Maxwell Boltzmann distribution


by vincisonfire
Tags: boltzmann, distribution, maxwell
vincisonfire
vincisonfire is offline
#1
Nov10-08, 01:29 PM
P: 4
1. The problem statement, all variables and given/known data
The energy difference between the first excited state of mercury and the ground state is 4.86 eV.
(a) If a sample of mercury vaporized in a flame contains 10^20
atoms in thermal equilibrium at 1600K, calculate the number of atoms in the n=1 (ground) and n=2 (first-excited) states. (Assume the Maxwell-Boltzmann distribution applies and that the n=1 and n=2 states have equal statistical weights.)
2. Relevant equations
Maxwell Boltzmann Distribution
3. The attempt at a solution
I thought that since they have the same statistical weight, there must be 5*10^19 particles in each state. But I don't think it is the good answer since we use this number for another exercice and it doesn't yield the good answer.
I don't know how to figure out these number using Maxwell Boltzmann distribution.
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Redbelly98
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#2
Nov11-08, 07:58 PM
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Maxwell-Boltzmann refers to the distributions of atom velocities and is not relevant here.

What are relative populations of 2 states, separated by energy ΔE and at a temperature T?
borgwal
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#3
Nov11-08, 08:10 PM
P: 370
"Statistical weight" refers to the number of different states there are with n=1 and n=2, respectively. In reality the statistical weight is certainly not the same for these states, but never mind: all you need is the energy difference between the two states, and then answer Redbelly's question.

[It's a bit unfortunate that *sometimes* by "statistical weight" one means exactly the Boltzmann factor you will have to calculate: so by that definition you are correct to say the populations in the states are equal. But clearly, the person who phrased the question did not have that meaning of "statistical weight" in mind.]


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