# Rotational Constant and Moment of Inertia of Fluorine gas

I have come across some homework question on thermodynamics which needs me to calculate $B$ of $F_2$

My attempt:

$B= \frac{h}{8\pi^2cI}$ where $I=\mu r^2=\frac{m_1m_2}{m_1+m_2} r^2$

Atomic mass of fluorine atom is $18.998$ $g$ ${mol}^{-1}$ and Bond length is $141.8\times10^{-12}$

$B=\frac{6.626\times 10^{-34}}{8 \pi^2 (3\times10^{10})\frac{\frac{18.998^2}{18.998\times2}}{6.02\times 10^{23} \times 1000}(141.8\times10^{-12})^2}\approx0.882$

which I think it is wrong because when I continue to calculate the rotational temperature $\Theta_R$, the result is way too low.

$\Theta_R=\frac{hcB}{k_B}=\frac{(6.63\times10^-34)(3\times10^{10})(0.882)}{1.38\times10^{-23}}=1.27 K$, $K$ for kelvin.

Actually, I don't think it matters because it should cancel in $\Theta_R$, BUT your mass is given in grams and not Kg. You need to double check all of your units here. –  sujeet Mar 25 '13 at 0:37
@sujeet I think we need to use $Kg$ as the unit of reduced mass here.. –  Paul Mar 25 '13 at 0:44
Paul, are you sure it's wrong? Rotational temperature for $\mathrm{CO}_2$ is $\sim0.5\,\mathrm{K}$ according to en.wikipedia.org/wiki/Rotational_temperature –  sujeet Mar 25 '13 at 1:08