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1、Diffusion theory does a pretty good job of describing the attachment of small ions to uncharged particles. The first step will be to solve the diffusion equationThe geometry and the boundary conditions are shown belowWe'll consider just a single uncharged particle surrounded by small ions.
2、 The particle has a radius of a. The small ion concentration zero at the surface of the particle where the small ions are collected. Far from the particle the small ion concentration is nWe assume a steady state small ion concentrationWe'll use spherical geometry for this probl
3、em with the coordinate axes centered on the particle. Because of the spherical symmetry there is no or dependence, we need to consider only the r dependence. The Laplacian becomesThis is an easy equation to solveNow we make use of the two boundary conditions. First, far from the pa
4、rticle (r approaches infinity) the small ion concentration should be n.Then at the surface of the particle the small ion concentration is zeroSo the solution isNow that we know n(r) we can determine the flux (particles passing through unit area per unit of time) of small ions toward the particle.
5、60; This depends on the gradient of n(r)Next we find the total number of particles moving inward toward the particle by multiplying the flux by the area of a sphere with radius r surrounding the particleWe assume these small ions are all collected by the particle. We've only considered a s
6、ingle particle at this point, so to determine the total rate of loss of small ions we need to multiply by the concentration No of uncharged particles.Now let's go back and write down the small ion balance equation againBy comparing the small ion loss expression we've just derived with the co
7、rresponding term in the small ion balance equation we can see that o = 4 a DAn expression for the diffusion coefficient, D, can be derived using principles from kinetic theory and isThe top expression above is also know as the Einstein (or Einstein-Smoluchowski) Relation.Here's an ex
8、ample calculation where we assume a particle radius of 0.1 m (10-7 m) and an electrical mobility of 1 x 10-4 (m/s)/(V/m).Now we will look at the collection of small ions by a charged particle. Diffusion will still be involved but we will need to add an additional term that accounts for the mov
9、ement of small ions caused by the influence of the electrical field surrounding the particle.First the E field surrounding a charged particle (the charge on the particle can only be multiples of an electronic charge, we assume just a single electronic charge in the figure below)The drift velocity du
10、e to this E field isNow we will multiply n, the concentration of small ions, by the area of a sphere of radius r surrounding the charged particle and by vdrift t, the distance the small ions will move in time t. This will be the number of small ions that pass through the sphere in time t.Now w
11、e'll multiply by N, the concentration of charged particles and divide by t to get the rate of attachment of small ions to charged particles due to the influence of the particle's electric field.To find the total rate of collection by the charged particles we need to add in the diffusive loss
12、 term. So we haveTogether the two terms in this expression correspond to the 1nN term in the small ion balance equation. From that we get the following expression for 1 One last item before we leave this supplementary lecture - a table comparing values of 0 and 1 particle radius (m)o (cm3/sec)1 (cm3/sec) 0.013.25 x 10-72.12 x 10-60.051.62 x 10-6 3.42 x 10-6 0.13.25 x
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