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JEE Advance - Physics (2010 - Paper 2 Offline - No. 16)

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.
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.
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 $${4 \over \pi } \times {10^{11}}$$ Hz. Then, the moment of inertia of CO molecule about its centre of mass is close to (Take h = 2$$\pi$$ $$\times$$ 10$$-$$34 J-s)
2.76 $$\times$$ 10$$-$$46 kg m2
1.87 $$\times$$ 10$$-$$46 kg m2
4.67 $$\times$$ 10$$-$$47 kg m2
1.17 $$\times$$ 10$$-$$47 kg m2

Explanation

The energy of photon is equal to the energy difference between the ground level and first excited level.

$$hv = {E_2} - {E_1}$$

$$hv = {{(4 - 1){h^2}} \over {8{\pi ^2}I}} \Rightarrow I = {{3h} \over {8{\pi ^2}v}}$$

$$I = {{3 \times 2\pi \times {{10}^{ - 34}}} \over {[(8{\pi ^2}4)/\pi ] \times {{10}^{11}}}} = {3 \over {16}} \times {10^{ - 45}}$$ kg-m2 = 1.87 $$\times$$ 10$$-$$46 kg-m2

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