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. , is close to [2010]

Question

The key feature of Bohr's theory of the spectrum of the hydrogen atom is the quantization of angular momentum when an electron revolves around a proton. We will extend this to a general rotational motion to find the quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition. [2010]

Q. 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. $= \frac{5}{3} \times 10^{- 27} kg$ , is close to

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(3)

Moment of inertia of CO molecule,

$= \frac{5}{3} \times 10^{- 27} kg$ $\Rightarrow r^{2} = \frac{I}{μ}$

where, $μ$ = reduced mass of the CO molecule and $r$ = distance between C and O

Reduced mass of the CO molecule,

$μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}} = [\frac{(12) (16)}{12 + 16}] \times \frac{5}{3} \times 10^{- 27} kg$

But, $I = 1.87 \times 10^{- 46} {kg m}^{2}$ (from the above question)

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

$∴$ $r = 1.3 \times 10^{- 10} m$

Solution

1 $2.4 \times 10^{- 10} m$

2 $1.9 \times 10^{- 10} m$

3 $1.3 \times 10^{- 10} m$

4 $4.4 \times 10^{- 11} m$

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Solution

1 2.4×10-10 m

2 1.9×10-10 m

3 1.3×10-10 m

4 4.4×10-11 m

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1 $2.4 \times 10^{- 10} m$

2 $1.9 \times 10^{- 10} m$

3 $1.3 \times 10^{- 10} m$

4 $4.4 \times 10^{- 11} m$