A (big) crystal cools down from its melting point T_{m} to room temperature T_{r} ( about 0^{o} C) with T = T_{m} · exp – (l · t). The point defects present have a diffusion coefficient given by D = D_{0} · exp – (E_{m}/kT).  
How large is the average distance L that they cover during cooling down from some temperature T to T_{r}?  
This is not an easy question. What you should do is:  
Use the Einstein relation for the diffusion length (and forget about lattice factors), but consider that the diffusion coefficient is a function of time, i.e.  


Proceed by first finding the values of l for initial cooling rates at the melting point of 1 ^{o}C/s, 10 ^{o}C/s, 50 ^{o}C/s and, for fun, 10^{4} ^{o}C/s.  
Using the following substitution will help with the integration  


The integral now runs from u_{0} corresponding to t'_{0} to whatever value of u corresponds to t' = ¥.  
You will obtain the following integral:  


This integral cannot be solved analytically. In order to get a simple and good approximation, you may use the linear Taylor expansion for 1/u around u_{0}.  
Show that for realistic u_{0} values you can replace 1/u by 1/u_{0} in a decent approximation and that you now can do the integral.  
Now use typical values for melting temperatures, migration activation energies E_{m}, and D_{0}; e.g. from the backbone, two tables or diagrams given here. For missing values (e.g. D_{0}), make some reasonable assumptions.  
Plot L as a function of T for activation energies E = 1.0 eV, E = 2.0 eV, and E = 5 eV with the four cooling rates given above as parameter.  
Play around a bit and draw some conclusions, e.g. with respect to


Link to the Solution 

4.2.1 Point Defects in NonEquilibrium
Numbers for Point Defect Diffusion
© H. Föll (Defects  Script)