Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 1 / A3 / a / iii

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- Solution iii

Solution

For an axis through the center of mass perpendicular to the bond, the point-mass moment of inertia is

I = m_{C} r_{1}^{2} + m_{S} r_{2}^{2} = \frac{m _{C} m _{S}^{2} + m _{S} m _{C}^{2}}{( m _{C} + m _{S} ) ^{2}} r^{2} = \frac{m _{C} m _{S}}{m _{C} + m _{S}} r^{2} .

The prefactor is the reduced mass

μ

; this is the standard reduction of a two-body linear rigid rotor to

I = μ r^{2}

.

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