# 10 Oscillators & 8 Quanta of Energy, Dominant Configuration?

1. Apr 22, 2012

### TChill

Hello, This is my first time using this forum, I just have a quick question that I'm trying not to get too held up on by re-reading (skimming) the chapter several times. Anyways:

1. The problem statement, all variables and given/known data

"Consider the case of 10 oscillators and eight quanta of energy. Determine the dominant configuration of energy for this system by identifying energy configurations and calculating the corresponding weights. What is the probability of observing the dominant configuration?"

2. Relevant equations

The Weight (W): W = N!/[(N0!)(N1!)(N2!)(N....!)]

3. The attempt at a solution

I know that the answer is 10!/[(1!)(1!)(3!)(5!)] = 5040 for the dominant weight, but when I used (instead of 3! & 5!, I used) 10!/[(4!)(4!)] which equals 6300. Why is 3! & 5! the dominant energy configuration used then? I assume you can't use two of the same N (populations/states)? Is that right?

I also got the answer to the rest of the question too [WTOTAL=20170, & Probability = W/Wtotal = 5040/20170 = 0.25], but I couldn't get passed the very first part of figuring out the dominant energy configurations for this system.

-Thanks

***NOTE: The text that I'm given is Physical Chemistry 9th Ed. by Atkins. I know that the question is from some Statistical Thermodynamics and Kinetics book by Engel & Reid. If it matters.. I feel like the book I have is a little difficult to follow

2. Apr 25, 2012

### diazona

If I'm understanding the problem correctly, the numbers in the denominator of the weight are the number of oscillators with that many quanta of energy. In particular there are $N_0$ oscillators in the ground state, $N_1$ oscillators in the first excited state, and so on. Given that interpretation, try computing the total amount of energy in each of the configurations you've discussed. You should find that
$$N_0 = 5, N_1 = 3, N_2 = 1, N_3 = 1,\ldots$$
contains the correct amount of energy, but
$$N_0 = 4, N_1 = 4, N_2 = 1, N_3 = 1,\ldots$$
does not.

3. Apr 25, 2012

### TChill

Thank you for a reply! However, I'm not understanding why the correct configuration is correct. I assumed the larger weight (W) would be dominant & therefore the answer, but it is not.

Why is that configuration the correct one?

4. Apr 26, 2012

### diazona

My point was that, of the two alternatives you mentioned in your question, the one with the larger weight is not a valid configuration for this physical system. My last post explains why that is the case.

Did you try computing the total amount of energy in each configuration, as I suggested?

5. Apr 26, 2012

### TChill

So what I believe wasn't clear about the problem was that there were only 4 energy levels (ε0, ε1, ε2, ε3). Silly mistake. But it wasn't exactly obvious from the textbook we are given that the occupation numbers were used to describe how many units occupy a given energy level.

Now it is clear why (1!)(1!)(4!)(4!) didn't work because it would correspond with 9 quanta of energy [E=ƩNii] therefore (0e*4)+(1e*4)+(2e*1)+(3e*1) = 0+4+2+3 = 9 which is what you meant by an invalid configuration because it over 8. It was difficult to understand what you meant without these connections, so it wasn't very clear. Thanks anyways though, appreciate it.