The atmosphere of the sun consists mostly of hydrogen atoms (not molecules) at a temperature of $6000k$. What are (a) the average translational kinetic energy per atom and (b) the rms speed of the atoms?

(a) The average translational kinetic energy of a molecule with mass m and velocity v is The average translational kinetic energy of a molecule is affected by its temperature, hence it is related to the temperature t per molecule in the manner

${ϵ}_{\mathrm{avg}}=\frac{3}{2}{k}_{\mathrm{B}}T$

Where $k_{\mathrm{B}}$is Boltzmann's constant and in $SI$unit its value is

$k_{\mathrm{B}}=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$

Plug the values for $k_{\mathrm{B}}$and $T$into equation (1) to get the energy for the atom

${ϵ}_{\mathrm{avg}}=\frac{3}{2}{k}_{\mathrm{B}}T$

$=\frac{3}{2}\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(6000\mathrm{K}\right)$

$=1.24\times {10}^{-19}\mathrm{J}$

(b)The transitional energy depends on the temperature only not the mass of the gas. The molecule with mass $m$and velocity $v$has an average translational kinetic energy and it is given by equation $(20.19)$in the form

${ϵ}_{\mathrm{avg}}=\frac{1}{2}m{v}_{\mathrm{rms}}^2$

As shown, both equations$\left(1\right)and\left(2\right)$have the same left side, so we can use these expressions to get an equation for root mean square velocity $v_{rms}$

$\begin{array}{c}\frac{1}{2}m{v}_{\mathrm{rms}}^2=\frac{3}{2}{k}_{\mathrm{B}}T\\ v_{\mathrm{rms}}^2=\frac{3k_{\mathrm{B}}T}{m}\\ v_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm{B}}T}{m}}\end{array}$

The mass of the hydrogen atom is the same for one proton $m_p=1.67\times {10}^{-27}\mathrm{kg}$. Now, we plug the values for $K_{B}T$and $m_p$into equation (3) to get $v_{rms}$

$v_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm{B}}T}{m_{\mathrm{p}}}}$

$=\sqrt{\frac{3\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(6000\mathrm{K}\right)}{1.67\times {10}^{-27}\mathrm{kg}}}$

$=1.22\times {10}^4\mathrm{m}/\mathrm{s}$