If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength `lambda` 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. `(3)/(4)lambda`
C. `(1)/(2)lambda`
D. `(2)/(3)lambda`

Explanation

To solve the problem, we need to find the wavelength of light required for a photoelectron with a momentum of 1.5p, given that p is the momentum of the fastest electron ejected from a metal surface after irradiation with light of wavelength λ. We will assume that the kinetic energy of the ejected photoelectron is very high compared to the work function of the metal.

### Step-by-Step Solution:

1. **Understand the relationship between energy and momentum:**
The kinetic energy (KE) of the ejected photoelectron can be expressed using the momentum (p):
\[ KE = \frac{p^2}{2m} \]
where m is the mass of the electron.

2. **Use the photoelectric equation:**
The energy of the incident photon can be expressed as:
\[ E = h\nu = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the light.

3. **Set up the energy balance equation:**
According to the photoelectric effect, the energy of the photon is equal to the kinetic energy of the ejected electron plus the work function (\(\phi\)):
\[ \frac{hc}{\lambda} = KE + \phi \]
Given that the kinetic energy is very high compared to the work function, we can approximate:
\[ \frac{hc}{\lambda} \approx KE \]

4. **Substituting the expression for kinetic energy:**
From step 1, we substitute the kinetic energy:
\[ \frac{hc}{\lambda} \approx \frac{p^2}{2m} \]

5. **For the momentum of 1.5p:**
Now, we consider the case where the momentum of the photoelectron is \(1.5p\). Thus, the kinetic energy for this scenario becomes:
\[ KE' = \frac{(1.5p)^2}{2m} = \frac{2.25p^2}{2m} = \frac{1.125p^2}{m} \]

6. **Set up the new energy balance equation:**
For the new momentum, we have:
\[ \frac{hc}{\lambda'} = KE' \approx \frac{2.25p^2}{2m} \]

7. **Divide the two equations:**
We can now divide the first equation (for momentum \(p\)) by the second equation (for momentum \(1.5p\)):
\[ \frac{\frac{hc}{\lambda}}{\frac{hc}{\lambda'}} = \frac{\frac{p^2}{2m}}{\frac{2.25p^2}{2m}} \]
This simplifies to:
\[ \frac{\lambda'}{\lambda} = \frac{1}{2.25} \]

8. **Solve for \(\lambda'\):**
Rearranging gives:
\[ \lambda' = \frac{\lambda}{2.25} \]
Since \(2.25 = \frac{9}{4}\), we can express this as:
\[ \lambda' = \frac{4}{9}\lambda \]

### Final Answer:
The wavelength of the light required for a photoelectron with momentum \(1.5p\) is:
\[ \lambda' = \frac{4}{9}\lambda \]