= Solution
The combined <boron trifluoride> series gives
$$
2\widetilde B=0.686\ \mathrm{cm}^{-1},\qquad
\boxed{\widetilde B=0.343\ \mathrm{cm}^{-1}=34.3\ \mathrm m^{-1}.}
$$
Using $I_b=h/(8\pi^2c\widetilde B)$ with the wavenumber in inverse metres,
$$
I_b=\frac{6.62607015\times10^{-34}}
{8\pi^2(2.99792458\times10^8)(34.3)}
=8.16115\times10^{-46}\ \mathrm{kg\,m^2}.
$$
With integer fluorine mass $m_{\rm F}=19u$ and $I_b=3m_{\rm F}R^2/2$,
$$
R=\sqrt{\frac{2I_b}{3(19u)}}
=1.31319\times10^{-10}\ \mathrm m,
\qquad\boxed{R\simeq131\ \mathrm{pm}.}
$$
Using $4\widetilde B$ for the observed spacing would incorrectly treat BF$_3$ as a linear diatomic rotor and overestimate the bond length by $\sqrt2$.
Back to article page
