# Avrami Equation

1. Jan 20, 2013

### Stat313

1. The problem statement, all variables and given/known data
A two dimensional crystal process is modelled by an Avrami equation phi(t)= 1-exp(-k*t^2), where k>0. Experimental observation suggests that the inflection point of phi occurs at t=2 hours. Find the dimension and numerical value of k.

2. Relevant equations

I know how to take the log-log of this equation in order to linearize it.

3. The attempt at a solution

I tried to have a system of two equations with 2 unknowns, but failed because I don't know the value of phi at any time t. Any help would be greatly appreciated.

2. Jan 20, 2013

### Staff: Mentor

Inflection points are usually found using derivatives of the curve.

3. Jan 20, 2013

### Stat313

I computed the second derivative with respect to t, I got
phi^2(t)= 2*k*exp(-k*t^2)-4*k^2*t^2*exp(-k*t^2)

Since we know that an inflection point occurs at t = 2, I solved for phi^2(2) = 0, which gives me k=0,1/8. so we choose k>0. Is my answer correct?

4. Jan 20, 2013

### Staff: Mentor

Your method is fine. If the magnitude of k is 1/8, what units will k have?

5. Jan 20, 2013

### Stat313

6. Jan 20, 2013

### Staff: Mentor

The equation contains the exponential term
$$e^{-k\;t^2}$$
What units should be associated with k in order to make this term correct (mathematically, so that when evaluated numerically will yield a real number).

7. Jan 20, 2013

### Stat313

Is't k just a constant?

8. Jan 20, 2013

### Staff: Mentor

Sure, but even constants have units if equations using them are to balance.

9. Jan 20, 2013

### Stat313

When I compute phi(2), it gives me 0.39 which is a real number.

10. Jan 20, 2013

### Stat313

Any hints?

11. Jan 20, 2013

### Staff: Mentor

Hint: The argument x of the exponential function $e^x$ must be a pure number without units (all units in the exponent must cancel).

12. Jan 20, 2013

### Stat313

Is the dimension of k is (1/time(t)), in this case we have t in hours?

13. Jan 20, 2013

### Staff: Mentor

Test it. If k = 1/(8*hr), does the exponent yield a unitless number if you plug in t = 2 hr?

14. Jan 20, 2013

### Stat313

No, it is not gonna cancel, but if I have k 1/time in (half hours), it does cancel? By the way, I am not a physics major, so bare with me please.

15. Jan 20, 2013

### haruspex

t has dimension time, so what dimension does t2 have? So what dimension do you need k to have in order to produce a dimensionless kt2?

16. Jan 20, 2013

### Staff: Mentor

In the exponent you have the expression $-k t^2$. That means the time is squared, so if t is in hours, the units of $t^2$ are $hr^2$. To make the units of k cancel this, it should have units of $1/(hr^2)$, or $hr^{-2}$.

17. Jan 20, 2013

### Stat313

t^2 has (time)^2 and k has to have (1/(time)^2) ?

18. Jan 20, 2013

### haruspex

Yes. So the units would be?

19. Jan 20, 2013

### Stat313

The units will cancel, (time)^2/(time)^2=1

Thanks for the help,

20. Jan 20, 2013

### haruspex

No, I meant the units of k.