JEE Advanced 2017 · Paper 2 · Physics Q3

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PhysicsSingle correct+3 / -1Question 3 of 54

A photoelectric material having work-function $\phi_0$ is illuminated with light of wavelength $\lambda$ $\left(\lambda < \frac{hc}{\phi_0}\right)$ . 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

\lambda_d/\lambda

\lambda_d^2/\lambda^2

\lambda_d^3/\lambda

\lambda_d^3/\lambda^2

Answer and worked solution

Official answer: D

Fastest electron. Its kinetic energy is
$\frac{h^2}{2m\lambda_d^2}=\frac{hc}\lambda-\phi_0.$

Differentiate.
$-\frac{h^2}{m\lambda_d^3}\,d\lambda_d=-\frac{hc}{\lambda^2}\,d\lambda\implies\frac{d\lambda_d}{d\lambda}=\frac{mc}h\cdot\frac{\lambda_d^3}{\lambda^2}\propto\frac{\lambda_d^3}{\lambda^2}.$

(D)

Topics:de Broglie waves Photoelectric effect

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JEE Advanced 2017 · Paper 2, Physics question 3