Find the moment of inertia of a hydrogen molecule about its centre of mass if the mass of each hydrogen atom is $m$ and the distance between them is $R$.
Q 65
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We assume the rotation axis to be a transverse axis through the centre of mass of the linear molecule $\mathrm{H}_2$. Then, each of the hydrogen atom is a distance $\frac{1}{2} \mathrm{R}$ from the $\mathrm{CM}$. Therefore, the $\mathrm{Ml}$ of the molecule about this axis,
$
I=2 m\left(\frac{R}{2}\right)^2=\frac{1}{2} m R^2
$
Notes :
1. For a $\mathrm{H}_2$ molecule, $\mathrm{mH}=1.674 \times 10^{-27} \mathrm{~kg}$ and bond length $=7.774 \times 10^{-11} \mathrm{~m}$, so that $\mathrm{I}=5.065 \times 10^{-48} \mathrm{~kg} \cdot \mathrm{m}^2$.
2. As atoms are treated as particles, we do not consider rotation about the line passing through the atoms.
$
I=2 m\left(\frac{R}{2}\right)^2=\frac{1}{2} m R^2
$
Notes :
1. For a $\mathrm{H}_2$ molecule, $\mathrm{mH}=1.674 \times 10^{-27} \mathrm{~kg}$ and bond length $=7.774 \times 10^{-11} \mathrm{~m}$, so that $\mathrm{I}=5.065 \times 10^{-48} \mathrm{~kg} \cdot \mathrm{m}^2$.
2. As atoms are treated as particles, we do not consider rotation about the line passing through the atoms.
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