \( 5hv_0 \) will be \( 8 \times 10^6 \, m/s \).