Question

The average kinetic energy of molecules in a gas at temperature T is 1.5 kT. Find the temperature at which the average kinetic energy of the molecules of hydrogen equals the binding energy of its atoms. Will hydrogen remain in molecular from at this temperature? Take k = 8.62 × 10⁻⁵ eV K⁻¹.

Answer 1

Average kinetic energy $\left(K\right)$ of the molecules in a gas at temperature $\left(T\right)$ is given by
K = $\frac{3}{2}kT$
Here,
k = 8.62 × 10⁻⁵ eVK⁻¹
T = Temperature of gas

The binding energy of hydrogen atom is 13.6 eV.

According to the question,
Average kinetic energy of hydrogen molecules = Binding energy of hydrogen atom
$\therefore$1.5 kT = 13.6
$\Rightarrow$1.5 × 8.62 × 10⁻⁵ × T = 13.6
$$\begin{gathered} \Rightarrow T=\frac{13.6}{1.5\times 8.62\times {10}^{-5}} \\ =1.05\times {10}^5\mathrm{K} \end{gathered}$$
No, it is impossible for hydrogen to remain in molecular state at such a high temperature.