A photoelectric material having work-function $\phi_{0}$ is illuminated with light of wavelength $\lambda\left(\lambda<\frac{h c}{\phi_{0}}\right)$. The fastest photoelectron has a de Broglie wavelength $\lambda_{d}$. A change in wavelength of the incident light by $\Delta \lambda$ results in a change $\Delta \lambda_{d}$ in $\lambda_{d}$. Then the ratio $\Delta \lambda_{d} / \Delta \lambda$ is proportional to

(D) $\lambda_{d}^{3} / \lambda^{2}$

(B) $\lambda_{d}^{2} / \lambda^{2}$

(C) $\lambda_{d}^{3} / \lambda$

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Modern Physics

A photoelectric material having work-function

ϕ_{0}

is illuminated with light of wavelength

λ (λ < \frac{h c}{ϕ _{0}})

. The fastest photoelectron has a de Broglie wavelength

λ_{d}

. A change in wavelength of the incident light by

Δ λ

results in a change

Δ λ_{d}

λ_{d}

. Then the ratio

Δ λ_{d} /Δ λ

is proportional to

Answer: (D)

From photoelectric effect:

\frac{p ^{2}}{2 m} = \frac{h c}{λ} - ϕ_{0}

. Using de Broglie relation

p = \frac{h}{λ _{d}}

, we get

\frac{h ^{2}}{2 m λ _{d}^{2}} = \frac{h c}{λ} - ϕ_{0}

. Differentiating:

- \frac{h ^{2}}{m} \frac{1}{λ _{d}^{3}} d λ_{d} = \frac{h c}{λ ^{2}} d λ

. Therefore,

\frac{Δ λ _{d}}{Δ λ} \propto \frac{λ _{d}^{3}}{λ ^{2}}

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Single correct option