Find the moment of inertia of the hydrogen molecule about an axis passing through its centre of mass and perpendicular to the inter-nuclear axis. Given: mass of the hydrogen atom $= 1.7 \times 10^{- 27}$ kg, inter-atomic distance $= 4 \times 10^{- 10}$ m.

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Solution

Mass of hydrogen atom $= 1.7 \times 10^{- 27} k g$
Total mass of two molecules $= 2 \times 1.7 \times 10^{- 27} k g$
distance between the two atoms $= 4 \times 10^{- 10} m$
distance from the axis of rotation $= \frac{4 \times 10^{- 10} m}{2} = 2 \times 10^{- 10} m$
moment of inertia $= I = m r^{2}$

$I = 2 \times 1.7 \times 10^{- 27} k g \times {(2 \times 10^{- 10})}^{2}$

$I = 13.6 \times 10^{- 47} {k g m}^{2}$

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Find the moment of inertia of the hydrogen molecule about an axis passing through its centre of mass and perpendicular to the inter-nuclear axis. Given: mass of the hydrogen atom =1.7×10−27 kg, inter-atomic distance =4×10−10m.

Mass of hydrogen atom=1.7×10−27kgTotal mass of two molecules =2×1.7×10−27kgdistance between the two atoms =4×10−10mdistance from the axis of rotation =4×10−10m2=2×10−10mmoment of inertia =I=mr2I=2×1.7×10−27kg×(2×10−10)2 I=13.6×10−47kgm2

Find the moment of inertia of the hydrogen molecule about an axis passing through its centre of mass and perpendicular to the inter-nuclear axis. Given: mass of the hydrogen atom $= 1.7 \times 10^{- 27}$ kg, inter-atomic distance $= 4 \times 10^{- 10}$ m.