Asked by Anon
So I have worked out the 'allowed' configurations for four particles which can occupy 5 different energy levels (1, 3, 5, 7, 9) and now I need to work out the average number of particles in a particular energy level (i.e 3).
1) 9,5,1,1
2) 9,3,3,1
3) 7,7,1,1
4) 7,3,3,3
5) 7,5,3,1
6) 5,5,5,1
7) 5,5,3,3
So there are 7 total configurations and three appears this number of times(in brackets):
1) 9,5,1,1 (0)
2) 9,3,3,1 (2)
3) 7,7,1,1 (0)
4) 7,3,3,3 (3)
5) 7,5,3,1 (1)
6) 5,5,5,1 (0)
7) 5,5,3,3 (2)
So would I say that 3 appears 8 times if all the sequences were counted to a total number of values of 28.
Am I on the right track or am I over complicating this?
1) 9,5,1,1
2) 9,3,3,1
3) 7,7,1,1
4) 7,3,3,3
5) 7,5,3,1
6) 5,5,5,1
7) 5,5,3,3
So there are 7 total configurations and three appears this number of times(in brackets):
1) 9,5,1,1 (0)
2) 9,3,3,1 (2)
3) 7,7,1,1 (0)
4) 7,3,3,3 (3)
5) 7,5,3,1 (1)
6) 5,5,5,1 (0)
7) 5,5,3,3 (2)
So would I say that 3 appears 8 times if all the sequences were counted to a total number of values of 28.
Am I on the right track or am I over complicating this?
Answers
Answered by
bobpursley
I am thinking you missed the point. You have four particles, and each of them can be in energy level a,b,c,d,e
so you need to take those energy levels, choosing four at a time to get your possible combinations. You job is to find the average number of any letter.
a,a,a,b
a,a,b,a
a,b,a,a and so on (it is a long list). Seems a combination formula will get you the results quicker. Perhaps I am not understanding the problem.
so you need to take those energy levels, choosing four at a time to get your possible combinations. You job is to find the average number of any letter.
a,a,a,b
a,a,b,a
a,b,a,a and so on (it is a long list). Seems a combination formula will get you the results quicker. Perhaps I am not understanding the problem.
Answered by
Anon
Hi Bob,
Thanks for answering, I have gone the long way around with this question as im still not sure how I can end up with the average number of particles in N_3 can be 8/7.
But according to my textbooks the answer should be arrived at as follows:
<N_3>= (0*1/7)+(2*1/7)+(0*1/7)+(3*1/7)+(1*1/7)+(0*1/7)+(2*1/7)=8/7
Thanks for answering, I have gone the long way around with this question as im still not sure how I can end up with the average number of particles in N_3 can be 8/7.
But according to my textbooks the answer should be arrived at as follows:
<N_3>= (0*1/7)+(2*1/7)+(0*1/7)+(3*1/7)+(1*1/7)+(0*1/7)+(2*1/7)=8/7
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