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Physics · Atomic Physics

JEE Main 2026 — 8 April, Evening Shift — Question 10

A monochromatic source of light operating at 15kW15\mathrm{kW} emits 2.5×10222.5\times10^{22} photons/s. The region of the electromagnetic spectrum to which the emitted electromagnetic radiation belongs to (Take h=6.6×10−34J.s\mathrm{h} = 6.6\times10^{-34}\mathrm{J.s}, c=3×108m/s\mathrm{c} = 3\times10^{8}\mathrm{m/s})

  1. Option A:

    Microwave

  2. Option B:

    Infrared

  3. Option C:

    Visible

  4. Option D:

    Ultraviolet

    Correct

Answer: D

Step-by-step solution

Energy per photon = 15×1032.5×1022=6×10−19\frac{15\times10^3}{2.5\times10^{22}} = 6\times10^{-19} J. Wavelength λ=hcE=6.6×10−34×3×1086×10−19=3.3×10−7\lambda = \frac{hc}{E} = \frac{6.6\times10^{-34}\times3\times10^8}{6\times10^{-19}} = 3.3\times10^{-7} m = 330 nm, which is ultraviolet.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
8 April, Evening Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Photon Theory of Light and Radiation Pressure

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