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Physics · Atomic Physics

JEE Main 2026 — 22 January, Evening Shift — Question 40

Light is incident on a metallic plate having work function 110×10−20 J110 \times 10^{-20} \mathrm{~J}. If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is ____\_\_\_\_ rad/s.(h=6.63×10−34 J.s)\mathrm{rad} / \mathrm{s} .\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)

  1. Option A:

    1.04×10161.04 \times 10^{16}

    Correct
  2. Option B:

    1.04×10131.04 \times 10^{13}

  3. Option C:

    1.66×10161.66 \times 10^{16}

  4. Option D:

    1.66×10151.66 \times 10^{15}

Answer: A

Step-by-step solution

ϕ=hv\phi=\mathrm{h} v v=ϕhv=\frac{\phi}{\mathrm{h}} ω=2πν=2πϕ h=2×3.14×110×10−26.63×10−34\omega=2 \pi \nu=\frac{2 \pi \phi}{\mathrm{~h}}=\frac{2 \times 3.14 \times 110 \times 10^{-2}}{6.63 \times 10^{-34}} ω=1.04×1016rad/sec\omega=1.04 \times 10^{16} \mathrm{rad} / \mathrm{sec}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
22 January, Evening Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Photoelectric Effect

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