Logo
Jeeify
MCQEasyJEE 2025Bohr's Model & Hydrogen Spectrum

JEE Physics 2025 Question with Solution

Given below are two statements:

Statement (I) : The dimensions of Planck’s constant and angular momentum are same.

Statement (II) : In Bohr’s model, electron revolves around the nucleus in those orbits for which angular momentum is an integral multiple of Planck’s constant.

In the light of the above statements, choose the most appropriate answer from the options given below:

  • A

    Both Statement I and Statement II are correct

  • B

    Statement I is incorrect but Statement II is correct

  • C

    Statement I is correct but Statement II is incorrect

  • D

    Both Statement I and Statement II are incorrect

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Two statements about Planck’s constant and Bohr’s model are to be checked.

Find: Which option correctly identifies the truth values of Statement I and Statement II.

For Statement I, Planck’s constant hh has the same dimensions as angular momentum.

[h]=[ML2T−1][h] = [M L^2 T^{-1}]

Angular momentum is

L=mvrL = mvr

So its dimensional formula is

[L]=[MLT−1⋅L]=[ML2T−1][L] = [M L T^{-1} \cdot L] = [M L^2 T^{-1}]

Hence, Statement I is correct.

For Statement II, Bohr’s quantization condition is

L=nh2πL = \frac{nh}{2\pi}

where nn is a positive integer. Therefore, angular momentum is an integral multiple of h2π\frac{h}{2\pi}, not of hh itself.

Hence, Statement II is incorrect.

Therefore, the correct option is C. The statement "Statement I is correct but Statement II is incorrect" is the right answer.

Using the Hint and Bohr Quantization

The hint states that Planck’s constant hh and angular momentum have the same dimensional formula, and that in Bohr’s model angular momentum is quantized in integer multiples of h2π\frac{h}{2\pi}.

So:

  1. Compare dimensions of hh and angular momentum. They match, so Statement I is true.
  2. Check the Bohr condition carefully. The allowed orbits satisfy
mvr=nh2πmvr = \frac{nh}{2\pi}

Thus the orbit condition is not an integral multiple of hh, but of h2π\frac{h}{2\pi}.

So Statement II is false.

Therefore, the correct option is C.

Common mistakes

  • Students often mark Statement II as correct because they remember angular momentum quantization but forget the factor 12π\frac{1}{2\pi}. This is wrong because Bohr’s condition is L=nh2πL = \frac{nh}{2\pi}, not L=nhL = nh. Always write the full quantization formula before judging the statement.

  • Some students confuse the symbol LL used for angular momentum with the dimensional symbol for length. This can lead to incorrect dimensional analysis. Use the physical expression L=mvrL = mvr to obtain dimensions instead of relying only on symbols.

  • A common error is trusting the raw option numbering or a contradictory secondary solution without checking the main solution and formula. This is wrong because the correct physics criterion is the Bohr quantization relation. Verify each statement directly from the standard formula before choosing an option.

Practice more Bohr's Model & Hydrogen Spectrum questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Bohr's Model & Hydrogen Spectrum

Related questions