Question

4. The first excited state of E2 of hydrogen atom is 10.2 eV above the ground state E1. The degeneracy of the ground and the first excited states are 2 and 8 respectively. Determine the ratio of no. of atoms in the first excited state to the no. in the ground state at T=6000K.

Answer 1

Write the Boltzmann Equation:

$$\frac{N_b}{N_a}=\frac{g_b}{g_a}\text{exp}\bigg(\frac{E_a-E_b}{kT}\bigg)=\\ \space\\ =\frac{8}{2}\text{exp}\bigg[\frac{[-13.6-(-10.2)]1.602\cdot10^{-19}}{1.38\cdot10^{-23}\cdot6000}\bigg]=5.57\cdot10^{-3}.$$