Physics · Atomic Physics

JEE Main 2026 — 8 April, Evening Shift — Question 14

$K_1$ and $K_2$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $\lambda_1$ and $\lambda_2$ respectively. If $\lambda_1 = 2\lambda_2$ then the work function of material is given by:

Option A:
$K_2 + 2K_1$
Option B:
$2K_2 - K_1$
Option C:
$K_1 - 2K_2$
Option D:
$K_2 - 2K_1$
Correct

Answer: D

Step-by-step solution

From photoelectric equation: $K_1 = \frac{hc}{\lambda_1} - \phi$ , $K_2 = \frac{hc}{\lambda_2} - \phi$ . With $\lambda_1 = 2\lambda_2$ , then $K_1 = \frac{hc}{2\lambda_2} - \phi$ and $K_2 = \frac{hc}{\lambda_2} - \phi$ . Eliminating $\frac{hc}{\lambda_2}$ gives $\phi = K_2 - 2K_1$ .

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Exam: JEE Main 2026
Paper: 8 April, Evening Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Photoelectric Effect

A monochromatic source of light operating at 15 kW emits 2.5times10^22 photons/s. The region of the electromagnetic spectrum to which the…

Answer: D

Step-by-step solution

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