# Radioactive decay: Solve for delta t

1. Dec 10, 2015

### Lagraaaange

1. The problem statement, all variables and given/known data
Two isotopes N1, N2. You are given the value of the proportions N1(t1)/N2(t1) and N1(t2)/N2(t2). From this calculate Δt.
2. Relevant equations
dN/dt = -λN

3. The attempt at a solution

Solve dN1/N = -λdt and dN2/N = -λdt
obtain: N1(t2) = N1(t1)exp(-λ1(t2-t1)) (1)
and N2(t2) = N2(t1)exp(-λ2(t2-t1)). (2)

Divide (1) by (2) and take ln of both sides
=> ln[(N1(t2)/N2(t2))/(N1(t1)/N2(t1))] = (-λ1+λ2)(Δt)
Okay so if I'm given the proportions (say X) I can get a numerical result for left hand side
ln(X)/(-λ1+λ2) = Δt

But if I'm not given the λ's how can I obtain them? I feel like I'm missing some relationships between λ and the proportionality constants.

2. Dec 10, 2015

### haruspex

I'm not sure whether there are supposed to be two lambdas or one. Maybe the only process is the conversion of the one isotope to the other?
But either way, you're right: you do have to know some rate. This is very easily proved by dimensional analysis. If the only info you have consists of two dimensionless numbers, no valid physical equation is going to produce an answer that has a physical dimension, such as time.

3. Dec 10, 2015

### Lagraaaange

1. Is this "right" in terms of variables; that is, are the mathematics and physics ok?
2. In the case of one isotope being converted to the other I would have one λ? I would still need the constant to be provide, right? I mean I'd need a half life or something if I don't have λ given.

4. Dec 10, 2015

### haruspex

Yes, what you wrote is correct for two separate isotopic decays. And yes, even if it is conversion from one to the other you still need to be given the rate.
Can you see how to change your result for λ1 and λ2 very simply to get the conversion case result?

5. Dec 10, 2015

### Lagraaaange

I don't see it. If I change it to a conversion of one to the other wouldn't I have a differential equation of the form: d(N2-N1) = -λdt

6. Dec 10, 2015

### haruspex

No, it would not be of that form. The equation for the decaying isotope does not change. The quantity of the decay product follows easily from that.
But the "very simple" modification I was thinking of doesn't work either.

7. Dec 15, 2015

### Lagraaaange

If the rate is for one isotope to another, wouldn't the lambdas just cancel out? I mean would giving 1 lambda even make sense given two rates at two times?

8. Dec 15, 2015

### haruspex

It would cancel in terms of the total number of atoms, but not in terms of the ratio. One will gain what the other loses.

9. Dec 15, 2015

### Lagraaaange

Consider the last line in my outline. It would be -λ+λ = 0. This would result in a nonsensical answer. Also this tells me that λ2 must be greater than λ1 unless the ratios result in less than 1 in which case the log would be negative and I'd be ok. I have an exam on a question of this type so I'm trying to make sense of this.

10. Dec 15, 2015

### haruspex

One isotope decaying into the other is different from both decaying at the same rate. It is not -λ+λ.

Last edited: Dec 15, 2015
11. Dec 15, 2015

### Lagraaaange

How does one obtain the conversion result, say where one isotope is converted to another?

12. Dec 15, 2015

### haruspex

You know the decay from the one. What that loses, in terms of number of atoms, the other gains.

13. Dec 16, 2015

### Lagraaaange

Suppose at time t a sample N1 goes to N2 with rate λN1.
Then N1(t) = Kexp(-λt). Let us choose t=0 to be when no particles decayed. Let N0 be total amount of particles, (N1+N2), thus N1(0) = N0
=> N1(t) = N0exp(-λt).
Note N0-N1(t) = N2(t) = N0(1-exp(-λt))
So N1(t)/N2(t) = N0exp(-λt)/N0(1-exp(-λt)) = 1/exp(λt)-1
if N1/N2 = R1, 1/exp(λt1)-1 = R1
=> exp(λt1) = 1+1/R1
Repeating we get, exp(λt2) = 1+1/R2
Dividing: exp(λΔt) = R1(R2+1)/R2(R1+1) and simplifying,

Δt=ln[(R1(R2+1))/(R2(R1+1))] / λ

What do you think?

14. Dec 16, 2015

### haruspex

Looks right.

15. Dec 16, 2015

### Lagraaaange

Thanks. Let's hope my professor agrees.

16. Dec 16, 2015

### haruspex

Bear in mind that we do not know for sure which scenario is intended: one isotope decaying into another, according to one decay rate, or two isotopes decaying into two other isotopes, at two distinct decay rates. Since no decay rates were provided, we can't tell.

17. Dec 16, 2015

### Lagraaaange

But I've got both case down, right? The former being for two isotopes into two and the latter for one to another?

18. Dec 16, 2015

Yes.