JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Modern Physics

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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

I

. By Bohr's quantization condition its rotational energy in the

n^{th}

level (

n = 0

is not allowed) is

(A) $\frac{1}{n ^{2}} (\frac{h ^{2}}{8 π ^{2} I})$
(B) $\frac{1}{n} (\frac{h ^{2}}{8 π ^{2} I})$
(C) $n (\frac{h ^{2}}{8 π ^{2} I})$
(D) $n^{2} (\frac{h ^{2}}{8 π ^{2} I})$ ✓ Correct

Answer: (D)

Solution

By Bohr's quantization condition:

L = \frac{nh}{2 π}

The rotational kinetic energy is:

E = \frac{L ^{2}}{2 I} = \frac{1}{2 I} (\frac{nh}{2 π})^{2} = \frac{n ^{2} h ^{2}}{8 π ^{2} I} = n^{2} (\frac{h ^{2}}{8 π ^{2} I})

The answer is **(D)

n^{2} (\frac{h ^{2}}{8 π ^{2} I})

**.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Single correct option

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JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Modern Physics

Solution

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