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.

In a CO molecule, the distance between C (mass = 12 a.m.u.) and O (mass = 16 a.m.u.), where 1 a.m.u.

= \frac{5}{3} \times 1 0^{- 27} kg

, is close to

(A) $2.4 \times 1 0^{- 10} m$
(B) $1.9 \times 1 0^{- 10} m$
(C) $1.3 \times 1 0^{- 10} m$ ✓ Correct
(D) $4.4 \times 1 0^{- 11} m$

Answer: (C)

Solution

The moment of inertia about the center of mass is:

I = μ d^{2}

where

μ

is the reduced mass:

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} = \frac{12 \times 16}{12 + 16} = \frac{192}{28} = \frac{48}{7} a.m.u.

Converting to kg:

μ = \frac{48}{7} \times \frac{5}{3} \times 1 0^{- 27} = \frac{80}{7} \times 1 0^{- 27} kg

From the previous question,

I = 1.87 \times 1 0^{- 46}

kg·m²:

d = \frac{I}{μ} = \frac{1.87 \times 1 0 ^{- 46}}{\frac{80}{7} \times 1 0 ^{- 27}} \approx 1.3 \times 1 0^{- 10} m

The answer is **(C)

1.3 \times 1 0^{- 10}

m**.

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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