JEE Advance - Physics (2010 - Paper 2 Offline - No. 17)

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 amu) and O (mass = 16 amu), where 1 amu $$ = {5 \over 3} \times {10^{ - 27}}$$ kg, is close to :

2.4 $$\times$$ 10^$$-$$10 m

1.9 $$\times$$ 10^$$-$$10 m

1.3 $$\times$$ 10^$$-$$10 m

4.4 $$\times$$ 10^$$-$$11 m

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The moment of inertia of CO molecule is

I = $$\mu$$r² ....... (i)

where, $$\mu$$ = reduced mass of the CO molecule, r = distance between C and O or bond length

The reduced mass $$\mu$$ of the CO molecule is

$$\mu = {{{m_1}{m_2}} \over {{m_1} + {m_2}}} = \left[ {{{(12)(16)} \over {12 + 16}}} \right] \times {5 \over 3} \times {10^{ - 27}}$$ kg

But $$I = 1.87 \times {10^{ - 46}}$$ kg m² (From the above question)

From equation (i), we get

$${r^2} = {I \over \mu }$$

Substituting the values of I and $$\mu$$ in above equation, we get

$${r^2} = {{1.87 \times {{10}^{ - 46}}} \over {\left[ {{{12 \times 16} \over {28}} \times {5 \over 3} \times {{10}^{ - 27}}} \right]}}$$

or $${r^2} = {{1.87 \times {{10}^{ - 46}} \times 28 \times 3} \over {12 \times 16 \times 5 \times {{10}^{ - 27}}}}$$ or $$r = 1.3 \times {10^{ - 10}}$$ m

JEE Advance - Physics (2010 - Paper 2 Offline - No. 17)

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