1. Kinetics of Nucleation
(b)
Bimolecular reactions describing flux of atoms into and out of i sized-clusters
For eqn. (a)
forward Rx
backward Rx
Writing a similar set of eqns. For (b) and combining
(C)

2. Provides us with an eqn. describing how the number of clusters containing i atoms changes with time.
For each cluster size i there is a finite difference eqn. of the form of (C).
The equations are coupled and must be solved simultaneously in order to obtain ni as a function of time i.e., ni t
t – transient time to steady state

3. Inspecting of eqn (C) reveals that it has the finite difference form of the diffusion equation. The coupled finite – difference eqns. can be recast into a continuum form that looks just like the diffusion eqn! The continuity eqn for the time dependent concentration nit of i clusters is just :
(d)
where the cluster current is
where
is the equilibrium population of i –sized clusters.

4. What we want is an exact solution to (d) subject to the initial conditions.
all other ni = 0
and the boundary conditions:
as t  
No exact solutions to this formidable problem exist and literature treatments have made various approximations in order to obtain a solution to this problem.

5. Volmer model 1926 (equilibrium)
An equilibrium distribution of clusters is assumed to exist :
Gi i*
G* i i* i
Each of the bimolecular reactions is assumed to be in equilibrium. i.e.,
Forward rate = backward rate
Nucleation event
rest of distribution not used – unphysical!

6. Nucleation Rate :( # of nuclei per unit volume and time)
where the rate constant ki has been decomposed into a collision frequency ( # of atoms arriving per unit time and area) and the number of attachment sites or surface sites Oi*, i.e., ki = Oii, e.g.,
For nucleation from the gas phase

7. Becker – Doring model 1935 (steady state)
VW ( ) (equilibrium)
BD (steady state – distribution is found by clusters acquiring or losing individual atoms q i i*
Once again, consider
(e)
The nucleation rate ( formation of  i +1) is
and corresponds to the net forward reaction rate for a reaction such as (e).

8. At steady state Ii  const i.e.,
(f)
(* Note Volmer’s assumption was that Ii  0)
The cluster population is constant
Clusters of size (i-1) are also involved in the reaction.
for which
and
and

9. We want to evaluate I -
Since the rate constants are independent of concentration we can solve for one rate constant in terms of the equil. i.e.,
concentration of ( i – 1 ) clusters
# of surf atoms in ( i – 1) clusters
rate const for transition
in units of # of jumps per atom per second

10. @ steady state net rate = 0 :
and
Writing this in differential form :
@ steady state

11. (I)
where q is an arbitrarily large value of i such that nq = 0.
the RHS of ( I )
Then

12. Now exp (Gi / kT) has a sharp maximum around Gi = G
Now exp (Gi / kT) has a sharp maximum around Gi = G* as the value of the  will be dominated by the exponential. So take
i.e., only the i in the vicinity of i* will contribute.
An expression for Gi can be found by expanding in a Taylor series about G* :
where
Note the linear term is absent since
 defines G*

13. Change of variables let
If i is not too large, the limits of integration can be taken -  +

14. for i far away from i* , or
Then solving for I
Note
Zeldovich factor

15. What should be used for
Depends on  phase
Vapor :
area of i* cluster
Liquid :
jump rate per atom, across the interface
- Pure liquid
no atomic neighbor must change
 Collision limited
inter-atomic distance
- alloy : charge of neighbor necessary
 diffusion limited

16. Temperature dependence of I
where ga is an interface activation energy indep. of T and recall for spherical clusters as
reduction of driving force
as T  0 atomic motilities are reduced and I  0

17. Temperature dependence of I
Tmax Te I
Write I in the form
For solidification of metals :
For I ~ 1 nucleus cm-3 sec-1
For I ~ 106 nuclei cm-3 sec-1

18. Since
in order to increase I from 1 to 106
we need to increase T by a factor of (76 / 62)1/2  1.1  10 % increase.
so I increases very quickly with T !
Number of critical nuclei:
Subs. For G*;
is a T = 1/3Tm

19. For Al
@ T = 1/3Tm = 311o K
 very small
For a 50oK undercool
Can actually only undercool by ~ 5oK before solidification occurs  catalytic sites heterogeneous nucleation.

20. Comparison with experiment
Review : D. Turnbull, Solid State Physics, 3, 226 – 301 (1956). Hg
3 -9 nm
homogeneous
heterogeneous
Disperse in oil
with surfactant
oil
Hg droplets suspended in oil that didn’t catalyze nucleation.

21. Turnbull was able to undercool by 79oK .
Theory says that
Experiment values for
The disagreement of ~ 107 is outside experimental error !
Turnbull was able to account for the discrepancy by considering the small temperature dependence of .
However, its not clear that using the bulk value of  is appropriate.
because of the strong dependence of I on T
solid state nucleation theory is semi-quantitative
condensation is in much better shape
( J. L. Katz etal. J. Chem. Phys. 62, 448 (1972)