An electromagnetic wave of wavelength ‘λ’ is incident on a photosensitive surface of negligible work function. If ‘m’ mass is of photoelectron emitted from the surface has de-Broglie wavelength λ_d, then :

When light of wavelength $\lambda$ strikes a photosensitive surface whose work function is negligible, the entire photon energy is transferred to the emitted electron as kinetic energy.

1. **Photon energy**

$\mathrm{E\_\gamma }=\frac{h}{c}$

2. **Kinetic energy of the photoelectron** (since work function �� 0)

$K=\frac{1}{2}m{v}^2$

3. **De Broglie relation for the electron**

$p=mv=\frac{h}{{\lambda }_d}$

From this, the electron speed is

$v=\frac{h}{m}$

4. **Insert v into the kinetic‑energy expression**

$K=\frac{1}{2}m{\frac{h}{m}}^2=\frac{h^2}{2}$

5. **Equate photon energy to kinetic energy**

$\frac{h}{c}=\frac{h^2}{2}$

6. **Solve for the incident wavelength λ**

$\lambda =\frac{2}{h}mc{{\lambda }_d}^2$

Thus the relationship between the wavelength of the incident electromagnetic wave and the de‑Broglie wavelength of the emitted photoelectron is

$\lambda =\left(\frac{2mc}{h}\right){{\lambda }_d}^2$

This matches the given correct option.