JEE Challenger
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PhyMedAdv 2010 • Paper 2

Rotational Energy Levels of Diatomic Molecule

Comprehension Passage

The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.

A diatomic molecule has moment of inertia II. By Bohr's quantization condition its rotational energy in the nthn^{\text{th}} level (n=0n = 0 is not allowed) is

Options

A

1n2(h28π2I)\frac{1}{n^2}\left(\frac{h^2}{8\pi^2 I}\right)

B

1n(h28π2I)\frac{1}{n}\left(\frac{h^2}{8\pi^2 I}\right)

C

n(h28π2I)n\left(\frac{h^2}{8\pi^2 I}\right)

D

n2(h28π2I)n^2\left(\frac{h^2}{8\pi^2 I}\right)

Correct

Topics & Concepts

AtomsBohr Model

Step-by-Step Solution

To determine the rotational energy of a diatomic molecule according to Bohr's quantization condition, we proceed as follows:

  1. Bohr's Quantization Condition:
    According to Bohr's postulate, the angular momentum LL of a rotating system is quantized as integral multiples of h2π\frac{h}{2\pi}:
    L=nh2πL = \frac{nh}{2\pi}
    where:
  • n=1,2,3,…n = 1, 2, 3, \ldots represents the quantum number (n=0n = 0 is not allowed),
  • hh is Planck's constant.

  1. Rotational Kinetic Energy:
    The rotational kinetic energy EE of a rigid body having moment of inertia II and angular momentum LL is given by:
    E=L22IE = \frac{L^2}{2I}

  2. Quantized Energy Levels:
    Substituting the quantized angular momentum LL into the expression for rotational kinetic energy:
    En=12I(nh2π)2=n2h22I⋅4π2=n2(h28π2I)E_n = \frac{1}{2I} \left(\frac{nh}{2\pi}\right)^2 = \frac{n^2 h^2}{2I \cdot 4\pi^2} = n^2 \left(\frac{h^2}{8\pi^2 I}\right)

Thus, the rotational energy in the nthn^{\text{th}} level is:
En=n2(h28π2I)E_n = n^2 \left(\frac{h^2}{8\pi^2 I}\right)

Therefore, the correct option is D.

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