Various posts have talked about the statistics of
isotopes in molecules.
In an earlier msg I showed how to set up the
conventional calculation, but warned that it was
not the whole story. Here is the next level of
detail (but still not the whole story).
Suppose we start with known probabilities for
the individual isotopes of a given atom. For
simplicity, suppose we conjure up a mixture
that is 1/3rd H, 1/3rd D, and 1/3rd T on an
atom-by-atom basis. Then the question is, what
are the probabilities for the various possible
diatomic molecules?
We can set up the problem in a way that is nice,
robust and convenient ... by tabulating the
_joint probabilities_.
In particular, we will not be assuming that the
various probabilities are independent. That is,
we will not be assuming that
P(a,b) = P(a) * P(b) [deprecated]
Let's consider various scenarios. We start with
a scenario where there is self-avoidance:
H D T
H 0 .16 .16
D .16 0 .16 [self-avoidance]
T .16 .16 0
Note that the tableau is necessarily symmetric, since
at the end of the day we can't tell HD from DH.
Note that in all these tableaux, I have truncated (not
rounded) repeating decimals, so that .16 is short for
.16666....
As an important check, we can calculate the _marginal
probabilities_ by summing each row and/or each column.
This gives us the single-particle probabilities, and
they come out to be 1/3rd in every case, as they should.
Here is another scenario: self-attraction:
H D T
H .33 0 0
D 0 .33 0 [self-attraction]
T 0 0 .33
Again it is good to check that the marginals come out
right.
Now the time is ripe to show the conventional scenario,
where the probabilities turn out to be independent:
H D T
H .11 .11 .11
D .11 .11 .11 [independence]
T .11 .11 .11
Again the marginals are as they should be.
Of course you can create additional scenarios by
taking linear combinations of the above, and even
that does not span the whole space of possibilities.
My point is that you cannot assume that the independence
scenario is what happens in real life. As long as the
marginal probabilities are correct, then all the
conservation laws are upheld (conservation of mass,
conservation of each isotopic species, etc.) and you
can't decide which scenario is right unless you know
the details of the processes by which the molecules
are formed.
The different scenarios have different entropies, so
there can be significant thermodynamic consequences.
The scenario that is favored at one temperature might
be disfavored at another, and there will be "latent heat"
and fancy osmotic-pressure effects as one set of
probabilities converts to another.
===
If you are doing mass spectrometry, you might classify
the molecules by their total mass, rather than by their
exact makeup. Such classes are formed by summing the
tableau along northeast-southwest diagonals. For example,
the probability of a mass=4 signal is .33 in each of the
three scenarios considered above ... while the probability
of a mass=3 signal is:
0.33 in the self-avoidance scenario,
zero in the self-attraction scenario, and
0.22 in the independence scenario.
These diagonals correspond to the "staggered rows" setup
I used in my previous msg.