Two photons of energies twice and thrice the work function of a metal are incident on the metal surface .Then, the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively ,is

Options

A. `sqrt2:1`
B. `sqrt3:3`
C. `sqrt3:sqrt2`
D. `1:sqrt2`

Answer

To solve the problem, we need to determine the maximum velocities of the photoelectrons emitted when two photons of different energies strike a metal surface. We will use the photoelectric effect principles to derive the required ratio.

### Step-by-Step Solution:

1. **Define the Work Function**:
Let the work function of the metal be denoted as \( \phi \).

2. **Photon Energies**:
- For the first photon, the energy is \( E_1 = 2\phi \).
- For the second photon, the energy is \( E_2 = 3\phi \).

3. **Maximum Kinetic Energy**:
The maximum kinetic energy (K.E.) of the emitted photoelectrons can be expressed using the equation:
\[ K.E. = E - \phi \]
where \( E \) is the energy of the incident photon.

4. **Calculate Kinetic Energy for Each Case**:
- For the first photon (energy \( 2\phi \)):
\[ K.E_1 = E_1 - \phi = 2\phi - \phi = \phi \]
- For the second photon (energy \( 3\phi \)):
\[ K.E_2 = E_2 - \phi = 3\phi - \phi = 2\phi \]

5. **Relate Kinetic Energy to Velocity**:
The kinetic energy of a photoelectron can also be expressed in terms of its velocity:
\[ K.E = \frac{1}{2}mv^2 \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

6. **Express Velocities in Terms of Kinetic Energy**:
- For the first case:
\[ K.E_1 = \frac{1}{2}mv_1^2 \implies \phi = \frac{1}{2}mv_1^2 \implies v_1^2 = \frac{2\phi}{m} \implies v_1 = \sqrt{\frac{2\phi}{m}} \]
- For the second case:
\[ K.E_2 = \frac{1}{2}mv_2^2 \implies 2\phi = \frac{1}{2}mv_2^2 \implies v_2^2 = \frac{4\phi}{m} \implies v_2 = \sqrt{\frac{4\phi}{m}} \]

7. **Calculate the Ratio of Velocities**:
Now we can find the ratio of the maximum velocities \( v_1 \) and \( v_2 \):
\[ \frac{v_1}{v_2} = \frac{\sqrt{\frac{2\phi}{m}}}{\sqrt{\frac{4\phi}{m}}} = \frac{\sqrt{2\phi}}{\sqrt{4\phi}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \]

8. **Final Ratio**:
Therefore, the ratio of the maximum velocities of the photoelectrons emitted in the two cases is:
\[ v_1 : v_2 = 1 : \sqrt{2} \]

### Conclusion:
The correct answer is option D: \( 1 : \sqrt{2} \).

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