Most Probable State Determination
Assuming Harmonic Oscillator energy levels (i.e. equi-spaced), with separation A, so that the energies are 0, A, 2A, 3A, 4A, etc., and assuming we have 6 oscillators to distribute amongst these states, obtain the populations of each of the states in the most probable `complection/configuration', i.e., the one where S = k ln(W) is defined. Assume that the total energy is 12A.
(See page 10, Dykstra's 'Physical Chemistry, A Modern Introduction')
Number in zero state (E=0)=
0
1
2
3
4
5
6
7
8
9
Number in first excited state (E=A) =
0
1
2
3
4
5
6
7
8
9
Number in second excited state (E=2A) =
0
1
2
3
4
5
6
7
8
9
Number in third excited state (E=3A) =
0
1
2
3
4
5
6
7
8
9
Number in fourth excited state (E=4A) =
0
1
2
3
4
5
6
7
8
9
Number in fifth excited state (E=5A) =
0
1
2
3
4
5
6
7
8
9
Number in sixth excited state (E=6A) =
0
1
2
3
4
5
6
7
8
9
Number in seventh excited state (E=7A) =
0
1
2
3
4
5
6
7
8
9
Query, are the above entries correct?
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Carl W. David
Department of Chemsitry
University of Connecticut
Storrs, Connecticut 06269-4060
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david@uconnvm.uconn.edu
Carl W. David