The moment of inertia of a diatomic molecule about an axis passing through its center of mass and perpendicular to the line joining the two atoms will be : ( $μ$ = Reduced mass of the system)

μ r^{2}

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\frac{μ r^{2}}{2}

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\frac{3}{4} μ r^{2}

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Solution

The correct option is A $μ r^{2}$
Let $m_{1}, m_{2}, r_{1}$ and $r_{2}$ be the masses and the position vectors. From the definition of center of mass we have $m_{1} r_{2} + m_{2} r_{2} = 0$
Thus
$r_{1} = - \frac{m_{2}}{m_{1} + m_{2}} r$ and $r_{2} = \frac{m_{1}}{m_{1} + m_{2}} r$
The moment of inertia of the molecule about an axis passing through center of mass and perpendicular to line joining the molecules
$I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2} = m_{1} (\frac{m_{2}}{m_{1} + m_{2}} r)^{2} + m_{2} (\frac{m_{1}}{m_{1} + m_{2}} r)^{2}$
or
$I = \frac{m_{1} m_{2}}{m_{1} + m_{2}} r^{2} = μ r^{2}$

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The moment of inertia of a diatomic molecule about an axis passing through its center of mass and perpendicular to the line joining the two atoms will be : (μ = Reduced mass of the system)

The moment of inertia of a diatomic molecule about an axis passing through its center of mass and perpendicular to the line joining the two atoms will be : ( $μ$ = Reduced mass of the system)