80 Diatomicgases
However, the truthis moreinterestingthan that, andthe resultisdifferentfor H 2 and
forD 2 .Let us concentrate upon the common gas H 2. The nuclei are just protons having
spin^12 .The nucleiare thusidenticalfermions, andthey must thereforebedescribed
byanantisymmetricwavefunction. In other wordsthe wavefunction must change
sign if the co-ordinates of the two protons are interchanged. The following discussion
bears similarity to our earlier treatment of bosons and fermions in section 5.2.
The totalwavefunction canbedecomposedinto theproduct ofa‘spacepart’ anda
‘spin part’. The space part describes the rotation, and it is found that thel= 0 , 2 ,4...
stateshave eveninterchange paritywhereas thel=1, 3...states are odd.Inthe
notation of Chapter 5 , the evenlstates areS(even, symmetric) whereas the oddl
states areA(odd, antisymmetric).
Considerfirst theevenlrotationalstates.Since these areS,theonlywayto obtain
atotal wavefunction which isAis for the spinpart to beA.And this means that the
two spins must be opposite. There is just one such spin state corresponding to the
totalspinS=0. Hence theevenlrotationalstates mustbe associatedwiththespin
singlet configuration only, and hence theyhave spin weighting1. These states are
often called ‘para-hydrogen’.
Ontheotherhandoddlstates areA,the totalwavefunctionfor H 2 istobeA,and
therefore the associated spinpart must now beS. This indicates theparallel spin states
(S=1),inwhichthere are three possiblealignments. Hence theoddlrotationalstates
arelinkedwiththespintriplet states, andtheyhaveaweightingof3. The states are
called ‘ortho-hydrogen’.
Theimplicationisclear. Assuming that ourH 2 isinthermalequilibrium, the
partitionfunction of(7.8) (althoughcorrectfor HD)isin error. Rather than summing
equally over the ortho- and para-states, we should have
ZZZrot=∑
even( 2 l+ 1 )exp[−l(l+ 1 )/T]+ 3
∑
odd( 2 l+ 1 )exp[−l(l+ 1 )/T] (7.11)where the even sumgoes overl =0, 2, 4 ...andtheoddsum overl = 1 ,3...
When (7.11) is used to computeCV, the result is curve B of Fig. 7.2, in evengreater
disagreement with experiment. However, the final twist to the story is uncovered
when the experimentisdone either veryslowly,orbetter still inthe presence of
acatalyst such as activated charcoal. Then it is indeed found that curve B fits the
experiment.
The reasonfor the poorfittothe experiment without the catalystisthat ‘ortho–para
conversion’ is very slow. Changinglby± 1 on collision requires a collisionprocess
whichchanges the totalspin, andthisimplies the presence ofathirdbody(i.e. a
surface) to take awaythespin. Thereisnosuchproblemfor collisionsinwhichl
changes by± 2. Now the way the experiments are done in practice is to prepare and
store theH 2 gas at room temperature,i.e. whenT.In thislimit, theoddandeven
sumsin equation (7.11) are equal(compare thediscussionleadingto equation (7.9)).