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Physics · Atomic Physics

JEE Main 2026 — 5 April, Morning Shift — Question 14

Light source having wavelength 331331 nm is used to generate photo-electrons whose stopping potential is 0.20.2 V. The work function of the used metal in the experiment is α×10−19\alpha \times 10^{-19} J. The value of α\alpha is (h = 6.62×10^{-34} J s, e = 1.6×10^{-19} C, c = 3×10^8 m/s)

  1. Option A:

    3.68

  2. Option B:

    4.68

  3. Option C:

    5.68

    Correct
  4. Option D:

    2.68

Answer: C

Step-by-step solution

ϕ=hcλ−eVs=6.62×10−34×3×108331×10−9−1.6×10−19×0.2=6.0×10−19−0.32×10−19=5.68×10−19\phi = \frac{hc}{\lambda} - eV_s = \frac{6.62\times10^{-34}\times3\times10^8}{331\times10^{-9}} - 1.6\times10^{-19}\times0.2 = 6.0\times10^{-19} - 0.32\times10^{-19} = 5.68\times10^{-19} J.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Morning Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Photoelectric Effect

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