Answer to Question #109301 in Atomic and Nuclear Physics for Shalini

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:

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