At a temperature T , the average kinetic energy of gas molecules is given by 1.5 kT . Determine the temperature at which the average kinetic energy of hydrogen molecules equals the binding energy of its atoms. Additionally, state whether hydrogen will remain in molecular form at this temperature. Take k = 8.62 10⁻⁵ eV K ⁻¹ .

Step-by-step solution

The binding energy of a hydrogen atom is 13.6 eV . The average kinetic energy of a gas molecule is given as 1.5 kT . Equating the average kinetic energy to the binding energy of the hydrogen atom: 1.5 kT = 13.6 eV Substituting the value of the Boltzmann constant k = 8.62 10⁻⁵ eV K ⁻¹ : 1.5 8.62 10⁻⁵ T = 13.6 1.293 10⁻⁴ T = 13.6 T = 13.6 1.293 10⁻⁴ 1.05 10^5 K At this extremely high temperature, the average kinetic energy of the molecules ( 13.6 eV ) is much greater than the bond dissociation energy of a hydrogen mo