IIT-JEE 2010 · Paper 2 · Physics Q55

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Paragraph for Questions 53 to 55

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.

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. $=\dfrac53\times10^{-27}$ kg, is close to

2.4\times10^{-10}

1.9\times10^{-10}

1.3\times10^{-10}

4.4\times10^{-11}

Answer and worked solution

Official answer: C

Reduced mass. $\mu=\dfrac{12\times16}{28}=6.86$ a.m.u. $=1.14\times10^{-26}$ kg.

Bond length. $I=\mu d^2$ :
$d=\sqrt{\frac{1.87\times10^{-46}}{1.14\times10^{-26}}}=\sqrt{1.64\times10^{-20}}\approx1.3\times10^{-10}\text{ m}.$

(C)

Topics:Moment of inertia

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IIT-JEE 2010 · Paper 2, Physics question 55