A light of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. The photoelectrons emitted from the surface have de-Broglie wavelength $\lambda_1$. Then ratio $\lambda: \lambda_1^2$ is ( $\mathrm{h}=$ Planck's constant, $\mathrm{c}=$ velocity of light, $\mathrm{m}=$ mass of electron)

The incident photon energy $E = \frac{hc}{\lambda}$ becomes the photoelectron's kinetic energy since the work function is negligible, so $KE = \frac{hc}{\lambda}$.

The photoelectron's kinetic energy also equals $\frac{p^2}{2m}$ where $p$ is its momentum. Equating these expressions:

$\frac{hc}{\lambda} = \frac{p^2}{2m}$

Using the de Broglie relation $\lambda_1 = \frac{h}{p}$, we substitute $p = \frac{h}{\lambda_1}$ into the energy equation:

$\frac{hc}{\lambda} = \frac{(h/\lambda_1)^2}{2m} = \frac{h^2}{2m\lambda_1^2}$

Solving for the ratio $\lambda : \lambda_1^2$:

$\frac{\lambda}{\lambda_1^2} = \frac{2mc}{h}$

Thus $\lambda : \lambda_1^2 = 2mc : h$, matching option C.