Question

Find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to n = 3 state. Hydrogen can now emit red light of wavelength 653.1 nm. Because of Maxwellian distribution of speeds, a hydrogen sample emits red light at temperatures much lower than that obtained from this problem. Assume that hydrogen molecules dissociate into atoms.

Answer 1

Given:
Wavelength of red light, $\lambda$ = 653.1 nm = 653.1$\times$10^$-9$ m

Kinetic energy of H₂ molecules $\left(K\right)$ is given by
$K=\frac{3}{2}kT$ ...(1)
Here, k = 8.62 × 10⁻⁵ eV/K
T = Temperature of H₂ molecules
Energy released $\left(E\right)$ when atom goes from ground state to n = 3 is given by
$E=13.6\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$
For ground state, n₁ = 1
Also, n₂ = 3
$$\begin{gathered} \therefore E=13.6\left(\frac{1}{1}-\frac{1}{9}\right) \\ =13.6\left(\frac{8}{9}\right)...\left(2\right) \end{gathered}$$

Kinetic energy of H₂ molecules = Energy released when hydrogen atom goes from ground state to n = 3 state
$$\begin{gathered} \therefore \frac{3}{2}\times 8.62\times {10}^{-5}\times T=\frac{13.6\times 8}{9} \\ \Rightarrow T=\frac{13.6\times 8\times 2}{9\times 3\times 8.62\times {10}^{-5}} \\ =9.4\times {10}^4\mathrm{K} \end{gathered}$$