If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength i, then for 1.5 p momentum of the photoelectron, the wavelength of the light should be (assume kinetic energy of ejected photoelectron to be very high in comparison to work function):

Options

A. `4/9 lambda`
B. `2/3 lambda`
C. `3/4 lambda`
D. `1/2 lambda`

Answer

To solve the problem, we need to find the wavelength of light when the momentum of the photoelectron is 1.5 times the initial momentum \( p \). We will use the relationship between energy, momentum, and wavelength in the context of the photoelectric effect.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the initial momentum of the fastest electron be \( p \).
- The initial wavelength of the light is \( \lambda \).
- The final momentum of the photoelectron is \( 1.5p \).

2. **Relate Energy to Momentum**:
- The total energy \( E \) of the photon can be expressed as:
\[ E = \frac{hc}{\lambda} \]
- The kinetic energy \( KE \) of the ejected electron can be expressed in terms of momentum:
\[ KE = \frac{p^2}{2m} \]
- According to the problem, the kinetic energy is much greater than the work function, so we can ignore the work function in our calculations.

3. **Set Up the Energy Equation**:
- Since the total energy is equal to the kinetic energy of the ejected electron:
\[ \frac{hc}{\lambda} = KE \]
- Substituting for kinetic energy:
\[ \frac{hc}{\lambda} = \frac{p^2}{2m} \]

4. **Relate Initial and Final Momentum**:
- For the final momentum \( p_2 = 1.5p \):
\[ KE_2 = \frac{(1.5p)^2}{2m} = \frac{2.25p^2}{2m} = \frac{1.125p^2}{m} \]
- The energy corresponding to this momentum is:
\[ \frac{hc}{\lambda_2} = \frac{1.125p^2}{m} \]

5. **Establish the Relationship Between Wavelengths**:
- We can now set up the ratio of the energies for the two cases:
\[ \frac{hc}{\lambda} = \frac{p^2}{2m} \quad \text{and} \quad \frac{hc}{\lambda_2} = \frac{1.125p^2}{m} \]
- Dividing the two equations:
\[ \frac{\frac{hc}{\lambda}}{\frac{hc}{\lambda_2}} = \frac{\frac{p^2}{2m}}{\frac{1.125p^2}{m}} \]
- Simplifying gives:
\[ \frac{\lambda_2}{\lambda} = \frac{2}{1.125} = \frac{16}{9} \]

6. **Final Calculation**:
- Rearranging gives:
\[ \lambda_2 = \frac{4}{9} \lambda \]

### Conclusion:
The wavelength of the light when the momentum of the photoelectron is \( 1.5p \) is:
\[ \lambda_2 = \frac{4}{9} \lambda \]