Partition Function for a helium atom

[1v1Dota2RightMeow]

Homework Statement

The first excited state of the helium atom lies at an energy 19.82 eV above the ground state. If this excited state is three-fold degenerate while the ground state is non-degenerate, find the relative populations of the first excited and the ground states for helium gas in thermal equilibrium at 10,000K.

Homework Equations

$$Z=\sum_i e^{-\varepsilon /kT}$$

The Attempt at a Solution

$$Z=\sum_i e^{-\varepsilon /kT}$$
$$Z=3e^{-(19.82eV) /(8.617e-5eV/K)(10000K)}$$
$$Z=3.074e-10$$

I found the partition constant for the first excited state, but then I'm not sure what to do. What do they mean by "the relative populations"?

 

[DrClaude]

I found the partition constant for the first excited state,

That statement doesn't make sense. The partition function is a sum over all states. You can't find the partition function for a single state.

What do they mean by "the relative populations"?

$P_2/P_1$, where $P_1$ is the population of the ground state and $P_2$ the population of the excited state.

 

[1v1Dota2RightMeow]

That statement doesn't make sense. The partition function is a sum over all states. You can't find the partition function for a single state.$P_2/P_1$, where $P_1$ is the population of the ground state and $P_2$ the population of the excited state.

DrClaude, thank you for responding. What exactly is the ''population'' of a ground state of an atom?

 

[DrClaude]

DrClaude, thank you for responding. What exactly is the ''population'' of a ground state of an atom?

It's the number of atoms in the ground state. The term comes from the idea that you have an ensemble of identically prepared systems: if you have 100 atoms with a probability of .9 of being in the ground state, then the population of the ground state is 90, and the relative population is .9.

 

[1v1Dota2RightMeow]

It's the number of atoms in the ground state. The term comes from the idea that you have an ensemble of identically prepared systems: if you have 100 atoms with a probability of .9 of being in the ground state, then the population of the ground state is 90, and the relative population is .9.

Wouldn't the relative population then be $$P2/P1 = 90/10 = 9$$

 

[DrClaude]

Wouldn't the relative population then be $$P2/P1 = 90/10 = 9$$

It's a question of interpretation. Re-reading the problem as you stated it in the OP, I think indeed that you need to calculate the relative population of the excited state as $P_2/P_1$, which in my example would give $0.1/0.9 \approx 0.111$.