A diatomic molecule has moment of inertia . By Bohr's quantization condition its rotational energy in the level ( n=0 is not allowed) is [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. 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

Smart Achievers · Accepted Answer

(4)

Rotational kinetic energy, $K . E ._{rot} = \frac{1}{2} I ω^{2} = \frac{1}{2} I {(\frac{L}{I})}^{2}$ $[∵ L = I ω]$

And according to Bohr's quantisation principle, $L = \frac{n h}{2 π}$

$∴$ $K . E ._{rot} = \frac{1}{2} \frac{L^{2}}{I} = \frac{1}{2 I} \times \frac{n^{2} h^{2}}{4 π^{2}} = n^{2} (\frac{h^{2}}{8 π^{2} I})$

Solution

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

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

3 $n (\frac{h^{2}}{8 π^{2} I})$

4 $n^{2} (\frac{h^{2}}{8 π^{2} I})$

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Solution

1 1n2(h28π2I)

2 1n(h28π2I)

3 n(h28π2I)

4 n2(h28π2I)

Ans. (4) Rotational kinetic energy, K.E.rot=12Iω2=12I(LI)2 [∵ L=Iω] And according to Bohr's quantisation principle, L=nh2π ∴ K.E.rot=12L2I=12I×n2h24π2=n2(h28π2I)

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

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

3 $n (\frac{h^{2}}{8 π^{2} I})$

4 $n^{2} (\frac{h^{2}}{8 π^{2} I})$