The lowest-frequency pure-rotational absorption line of ${ }^{12} \mathrm{C}^{32} \mathrm{~S}$ occurs at $48991.0 \mathrm{MHz}$. Find the bond distance in ${ }^{12} \mathrm{C}^{32} \mathrm{~S}$.
The lowest-frequency rotational absorption is the $J=0 \rightarrow 1$ line. Equations (1.4), $(6.52)$, and $(6.51)$ give
$$
h \nu=E_{\mathrm{upper}}-E_{\mathrm{lower}}=\frac{1(2) \hbar^2}{2 \mu d^2}-\frac{0(1) \hbar^2}{2 \mu d^2}
$$
which gives $d=\left(h / 4 \pi^2 \nu \mu\right)^{1 / 2}$. Table A.3 in the Appendix gives
$$
\mu=\frac{m_1 m_2}{m_1+m_2}=\frac{12(31.97207)}{(12+31.97207)} \frac{1}{6.02214 \times 10^{23}} \mathrm{~g}=1.44885 \times 10^{-23} \mathrm{~g}
$$
The SI unit of mass is the kilogram, and
$$
\begin{aligned}
d=\frac{1}{2 \pi}\left(\frac{h}{\nu_{0 \rightarrow 1} \mu}\right)^{1 / 2} & =\frac{1}{2 \pi}\left[\frac{6.62607 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\left(48991.0 \times 10^6 \mathrm{~s}^{-1}\right)\left(1.44885 \times 10^{-26} \mathrm{~kg}\right)}\right]^{1 / 2} \\
& =1.5377 \times 10^{-10} \mathrm{~m}
\end{aligned}
$$
