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Maxwell-Boltzmann Energy distribution

  • Thread starter WrongMan
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  • #1
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Homework Statement


find the average energy of a system with n energy states (0, 1E, 2E, 3E...nE)

Homework Equations


P(E) = e-BE/Z - where B=1/KbT and Z= ∑e(-BE)n
<E>=∑(nE* (e-BE)n) /Z

The Attempt at a Solution


i feel like ive gone down the correct path - that is finding result of the sums.
Z - if S=R0+R1+...+Rn i do S-RS and get S=R0-Rn+1/1-R ... so Z= e0-e-BE(n+1)/1-eBE
now it gets tricky, i tried to evaluate the sum ∑(nE* (e-BE)n) in a similar way,
so S2=0R0+1R1+2R2...
and RS2=0R+1R2+2R3...
so S2-RS2=S-1 (remember S=Z; "-1" cause R0 is missng)
so S2=(1-e-BE(n+1)/(1-e-BE)2-1/(1-e-BE).

substituting into average fomula and simplifying my answer is: (1/1-e-BE) - (1/1-eBE(n+1))
is this correct?
 

Answers and Replies

  • #2
TSny
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now it gets tricky, i tried to evaluate the sum ∑(nE* (e-BE)n) in a similar way,
so S2=0R0+1R1+2R2...
and RS2=0R+1R2+2R3...
so S2-RS2=S-1 (remember S=Z; "-1" cause R0 is missng)
You're missing a term on the right hand side of the last equation. (Don't forget to consider the last terms in S2 and RS2).
 
  • #3
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You're missing a term on the right hand side of the last equation. (Don't forget to consider the last terms in S2 and RS2).
oh right RS2 ends with nRn+1 so S2 ends with "-nRn+1" (the missing term)
so my new answer is: (1/1-e-BE) + (-1-ne-BE(n+1)/1-e-BE(n+1))
 
  • #4
TSny
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OK, that looks right for S2 / Z. But that's not quite the answer for <E>.

When you have expressions like 1/1-e-BE , you should include parentheses around the entire denominator: 1/(1-e-BE).
 

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