Place the carbon and sulfur nuclei on a line separated by

r

m_{C}

and

m_{S}

. The center of mass lies between them and closer to the heavier sulfur nucleus. In the linear rigid rotor approximation,

r

remains fixed while the molecule rotates.

ii

Solution

x

from the carbon nucleus toward the sulfur nucleus. The definition of the center of mass gives

x_{cm} = \frac{m _{C} 0 + m _{S} r}{m _{C} + m _{S}} .

Therefore the requested distance is

r_{1} = \frac{m _{S}}{m _{C} + m _{S}} r .

The other distance is

r_{2} = r - r_{1} = m_{C} r / (m_{C} + m_{S})

, consistent with

m_{C} r_{1} = m_{S} r_{2}

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}

Using the stipulated integer isotope masses gives

μ = \frac{12 \times 32}{44} u = 8.72727 u, I = μ (1.53 \times 1 0^{- 10} m)^{2} = 3.39243 \times 1 0^{- 46} kg m^{2} .

B = \frac{h}{8 π ^{2} c I} = 0.8252 c m^{- 1} .

Here

c

must be expressed in

cm s^{- 1}

when

I

is in SI units. Equivalently calculate

B = h / (8 π^{2} I)

Hz

first, then divide by

c

cm s^{- 1}

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