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.

Linked Question 1

It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to $\frac{4}{π} \times 10^{11}$ Hz. Then the moment of inertia of CO molecule about its center of mass is close to (Take h = 2 $π$ × 10^-34 J s)

4.67 × 10^-47 Kg m²

1.87 × 10^-46 Kg m²

1.17 × 10^-47 Kg m²

2.75 × 10^-46 Kg m²

$\frac{3 \times h^{2}}{8 π^{2} λI} = hf = \frac{4}{π} \times 10^{11}$

$I = \frac{3 h}{8 π \times 4} \times 10^{11} \times 2 π \times 10^{- 34} = \frac{3}{16} \times 10^{- 45} {kgm}^{2}$

Linked Question 2

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

$\frac{1}{n^{2}} (\frac{h^{2}}{8 π^{2} I})$

$n (\frac{h^{2}}{8 π^{2} I})$

$\frac{1}{n} (\frac{h^{2}}{8 π^{2} I})$

$n^{2} (\frac{h^{2}}{8 π^{2} I})$

$Iω = \frac{nh}{2 π}$

$\frac{1}{2} {Iω}^{2} = \frac{I^{2} ω^{2}}{2 I} = \frac{n^{2} h^{2}}{8 π^{2} I}$

Linked Question 3

In a CO molecule, the distance between C (mass = 12 a.m.u.) and O (mass = 16 am.u.), where 1 a.m.u. $= \frac{5}{3} \times 10^{- 27}$ Kg, is close to

2.4 × 10^-10 m

1.9 × 10^-10 m

1.3 × 10^-10 m

4.4 × 10^-10 m

I_cm = µx²

^{$\frac{3}{16} \times 10^{- 45} = \frac{12 \times 16}{12 + 16} \times \frac{5}{3} \times 10^{- 27} \times x^{2} = \frac{12 \times 16}{28} \times \frac{5}{3} \times 10^{- 27} \times x^{2}$}

^{$x^{2} = \frac{3 \times 7}{256 \times 5} \times 10^{- 18} = \frac{21}{1280} \times 10^{- 18}$}

^{$x = \frac{1}{8} \times 10^{- 9} m = 1 .25 \times 10^{- 10} m$}