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.

It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to

\frac{4}{π} \times 1 0^{11} Hz

. Then the moment of inertia of CO molecule about its center of mass is close to (Take

h = 2 π \times 1 0^{- 34} J s

)

(A) $2.76 \times 1 0^{- 46} kg m^{2}$
(B) $1.87 \times 1 0^{- 46} kg m^{2}$ ✓ Correct
(C) $4.67 \times 1 0^{- 47} kg m^{2}$
(D) $1.17 \times 1 0^{- 47} kg m^{2}$

Answer: (B)

Solution

The energy difference between

n = 2

and

n = 1

states is:

Δ E = E_{2} - E_{1} = 4 (\frac{h ^{2}}{8 π ^{2} I}) - 1 (\frac{h ^{2}}{8 π ^{2} I}) = 3 (\frac{h ^{2}}{8 π ^{2} I})

This equals

h ν

h ν = 3 (\frac{h ^{2}}{8 π ^{2} I})

I = \frac{3 h}{8 π ^{2} ν}

Substituting

h = 2 π \times 1 0^{- 34}

J·s and

ν = \frac{4}{π} \times 1 0^{11}

Hz:

I = \frac{3 \times 2 π \times 1 0 ^{- 34}}{8 π ^{2} \times \frac{4}{π} \times 1 0 ^{11}} = \frac{6 π \times 1 0 ^{- 34}}{32 π \times 1 0 ^{11}} = \frac{6 \times 1 0 ^{- 45}}{32} \approx 1.87 \times 1 0^{- 46} kg\cdotm^{2}

The answer is **(B)

1.87 \times 1 0^{- 46}

kg·m²**.

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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