The Boltzmann Distribution
Ludwig Boltzmann (1844 – 1906).
Association
The intensity of transition from one quantum state to another resulting
from the interaction of electromagnetic radiation with a molecule is
not only dependent:
• on matching the energy of the radiation with differences in
molecular energy levels and
• selection rules governing allowed transitions;
but the intensity of transition is also dependent on the number of
molecules occupied different levels at initial state of the transition.
The ratio of molecules at the jth level to the others at different levels
shows an exponential variation
nj
N

exp  ε j / kT 
1.
 exp- εi / kT 
i
N   ni
i
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where N is the total number of molecules populated all the possible
levels, nj is the number of molecules on the jth level, k is the
Boltzmann constant, and T is the temperature in Kelvin.
Fig. 1. Electronic and vibrational energy levels. Waves symbolise the
appropriate wave function.
Concluding to two levels one of them is populated by n1 the other by
n2 molecules, and N = n1 + n2.
n2
exp  ε 2 / kT 

n1  n2 exp  ε1 / kT   exp  ε 2 / kT 
and after simplification
n1  n2 n1
exp  ε1 / kT   exp  ε2 / kT  exp  ε1 / kT 
 1 

1
n2
n2
exp  ε2 / kT 
exp  ε2 / kT 
and its reciprocal gives the n2/n1.ratio
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2.
n2
 exp   2  1  / kT   exp  ε / kT 
n1
3.
where Δε = ε2 - ε1 is the energy difference between levels.
In general, the Boltzmann distribution tells us that the ratio of
populations varies exponentially with the energy difference, and the
greater the level difference the smaller the population on the ith level.
 εi  ε j 
ni

 exp  
nj
kT


The dynamic modes into which the internal energy can be distributed
are:
translation-, rotation-, vibration- and electron-energy.
Taking the natural logarithm of equation
ln
εi  ε j
ni

nj
kT
The natural logarithm of the ratio of the particle number in two
different energy states is proportional to the negative of their energy
separation.
Energy level populations
We shall now work out population ratios for different kinds of energy
mode at room temperature, 25 oC (298.15 K). We calculate population
distributions for one mole. The denominator of exponent takes the
value
R=NA k = 8.314 JK-1mol-1, and RT = 298.15·8.314 = 24789 Jmol-1
RT = 2.48 kJmol-1 (rounding).
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Electronic level population
For an electronic level difference Δε is quite high. If we take a
hydrogen atom in gas phase Δε = 1000 kJmol-1. The population of the
2s excited state n2s relative to ground state, n1s is
n2 s
 1000 
 403
 exp  
 0 , so the number 2s H-atoms
e
n1s
 2.48 
n2 s  n1s  0
so that we can be very confident that there are no 2s hydrogen atoms
around room temperature.
Vibrational level population
Next we consider vibrational modes. If we take CO molecules, the
vibrational quantum levels have an energy separation of around Δε=25
kJ mol-1. Vibrational energy levels (v = 1, 2, 3…) are much more
closely spaced than electronic ones. Hence
n v 1
 25 
10
5
 exp  
  e  5  10
nv 0
 2.48 
If we have a system containing 20000 molecules, than
nv1  20000  5 105  1
It means that one out of every 20000 molecules is found at the first
vibration excited state in a time average.
Rotational level population
Rotational levels are much more closely spaced than vibrational
levels. For CO the rotational energy levels are around 0.05 kJ mol-1
apart. We can write
n J 1
 0.05 
 0.02
 3  exp  
 2.94
  3 e
nJ 0
2
.
48


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The factor of 3 in the equation above requires some explanation. It
arises because the first excited level J=1 is actually threefold
degenerate (i.e. there are 3 levels all with the same energy). This
degeneracy factor must be taken into account when assessing the
probability of finding a molecule in a given level.
Example: Temperature dependence of population.
A comparison of population density between temperatures 25 and
1000 oC.
Calculate the ratio of molecules in excited rotation, vibration and
electronic energy level to that in the lowest energy level.
T1 = 25 oC, T2 = 1000 oC
first excited energy levels are at 30 cm-1 rotation,
1000 cm-1 vibration,
40 000 cm-1 electronic,
above the lowest one.
Assume that, J = 4 for excited rotation level, and such a level is 2J+1
fold degenerate.
Vibration and electronic states have no degeneracies.
h = 6.626 10-34J/s, c = 3 1010 cm/s, k = 1.38 10-23 J/K, T1 = 298 K, T2 =
1273 K
Data:
Rotation
The energy difference is given in wavenumber, ~ in cm-1.
c
  h   h   h  c ~

4.
 hc~
 hc~
n upper

n upper

  7.8 and at T2
  8.7
 9  exp  
 exp  
n lower 
n lower 
 kT1 
 kT2 
In each set of ten molecules at T2 = 1000 oC only one is at the lowest
energy level.
Vibration
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 hc~
 hc~
n upper

n upper

  0.008 and at T2
  0.323
 exp  
 exp  


n lower 
kT
n
lower
kT


1 
2 
A 40-fold increment in population inversion is caused by the increase
in temperature.
Electronic
 hc~
n upper

