Fraction of Hydrogen Atoms in First Excited State

[burg25]

Homework Statement

1. Homework Statement
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.

Homework Equations

Kb = 1.38e-23 J/K
1/T = E/Kb

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

 

[nickjer]

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

 

[burg25]

What do you mean? Could you clarify?

 

[burg25]

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

 

[burg25]

Yeah I got it. Thanks

 

[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 $P(E_0)\approx 1$.