ABC26PH0109 · Molecular Spectroscopy

Subject: Physical Chemistry · Chapter: Molecular Spectroscopy · Topic: Molecular Spectroscopy – General · Exam: CSIR-NET DEC 2019 · Marks: 2 · Difficulty: Easy

The atomic masses of fluorine and hydrogen are 19.0 and 1.0 amu, respectively ($1\,\mathrm{amu}=1.67\times10^{-27}\,\mathrm{kg}$). The bond length of HF is $2.0\,\AA$. The moment of inertia of HF is
(a)$3.2 \times 10^{-47} \mathrm{kg} \mathrm{m}^{2}$
(b)$6.4\times10^{-47}\,\mathrm{kg\,m^{2}}$
(c)$9.6\times10^{-47}\,\mathrm{kg\,m^{2}}$
(d)$4.8\times10^{-47}\,\mathrm{kg\,m^{2}}$
Answer
Answer (as printed): B
Explanation
$I=\mu r_0^{2}$, where the reduced mass $\mu = \dfrac{m_F m_H}{m_F+m_H}$.
\[ \mu = \frac{19.0\times1.0}{19.0+1.0}\,\mathrm{amu} = 0.95\,\mathrm{amu} = 0.95\times1.67\times10^{-27}\,\mathrm{kg} \]
\[ I = \mu r_0^{2} = (0.95\times1.67\times10^{-27}\,\mathrm{kg})\times(2.0\times10^{-10}\,\mathrm{m})^{2} \approx 6.4\times10^{-47}\,\mathrm{kg\,m^{2}} \]

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