An EM wave of wavelength `lambda` is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de-Broglie wavelength `lambda_1`, prove that
`lambda=((2mc)/h)lambda_1^2`

Options

A. `lamda=(mc)/(h)lamda^('2)`
B. `lamda=(3mc)/(2h)lamda^('2)`
C. `lamda=(2mc)/(h)lamda^('2)`
D. `lamda=(5mc)/(h)lamda^('2)`

Explanation

To prove the relation \( \lambda = \frac{2mc}{h} \lambda_1^2 \), we will follow a systematic approach using the principles of photoelectric effect and de-Broglie wavelength.

### Step-by-Step Solution:

1. **Understanding the Energy of Incident Photons:**
The energy \( E \) of an incident photon can be expressed using the formula:
\[ E = h \nu \]
where \( h \) is Planck's constant and \( \nu \) is the frequency of the electromagnetic wave. The frequency is related to the wavelength \( \lambda \) by the equation:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light.

2. **Relating Photon Energy to Wavelength:**
Substituting the expression for frequency into the energy equation gives:
\[ E = h \left( \frac{c}{\lambda} \right) = \frac{hc}{\lambda} \]

3. **Kinetic Energy of Emitted Electrons:**
Since the work function is negligible, the kinetic energy \( KE \) of the emitted photoelectrons is equal to the energy of the incident photon:
\[ KE = E = \frac{hc}{\lambda} \]

4. **Using the de-Broglie Wavelength:**
The de-Broglie wavelength \( \lambda_1 \) of the emitted electrons can be expressed as:
\[ \lambda_1 = \frac{h}{p} \]
where \( p \) is the momentum of the electrons. The momentum can also be expressed in terms of kinetic energy:
\[ KE = \frac{p^2}{2m} \implies p = \sqrt{2m \cdot KE} \]

5. **Substituting Kinetic Energy:**
Substituting the expression for \( KE \) into the momentum equation gives:
\[ p = \sqrt{2m \cdot \frac{hc}{\lambda}} = \sqrt{\frac{2m \cdot hc}{\lambda}} \]

6. **Relating de-Broglie Wavelength to Momentum:**
Now substituting \( p \) into the de-Broglie wavelength equation:
\[ \lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{\frac{2m \cdot hc}{\lambda}}} \]
Squaring both sides:
\[ \lambda_1^2 = \frac{h^2 \lambda}{2m \cdot hc} \]

7. **Rearranging the Equation:**
Rearranging the above equation to express \( \lambda \):
\[ \lambda = \frac{2m \cdot c}{h} \lambda_1^2 \]

8. **Final Result:**
Thus, we have proved that:
\[ \lambda = \frac{2mc}{h} \lambda_1^2 \]