# Fraction of Hydrogen Atoms in First Excited State

1. Apr 28, 2010

### burg25

1. The problem statement, all variables and given/known data
1. The problem statement, all variables and given/known data
The temperature of the surface of a certain star is 8000 K. Most hydrogen atoms at the surface of the star are in the electronic ground state. What is the approximate fraction of the hydrogen atoms that are in the first excited state (and therefore could emit a photon)? The energy of the first excited state above the ground state is (-13.6/22 eV) - (-13.6 eV) = 10.2 eV = 1.632e-18 J.

2. Relevant equations
Kb = 1.38e-23 J/K
1/T = E/Kb

3. The attempt at a solution
I don't really know how to do this problem. My guess was to divide the energy in joules by the temperature and then divide by the boltzmann constant to make it unitless.

(1.632e-18/8000)/(1.38e-23)=14.8

2. Apr 28, 2010

### nickjer

Hmm.... You are right so far, but you might want to use the Boltzmann factor for this problem as well.

3. Apr 28, 2010

### burg25

What do you mean? Could you clarify?

4. Apr 28, 2010

### burg25

Wait I think I understand. Are you saying take e raised to this value?

5. Apr 28, 2010

### burg25

Yeah I got it. Thanks

6. Apr 28, 2010

### nickjer

Yes, using the Boltzmann factor you get:

$$\frac{P(E_1)}{P(E_0)} = e^{-\frac{E_1-E_0}{kT}}$$

where $P(E)$ is the probability the electron is in the state with energy $E$.

Since a vast majority of the electrons are in the ground state, you can say $P(E_0)\approx 1$.