Here, $$\lambda$$ is the wavelength of incident light and $$\lambda$$_d is the de Broglie wavelength of the fastest photoelectron. The fastest ejected photoelectron has the maximum kinetic energy which is given by

$${K_{\max }} = {{hc} \over \lambda } - {\phi _0}$$ .....(1)

The de Broglie wavelength of the photoelectron having kinetic energy $${K_{\max }} = {{{p^2}} \over {2m}}$$ is given by

$${\lambda _d} = {h \over p} = {h \over {\sqrt {2m{K_{\max }}} }}$$ ..... (2)

where m is the mass of the electron and p is its linear momentum. Eliminate K_max from equations (1) and (2) to get

$${{{h^2}} \over {2m\lambda _d^2}} = {{hc} \over \lambda } - {\phi _0}$$ ...... (3)

Differentiate equation (3) to get

$$( - 2){{{h^2}} \over {2m\lambda _d^3}}\Delta {\lambda _d} = ( - 1){{hc} \over {{\lambda ^2}}}\Delta \lambda $$,

which gives

$${{\Delta {\lambda _d}} \over {\Delta \lambda }} = {{mc} \over h}{{\lambda _d^3} \over {{\lambda ^2}}}$$.