Physics - Heat & Thermo Question with Solution | TestHub

For monatomic molecule, degree of freedom $=3$.

The formula for the average kinetic energy is given by

$K_{\mathrm{avg}}=\frac{1}{2}f {\mathrm{K}}_{\mathrm{B}}\mathrm{T} ...\left(1\right)$

From equation (1), it follows that

$$\begin{gathered} 0.414 \mathrm{eV}\times \frac{1.6\times {10}^{-19} \mathrm{J}}{1 \mathrm{eV}}=\frac{3}{2}\times 1.38\times {10}^{-23} \mathrm{J} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}\times T \\ \Rightarrow T=\frac{0.414 \mathrm{eV}\times \frac{1.6\times {10}^{-19} \mathrm{J}}{1 \mathrm{eV}}\times 2}{3\times 1.38\times {10}^{-23} \mathrm{J} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}} \\ =3200 \mathrm{K} \end{gathered}$$