Question

A hydrogen atom can be in either of two 1S states, whose energy we'll call 0, either of two 2S states, or any of 6 2P states. The 2S and 2P states have energies of 10.2 eV. There are other states with higher energy, but we'll ignore them for simplicity. The 2P states have distinctive optical properties, so we're interested in how many are present even when it's a small fraction of the total.a) What fraction of the H is in 2P states at T=5900 K, a typical Sun surface temperature?b) What fraction of the H is in 2P states at T= 4300 K, a typical sunspot temperature?

Answer 1

In order to solve this problem we must resort to Boltzmann's theory,

His theory describes how energy levels are populated within atoms. The Boltzmann equation gives ratios of level populations as a function of temperature as follow,

$\frac{N_2}{N_1} =\frac{g_2}{g_1}e^{-\frac{\Delta E}{k_B T}}$

Where,

g1/g2 are the states (statistical weights)

$\Delta E$ Energy

T temperature

$k_B$ Boltzmann constant

We have all the data, thus replacing,

A) The fraction of H in 2P state at 5900K is,

$\frac{N_2}{N_1} = \frac{6}{2} e^{\frac{-10.2(1.6*10^{-19})}{1.38*10^{-23}*5900}}$

$\frac{N_2}{N_1} = 5.92.10^{-9}$

Note that I did the convertion in energy, remember that

$1eV = 1.6*10^{-19}J$

The fraction of H in 2P state at 5900K is $5.92.10^{-9}$

B) The fraction of H in 2P state at 4300K is,

$\frac{N_2}{N_1} = \frac{6}{2} e^{\frac{-10.2(1.6*10^{-19})}{1.38*10^{-23}*4300}}$

$\frac{N_2}{N_1} = 3.41*10^{-12}$

The fraction of H in 2P state at 4300K is  $3.41*10^{-12}$