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Physics · Atomic Physics

JEE Main 2026 — 21 January, Morning Shift — Question 36

A light wave described by E=60sin⁡(3×1015)t+sin⁡(12×1015)\mathrm{E}=60 \sin \left(3 \times 10^{15}\right) \mathrm{t}+ \sin \left(12 \times 10^{15}\right) t] (in SI units) falls on a metal surface of work function 2.8 eV . The maximum kinetic energy of ejected photoelectron is (approximately) ____\_\_\_\_ eV. (h=6.6×10−34 J−s.\quad\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s} . \quad\right. and e=1.6×10−19C\mathrm{e}=1.6 \times 10^{-19} \mathrm{C} )

  1. Option A:

    5.1

    Correct
  2. Option B:

    3.8

  3. Option C:

    6

  4. Option D:

    7.8

Answer: A

Step-by-step solution

ω1=3×1015rad/sec\quad \omega_{1}=3 \times 10^{15} \mathrm{rad} / \mathrm{sec}

ω2=12×1015rad/sec∵v=ω2πEphoton =hv=6.6×10−34×1.91×1015=1.26×10−18 JEmax⁡=1.26×10−181.6×10−19≃7.9eV Kmax⁡=Emax⁡−ϕ0=7.9−2.8 Kmax⁡=5.1eV\begin{aligned} & \omega_{2}=12 \times 10^{15} \mathrm{rad} / \mathrm{sec} & \because \quad v=\frac{\omega}{2 \pi} & \begin{aligned} \mathrm{E}_{\text {photon }} & =\mathrm{h} v=6.6 \times 10^{-34} \times 1.91 \times 10^{15} & =1.26 \times 10^{-18} \mathrm{~J} \mathrm{E}_{\max } & =\frac{1.26 \times 10^{-18}}{1.6 \times 10^{-19}} \simeq 7.9 \mathrm{eV} \mathrm{~K}_{\max } & =\mathrm{E}_{\max }-\phi_{0} & =7.9-2.8 \mathrm{~K}_{\max } & =5.1 \mathrm{eV} \end{aligned} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Morning Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Photoelectric Effect

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