A photoelectric material having work-function ϕ _ 0 is illuminated...

According to the equation of photoelectric effect,

K{E}_\max = \dfrac{hc}{\lambda} - {ϕ}_{0}

Kinetic energy in terms of momentum is given as $KE = \dfrac{{p}^{2}}{2m}$

$\dfrac{{p}^{2}}{2m} = \dfrac{hc}{\lambda} - {ϕ}_{0}$

$\dfrac{{\left(\dfrac{h}{{\lambda}_{d}}\right)}^{2}}{2m} = \dfrac{hc}{\lambda} - {ϕ}_{0}$

Assuming small changes, differentiating both sides,

$\dfrac{{h}^{2}}{2m}\left(- \dfrac{2d{\lambda}_{d}}{{\lambda}_{d}^{3}}\right) = - \dfrac{hc}{{\lambda}^{2}}d\lambda$

$\dfrac{d{\lambda}_{d}}{d\lambda} ∝ \dfrac{{\lambda}_{d}^{3}}{{\lambda}^{2}}$