IIT-JEE 2010 · Paper 2 · Physics Q54

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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.

It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to $\dfrac4\pi\times10^{11}$ Hz. Then the moment of inertia of CO molecule about its center of mass is close to (Take $h=2\pi\times10^{-34}$ J s)

2.76\times10^{-46}

kg m²

1.87\times10^{-46}

kg m²

4.67\times10^{-47}

kg m²

1.17\times10^{-47}

kg m²

Answer and worked solution

Official answer: B

Transition 1 → 2. $\Delta E=(4-1)\dfrac{h^2}{8\pi^2I}=h\nu$ , so
$I=\frac{3h}{8\pi^2\nu}=\frac{3\times2\pi\times10^{-34}}{8\pi^2\times\frac4\pi\times10^{11}}=\frac{6}{32}\times10^{-45}=1.87\times10^{-46}\ \mathrm{kg\,m^2}.$

(B)

Topics:Bohr model and hydrogen spectrum Moment of inertia

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IIT-JEE 2010 · Paper 2, Physics question 54