# Spivak - Radioactivity and the Exponential Function

1. Jul 23, 2012

### AlwaysCurious

1. The problem statement, all variables and given/known data
A radioactive substance diminishes at a rate proportional to the amount present (since all atoms have equal probability of disintegrating, ...). If A(t) is the amount at time t, this means that A'(t)= p * A(t) for some p representing the probability that an atom will disintegrate in one unit of time.

Find A(t) in terms of the amount A(0) present at time 0.

I have solved the problem, and have checked the solutions manual and am correct. However, this at least seems to me to give the wrong physical result. I am definitely wrong here, and am curious why I am wrong.

2. Relevant equations

just the one mentioned above, A'(t) = p * A(t).

3. The attempt at a solution

My solution:

Rearranging the equation gives A'(t)/A(t) = log(A(t)) ' = p implies that log(A(t)) = p*t + K for some constant K. Then A(t) = exp(pt + K). To find K, we simply plug in t = 0, and K = log(A(0)), or A(0) = exp(K). Then A(t) = A(0) * exp(pt).

Mathematically, I do not see much wrong with this (except for the case where the initial amount is zero). What I am curious about is the formula itself: if the probability is a positive quantity less than 1 (say, 75%), and A(0) is say, one unit of mass, then the amount of substance seems to increase with time, as the function describing the behavior is exp(0.75t), which gets large as t gets large.

Where am I wrong? I certainly am not a crackpot who has "proved math wrong", so I am assuming that I have made a flaw somewhere (and it probably isn't the formula itself, as this is the one given by the book which I trust fairly well).

2. Jul 23, 2012

### AlwaysCurious

I see the problem - or at least I think I do. The original equation should be A'(t) = -pA(t), because the amount shrinks proportionately to the amount, not grows!

The solution to this new equation is A(0)e^-pt, one that at least I find somewhat beautiful and validates some of the intuition-y characterizations of e as something that involves "continuous self-referential growth".

3. Jul 23, 2012

### HallsofIvy

Staff Emeritus
Or your "p" is a negative number to begin with.

4. Jul 23, 2012

### AlwaysCurious

But isn't "p" supposed to be the probability that the particle decays? How could the probability be negative?

5. Jul 23, 2012

### HallsofIvy

Staff Emeritus
Yes, you are right. I had passed over that part of your original post.