A light of `4 eV` incident on metal surface of work function `phi eV` . Another light of `2.5 eV` incident on another metal surface of work function `phi_2 eV` . Find the ratio of maximum velocity of photo electrons `(v_1/v_2)`.

Options

A. `(sqrt(4+phi_1/(2.5-phi_2))`
B. `(sqrt(4-phi_1/(2.5-phi_2))`
C. `(sqrt(3-phi_1/(2.5+phi_2))`
D. `(sqrt(6+phi_1/(2.5-phi_2))`

Answer

To solve the problem, we need to find the ratio of the maximum velocities of photoelectrons emitted from two different metal surfaces when illuminated by light of different energies.

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**:
The energy of the incident light is used to overcome the work function of the metal and the remaining energy is converted into kinetic energy of the emitted photoelectrons.

2. **Write the Energy Conservation Equation for the First Metal**:
For the first metal with work function \( \phi_1 \) and incident light energy of \( 4 \, \text{eV} \):
\[ K.E. = E - \phi_1 \]
where \( K.E. \) is the kinetic energy of the emitted electrons. The kinetic energy can also be expressed in terms of velocity \( v_1 \):
\[ \frac{1}{2} m v_1^2 = 4 - \phi_1 \quad \text{(Equation 1)} \]

3. **Write the Energy Conservation Equation for the Second Metal**:
For the second metal with work function \( \phi_2 \) and incident light energy of \( 2.5 \, \text{eV} \):
\[ K.E. = E - \phi_2 \]
Similarly, we can express this in terms of velocity \( v_2 \):
\[ \frac{1}{2} m v_2^2 = 2.5 - \phi_2 \quad \text{(Equation 2)} \]

4. **Divide the Two Equations**:
To find the ratio \( \frac{v_1}{v_2} \), we can divide Equation 1 by Equation 2:
\[ \frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{4 - \phi_1}{2.5 - \phi_2} \]
The \( \frac{1}{2} m \) cancels out:
\[ \frac{v_1^2}{v_2^2} = \frac{4 - \phi_1}{2.5 - \phi_2} \]

5. **Take the Square Root**:
To find the ratio of velocities, take the square root of both sides:
\[ \frac{v_1}{v_2} = \sqrt{\frac{4 - \phi_1}{2.5 - \phi_2}} \]

### Final Result:
The ratio of the maximum velocities of photoelectrons emitted from the two metals is given by:
\[ \frac{v_1}{v_2} = \sqrt{\frac{4 - \phi_1}{2.5 - \phi_2}} \]