Boltzmann distribution

  • Thread starter shayan825
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  • #1
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A certain particle is interacting with a reservoir at 500 k and can be in any four possible states. The ground state has energy 3.1 eV and three excited states all have the same energy. what is the probability that the particle is in ground state? what is the probability that the particle is in a particular excited state? what is the probability that it is in a state of energy -3.0 eV? using Boltzmann distribution

I use the formula (e^E/kT)/sum of e^E/kT but get a wrong answer.
 

Answers and Replies

  • #2
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There is a factor of -1 missing from the exponent. Are you also including the degeneracy in the sum?
 
  • #3
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There is a factor of -1 missing from the exponent. Are you also including the degeneracy in the sum?

yes, for the first question(the probability that the particle is in the ground state) I get 0.032. e^(-3.1/500k) / e^(-3.1/500k) + 3e^(-3.0/500k). The answer should be 0.7725
 
  • #4
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Why would 3.1 eV be the energy of the ground state, and -3.0 eV be the energy of the excited state? Is it supposed to be the other way around?
 
  • #5
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Why would 3.1 eV be the energy of the ground state, and -3.0 eV be the energy of the excited state? Is it supposed to be the other way around?[/QUOTwell,

thats what the book says..ya
 
  • #6
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There seem to be a couple of problems here:
(1) The energy of the ground state should not be greater than the energy of an excited state. The ground state should have the lowest energy. I guess you meant that the ground state is -3.1 eV and the excited state is -3.0 eV, i.e. a 0.1 eV difference. (You were missing this minus sign before).
(2) The Boltzmann factor is Exp[-ΔE/kT], not Exp[ΔE/kT], where ΔE > 0.
(3) You seem to be mixing the usage of k and K. K denotes a temperature in Kelvin and is a unit, and k is the Boltzmann constant, a number. You say that the reservoir is at 500 k, when I think you meant 500 K. You then divide the energies in eV by 500k, but it is unclear whether you think this means 500*k or 500 K. (500*k would be right assuming you're using the value of k in eV/K).

It will be easiest, from a computational standpoint, if you redefine -3.1 eV to be the zero of the energy . Then the ground state has energy 0 eV, and the excited states have energy 0.1 eV.
 

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