A photoemissive 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)

λ e =sqrt frac h 2 m c (frac 1 λ i -λ 0 )

λ e =frac h sqrt 2 m c (frac 1 λ i -frac 1 λ 0 )

Physics · Atomic Physics

JEE Main 2025 — 7 April, Evening Shift — Question 58

A photoemissive substance is illuminated with a radiation of wavelength $\lambda_{i}$ so that it releases electrons with de-Broglie wavelength $\lambda_{e}$ . 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)

Option A:
$\lambda_{e}=\sqrt{\frac{h \lambda_{i}}{2 m c}}$
Option B:
$\lambda_{e}=\sqrt{\frac{h}{2 m c\left(\frac{1}{\lambda_{i}-\lambda_{0}}\right)}}$
Correct
Option C:
$\lambda_{e}=\sqrt{\frac{h \lambda_{0}}{2 m c}}$
Option D:
$\lambda_{e}=\frac{h}{\sqrt{2 m c\left(\frac{1}{\lambda_{i}}-\frac{1}{\lambda_{0}}\right)}}$

Answer: B

Step-by-step solution

$\frac{h c}{\lambda_{i}}=\phi+K E$

$\phi=\frac{h c}{\lambda_{0}}, \mathrm{KE}=\left(\frac{h}{\lambda_{e}}\right)^{2} \cdot \frac{1}{2 m}$

$\frac{h c}{\lambda_{i}}=\frac{h c}{\lambda_{0}}+\left(\frac{h}{\lambda_{e}}\right)^{2} \cdot \frac{1}{2 m}$

$\left(\frac{\lambda_{e}}{h}\right)^{2}=\frac{1}{2 m h c\left(\frac{1}{\lambda_{i}}-\frac{1}{\lambda_{0}}\right)}$

$\lambda_{e}=\sqrt{\frac{h}{2 m c\left(\frac{1}{\lambda_{i}}-\frac{1}{\lambda_{0}}\right)}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Evening Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Photoelectric Effect