A photo-emissive substance is illuminated with a radiation of wavelength $\lambda_{\mathrm{i}}$ so that it releases electrons with de-Broglie wavelength $\lambda_{\mathrm{e}}$. The longest wavelength of radiation that can emit photoelectron is $\lambda_{0}$. Expression for de-Broglie wavelength is given by : ( m : mass of the electron, h : Planck's constant and c : speed of light)

(A) $\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mc}\left(\frac{1}{\lambda_{\mathrm{i}}}-\frac{1}{\lambda_{0}}\right)}}$

(B) $\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h} \lambda_{0}}{2 \mathrm{mc}}}$

(C) $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mc}\left(\frac{1}{\lambda_{\mathrm{i}}}-\frac{1}{\lambda_{0}}\right)}}$

(D) $\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h} \lambda_{\mathrm{i}}}{2 \mathrm{mc}}}$

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Modern Physics

A photo-emissive substance is illuminated with a radiation of wavelength

λ_{i}

so that it releases electrons with de-Broglie wavelength

λ_{e}

. The longest wavelength of radiation that can emit photoelectron is

λ_{0}

. Expression for de-Broglie wavelength is given by :
( m : mass of the electron, h : Planck's constant and c : speed of light)

Answer: (A)

The photoelectron's kinetic energy is the photon energy less the work function,

K = \frac{h c}{λ _{i}} - \frac{h c}{λ _{0}}

, since the threshold wavelength fixes the work function as

h c / λ_{0}

. The de-Broglie wavelength is

λ_{e} = \frac{h}{2 m K} = \frac{h}{2 m h c ( \frac{1}{λ _{i}} - \frac{1}{λ _{0}} )}

. Pulling one factor of

h

under the root,

λ_{e} = \frac{h}{2 m c ( \frac{1}{λ _{i}} - \frac{1}{λ _{0}} )}

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option