The distribution of the energy of the molecules is also of importance in the kinetics of chemical reactions. The fraction of molecules with energy in the range between \(E\) and \(E+d E\) is given \(F(E) d E\), where \(F(E)\) is an energy distribution function. Show that

\[F(E)=\frac{2}{\sqrt{\pi}}\left(\frac{1}{k_{\mathrm{B}} T}\right)^{3 / 2} E^{1 / 2} \exp \left(-E /\left(k_{\mathrm{B}} T\right)\right)\]

Hint: the kinetic energy is \(E=m c^{2} / 2\) and hence the substitution \(d E=m c d c\) in the expression for the KE distribution can be used.

Note that an equation of this type is used for the study of the effect of temperature on the reaction rate constant.