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 λd , then :

The correct relation is:
$\lambda =\left(\frac{2mc}{h}\right){\lambda }_d^2$

Step-by-Step Derivation:

1. Energy of the Incident Photon:
For an electromagnetic wave of wavelength $\lambda$, the energy $E$ of each incident photon is given by Planck's relation:
$E=\frac{hc}{\lambda }$
where $h$ is Planck's constant and $c$ is the speed of light.

2. Kinetic Energy of the Photoelectron:
According to Einstein's photoelectric equation, the maximum kinetic energy $K$ of the emitted photoelectron is related to the photon energy $E$ and the work function of the surface $\Phi$ by:
$K=E-\Phi$
Since the work function $\Phi$ of the photosensitive surface is negligible ($\Phi \approx 0$), we have:
$K=E=\frac{hc}{\lambda }$

3. de Broglie Wavelength of the Photoelectron:
The de Broglie wavelength ${\lambda }_d$ of a photoelectron of mass $m$ and kinetic energy $K$ is given by:
${\lambda }_d=\frac{h}{\sqrt{2mK}}$

4. Establishing the Relation:
Squaring both sides of the de Broglie wavelength equation, we get:
${\lambda }_d^2=\frac{h^2}{2mK}$
Substitute the value of kinetic energy $K=\frac{hc}{\lambda }$ into the equation:
${\lambda }_d^2=\frac{h^2}{2m\left(\frac{hc}{\lambda }\right)}$
Simplify the fraction:
${\lambda }_d^2=\frac{h^2\lambda }{2mhc}$
${\lambda }_d^2=\frac{h\lambda }{2mc}$

Rearranging this equation to solve for the incident wavelength $\lambda$:
$\lambda =\left(\frac{2mc}{h}\right){\lambda }_d^2$