  1.24  10  84 and at T2
 exp  
n lower 
 kT1 
 hc~
n upper

  2.34  10  20
 exp  
n lower 
 kT2 
Practically, all the molecules are in ground electronic state at room
temperature.
The Boltzmann distribution tells us how the internal energy, U of a
system is distributed amongst the various energy levels of the system.
Let our system quantised with energy levels εi where i index is used to
label the levels. The nature of the levels is irrelevant. The lowest level
will be labelled with the index i =0, so that
ε0  ε1  ε2  ...εi
Giving that our molecules are in a state of constant thermal agitation
it is evident that a given molecule will not always be found in the
same state. It is true, however, that over a period of time there tends to
be the same number of molecules in a given state. If we label the
number of molecules in state i as ni, then the total internal energy of
the system is
U  n0 ε0  n1ε1  n2 ε2 ...   ni εi
5.
i
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Macrostate, microstate, multiplicity
If we use Newtonian mechanics, the state of the system is
characterized by the positions and velocities of all the particles in the
system. In thermodynamics, we use the expression microstate to refer
to such a state.
We use the expression macrostate when we only specify the values of
the macroscopic parameters of the system, e.g. an ideal gas which
state is characterized by T, p, n, V.
The number of microstates in a macrostate is called the multiplicity of
the microstate.
Molecules as Balls Energy Levels as Boxes
Suppose we have 7 atoms and we introduce quantum mechanics into
the picture, that is, the energies of these atoms are quantized. We can
think of these energy levels as boxes.
For simplicity, we will assume that the allowed energy levels are
between ε0… ε7, increased by increment one. E.g. ε0 contains zero
energy unit, while ε5 contains five units. Altogether, we have 8 boxes
of different energies the atoms can take up.
ε0n0 + ε1n1 + ε2n2+ ε3n3 + ε4n4 + ε5n5 + ε6n6 + ε7n7 = 7ε
6.
The number of atoms having the same energy and belonging to a
specific box is denoted by n. The index of n specifies the box, i.e. n1
belongs to box . ε1
The equation above shows that only 7 atoms may be found in the 8
boxes.
The total energy of the system is, however, restricted. It should
always be equal 7ε.
For example: let’s say we have 5 molecules with no energy, one with
3ε and one with 4ε. In details, see Table 1.
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Table 1.
Number of boxes Number of atoms in a
box
n0
5
n1
0
n2
0
n3
1
n4
1
n5
0
n6
0
n7
0
Σ
7
Number of energy
units
0ε*5=0
3ε*1=3
4ε*1=4
7
Sum: In the eight boxes we distributed 7 molecules with system’s
energy 7ε.
The macroscopic state of the system is defined by a distribution on the
microstates that are accessible to a system in the course of its thermal
fluctuations.
The number of microstates, W is given by the following equation:
W
N!
n0 ! n1! n2 !...
7.
and ni is the number of atoms found in a certain box. Do not forget 0!
is one.
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Table 2. The possible arrangements of 7 atoms and 8 allowed energy
levels:
Case n0 n1 n2 n3 n4 n5 n6 n7
W
1
6
1
7
2
5
1
1
42
3
5
1
1
42
4
5
1
1
42
5
4
2
1
105
6
4
1
1
1
210
7
4
1
2
105
8
4
2
1
105
9
3
3
1
140
10
3
2
1
1
420
11
3
1
3
140
12
2
4
1
105
13
2
3
2
210
14
1
5
1
42
15
7
1
Σ
1716
For case one:
n0 = 6
n1 = 0
n2 = 0
n3 = 0
n4 = 0
n5 = 0
n6 = 0
n7 = 1
N=7
W
7!
5040

7
6!0!0!0!0!0!0!1! 720
and for case 10, similarly, omitting zero factorials
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W
7!
5040

 420
1!1!2!3!
12
Boltzmann suggested that if one can observe such an assembly over a
long period of time, each microstate will occur with equal probability
and one will find that the number of occurrences for any particular set
of distribution is proportional to the number of microstates that
correspond to that set.
For example, in the above system, there are a total of 1716 ways or
microstates. (The sum of column W). The probability of case 1
occurring, for example, is 7 in 1716. The probability of case 2
happening is 42 in 1716. The probability that case 10 is 420 in 1716
(the highest of all).
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Additional reading
Macrostates and microstates
The state of a system is characterized by the values of all the variables
needed to compute the properties of the system at a later time. If we
use Newtonian mechanics, the state of the system is characterized by
the positions and velocities of all the particles in the system. In
thermodynamics, we use the expression microstate to refer to such a
state. The prefix micro stresses the fact that we know the positions and
velocities of all the microscopic components of the system. By
contrast, we use the expression macrostate when we only specify the
values of the macroscopic parameters of the system. Of course, this
does not characterize the system completely. There may be many
possible microstates that are consistent with a given macrostate. The
number of microstates in a macrostate is called the multiplicity of the
microstate. The following analogy will clarify these concepts.
Suppose we have a set of four colored balls: red (R), blue (B), yellow
(Y) and green (G). Suppose also that we have two boxes: box 1 and
box 2, where we place the balls. A microstate of the system is defined
by specifying exactly which ball goes to which box. For a macrostate,
we only provide partial information. For example, a macrostate could
be specified by determining how many balls are in each box, without
regard of its color. Let us construct a table with the possible
combinations.
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Notice that the multiplicity is highest (6) for the macrostate that
specifies an equal amount of balls in the two boxes. This trend is very
strongly reinforced if the number of particles in the system is high.
Suppose for example we have 20 equal balls. There is only one
microstate with 20 balls in box 1 and no balls in box 2. Hence the
macrostate (20,0) has amultiplicity of 1. On the other hand, the
multiplicity of the macrostate (10,10) is the number of possible ways
you can put 10 balls in box 1 and 10 balls in box 2. This number is
given by
20!
 184.756
10!10!
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