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.

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