Question

The ratio of average translational kinetic energy to rotational kinetic energy of a diatomic molecule at temperature T is

• A. $3$
• B. $\frac{7}{5}$
• C. $\frac{5}{3}$
• D. $\frac{3}{2}$

Answer 1

The correct option is D $\frac{3}{2}$

We know total $K.E=\frac{f}{2}kT$
$f\to$ degree of freedom
for diatomic gas
$f=f_T+f_R$
$f_T=3$
$f_R=2$
$(K.E{)}_T=\frac{f_T}{2}kT=\frac{3}{2}kT$
$(K.E{)}_R=\frac{f_R}{2}kT=\frac{2}{2}kT$
$\frac{(K.E{)}_T}{(K.E{)}_R}=\frac{3}{2}$