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Physics · Atomic Physics

JEE Main 2024 — 6 April, Shift 1 — Question 33

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :

  1. Option A:

    4:14: 1

    Correct
  2. Option B:

    1:21: 2

  3. Option C:

    1:41: 4

  4. Option D:

    2:12: 1

Answer: A

Step-by-step solution

1λ=Rz2(1n12−1n22)\frac{1}{\lambda}=\mathrm{Rz}^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)

1λL=Rz2(112)1λB=Rz⁡2(122)\frac{\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{Rz}^{2}\left(\frac{1}{1^{2}}\right)}{\frac{1}{\lambda_{\mathrm{B}}}=\operatorname{Rz}^{2}\left(\frac{1}{2^{2}}\right)}

λBλL=4:1\frac{\lambda_{B}}{\lambda_{\mathrm{L}}}=4: 1

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
6 April, Shift 1
Subject
Physics
Chapter
Atomic Physics
Topic
Hydrogen Spectrum

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