HandsOn 25 - Location of the Precipitation Point

The following derivation uses a full complement of mathematics. Its purpose is to clarify the usual derivations of the location of the precipitation point in a diffusion tube (found in many chemistry books, for example) are inexact in their procedure, though correct in the final outcome.

Normal distribution. The normal or Gaussian distribution describes the results of a random walk. This is the distribution observed in Chapter 2. The Gaussian distribution describes the probability P(x) that a walker who starts at x = 0 will be at position x after taking N steps randomly back and forth along a line, each step of average length Lstep

 P(x) = e-x2/(2Lstep2N), (6.12)

In our model of the diffusion chamber, molecules diffuse from the two ends. We can choose x = 0 for the origin of one molecule, and x = for the origin of the other molecule, where is the length of the diffusion chamber. Then the probability P1(x) of finding a molecule of the first type at position x is

 P1(x) = e-x2/(2Lstep2N1), (6.13)

and, similarly, probability P2(x) of finding a molecule of the second type at position x is

 P2(x) = e- ( - x)2/(2Lstep2N2). (6.14)

Here N1 and N2 are the number of steps taken by the different kinds of molecules in the same time, and we assume that the average length Lstep of one step is the same for both kinds of molecules.

There is precipitation of solid (dust) when two molecules of different kinds meet one another and react. The first precipitation occurs when the first molecules to meet each other, namely in the extreme tails of the two distributions. Hence, the probability of meeting the tails'' of the two distributions would be the product P1(x)  x  P2(x), which would be also the probability Ppre(x) of observing precipitation at position x

 Ppre(x) = P1(x) x P2(x) = e-x2/(2Lstep2N1)e- ( - x)2/(2Lstep2N2), (6.15)

where the product of the two square roots from (6.13) and (6.14) is denoted by c (a constant that does not depend on x). Next, applying the standard property of the exponential fraction

eAeB = eA + B,

we can write Eq. (6.15) in the form

 Ppre(x) = e- [x2/(2Lstep2N1) + ( - x)2/(2Lstep2N2)], (6.16)

We want to find the position x at which the precipitation is the most likely, that is, the peak of the probability given by Eq. (6.16). It occurs for that x for which the expression

- [x2/(2Lstep2N1) + ( - x)2/(2Lstep2N2)]

is least negative. If you are familiar with calculus, find this by taking the derivative of this precipitation probability Ppre(x) with respect to x and setting the result equal to zero. The primary outcome is the derivative of the exponent. When the derivative of the exponent is set equal to zero, it yields:

 = . (6.17)

This says that the ratio of distances of the peak from the two ends has the value:

 = . (6.18)

Now we make the argument that for a given average step length, the number of steps taken is proportional to the average velocity, which is inversely proportional to the square root of the mass [see Eq. (6.2)]. Thus the result is:

 = . (6.19)

Hence the initial precipitation occurs at a location whose ratio of distances from the two ends is inversely proportional to the square root of the ratio of the masses.

Here we note that the use of distributions (6.13) and (6.14) is correct only for the first precipitation. When the precipitation has depleted the molecules in the tails of the distributions, then the distributions are no longer Gaussian. In fact, it turns out that when a steady state has been reached (when precipitation removes molecules at the same rate as new molecules leave the two ends of the diffusion tube), the distributions become straight lines descending from each end to zero at the point x given by Eq. 6.19.

Previous: HandsOn 24 - Diffusion Chamber

Next: 6.2 - Periodic Precipitation: Liesegang Rings