and fifth energy levels. At what energy level can we truncate the summation
for the electronic partition function? Assume standard temperature of 25°C.Solution
We will evaluate each negative exponential and add successive terms to our
previous value for the partition function, and see how much qelectchanges.
Up to the first term, the partition function is easy: it is simply equal to the
degeneracy of the ground electronic state:
qelect,1 1
Up to the second term, we need to evaluate one negative exponential. (The
exponential function exp, introduced in Chapter 17, is the same as eexcept it
allows long or complex exponents to appear in a more legible form.)3 exp 2.77
Therefore, to two terms,qelectis
qelect,2 1 2.77 3.77
Notice that the inclusion of just the first excited electronic state almost
quadrupled the value for qelect. Including the third term:5 exp 4.05
the overall qelectis now 1 2.77 4.05 7.82. Now, for the fourth state:5 exp 2.139
10 ^21
which is 21 orders of magnitude smaller than the previous term. We won’t
even check the last term, because it is obvious that adding this last term to
qelectwill not substantially change its value. Therefore, we can say that qelect
for C is 7.82 and can be calculated using only the first three electronic states.Of course, the temperature of the system will have an effect on which elec-
tronic states will contribute to the electronic partition function. (So will the de-
generacy of the state, but it has a smaller potential effect on qelect.) Generally, if
the ratio E/Thas a value of about 10^22 J/K or larger,the negative exponential
is about 0.0007, which is negligible with respect to 1 (the minimum value of
qelect). Usually, these and larger terms can be safely ignored in the calculation of
qelect. Otherwise, that particular electronic state should be included explicitly.Example 18.2
Nickel atoms have a low-lying excited state at about 200 cm^1 (3.97
10 ^21 J).
Assuming that both electronic states have a degeneracy of 3 and that no
additional low-lying excited states contribute significantly to the electronic
partition function, calculate qelectat 1000 K.Solution
Checking quickly, we find that the ratio E/Tequals (3.97
10 ^23 J)/(298 K)
1.33
10 ^25 J/K, which is smaller than our estimated cutoff value of 10^22 J/K,2.0249^10 ^19 J
(1.381
10 ^23 J/K)(298 K)8.64
10 ^22 J
(1.381
10 ^23 J/K)(298 K)3.26^10 ^22 J
(1.381
10 ^23 J/K)(298 K)620 CHAPTER 18 More Statistical Thermodynamics