- #1

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

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- Thread starter shayan825
- Start date

- #1

- 16

- 0

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

- #2

- 49

- 7

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

- #3

- 16

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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

- 49

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- #5

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groundstate, and -3.0 eV be the energy of theexcitedstate? Is it supposed to be the other way around?[/QUOTwell,

thats what the book says..ya

- #6

- 49

- 7

(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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