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Physics · Atomic Physics

JEE Main 2024 — 29 January, Shift 2 — Question 31

Two sources of light emit with a power of 200 W . The ratio of number of photons of visible light emitted by each source having wavelengths 300 nm and 500 nm respectively, will be :

  1. Option A:

    1:51: 5

  2. Option B:

    1:31: 3

  3. Option C:

    5:35: 3

  4. Option D:

    3:53: 5

    Correct

Answer: D

Step-by-step solution

n1×hcλ1=200\mathrm{n}_{1} \times \frac{\mathrm{hc}}{\lambda_{1}}=200

n2×hcλ2=200\mathrm{n}_{2} \times \frac{\mathrm{hc}}{\lambda_{2}}=200

n1n2=λ1λ2=300500\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{300}{500} n1n2=35\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{3}{5}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 2
Subject
Physics
Chapter
Atomic Physics
Topic
Photon Theory of Light and Radiation Pressure

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