Temperature at which ground state and first excited state have equal populations

[Benzoate]

Homework Statement

Consider a pure hydrogen hydrogen gas ata tempeature of 10080 K. What is the ratio of the populations of the ground state(n=1) to the first excited state(n=2). Note that the energy difference is 1.634e-18 joules between these two states. At what temperature would both levels have equal populations?

Homework Equations

Boltzmann equation: N(B)/N(A)= g(b)/g(a) exp[(E(A)-E(B))/kT
g=2*n^2

The Attempt at a Solution

finding N(B)/N(A) was easy. N(2)/N(1)=2(2)^2/2(1)^2*exp(1.634e-18 joules)/(10080 K)(1.380 *10^-23))= .000031665

I had a difficult time finding the Temperature when both populations are equal. When both populations are equal, does that imply N(a)/N(b) = 1 because N(a)=N(B)?

Now finding the temperature is simple : 1=4 exp(1.634e-18 joules/(T(1.380e-23)) => T= -85411.7 Kelvins.

The only problem is I don't know if I should assume N(B)/N(A) = 1 , just because the temperatures are equal

 

[CompuChip]

No, you shouldn't assume N(B)/N(A) = 1 because the temperatures are equal.
You should assume it because you assume that the populations are equal; then you fill in the formula and solve for the temperature where this occurs.

So you have obtained the right answer with the right method but with a wrong way of thought

 

[Benzoate]

No, you shouldn't assume N(B)/N(A) = 1 because the temperatures are equal.
You should assume it because you assume that the populations are equal; then you fill in the formula and solve for the temperature where this occurs.

So you have obtained the right answer with the right method but with a wrong way of thought

sorry, I meant to say that the populations are equal, so should I assume that N(a)/N(b)=1 since N(a)=N(b)

 

[CompuChip]

Then you got the right answer by the right method and a typo :)