# A half-life question, involving 2 elements

1. May 2, 2006

### raul_l

Element A has a half-life of $$T_{1}$$ and decays into element B, which has a half-life of $$T_{2}$$, while $$T_{2}<T_{1}$$. When is the mass of element B the greatest? (i.e. after how many doublings?)

I have been trying to derive a formula to calculate this, but have so far been unsuccessful.

2. May 2, 2006

### Tom Mattson

Staff Emeritus
Can we see one of your attempts?

3. May 3, 2006

### raul_l

Ok.
Lets say elements A and B have masses $$m_{1}$$ and $$m_{2}$$. The equation to calculate the mass of a decaying element is $$m=m_{0}\times2^{-\frac{t}{T/2}}$$, right?
For element A it should be $$m=m_{1}\times2^{-\frac{t}{T_{1}}}$$ and for element B $$m=m_{2}\times2^{-\frac{t-T_{1}}{T_{2}}}$$. I think it should be $$t-T_{1}$$, because it starts decaying after the first doubling of element A. And since element B depends on the mass of element A, its mass over time should be calculated using this equation:
$$m=(m_{1}\times2^{-\frac{t}{T_{1}}})\times2^{-\frac{t-T_{1}}{T_{2}}}$$
Am I making any sense at all? It feels like I'm not taking everything into account in the previous equation. Of course, after I get the equation right, it should be easy to calculate the greatest mass B can have (by differentiating the equation and so on).
Also, it was mentioned in the task, that $$T_{1}>T_{2}$$. I'm still not sure, what difference does it make :).

Last edited: May 3, 2006
4. May 3, 2006

### Curious3141

I would not consider this an elementary problem. It involves solving two first order differential equations simultaneously and it's fairly involved. Since the orig. poster has shown some work, I think I'll pitch in with how I think the problem should be approached; however, this approach seems much too complicated for the estimated level. Maybe someone will find an easier approach.

The problem is easiest when viewed in terms of flux between 3 compartments : A, B and C (C is the decay product of B).

A can only decrease over time as decay occurs. B has both influx (due to decay of A) and efflux (due to decay of B to C) occuring simultaneously.

Let $$N_0, N_A, N_B, T_A, T_B, \lambda_A, \lambda_B$$ represent respectively initial total number of A-atoms, number of A-atoms present at time t, number of B-atoms at time t, half-life of A, half-life of B, the decay constant of A and decay constant of B. The decay constants are related to the respective half-life by $$\lambda = \frac{ln 2}{T}$$.

Then the process can be modelled by the simult. d.e.s :

$$\frac{dN_A}{dt} = -\lambda_AN_A$$ ---(1)
$$\frac{dN_B}{dt} = -\frac{dN_A}{dt} - \lambda_BN_B = \lambda_AN_A - \lambda_BN_B$$---(2)

Solving those is not difficult, but isn't elementary either. (1) is easily solvable by separation of variables, then the result is substituted into (2). This requires an integrating factor to solve.

The final expression I get for $$N_B$$ as a function of t is :

$$N_B = (N_0)(\frac{\lambda_A}{\lambda_B - \lambda_A})(e^{-\lambda_At} - e^{-\lambda_Bt})$$

or equivalently,

$$N_B = (N_0)(\frac{T_B}{T_A - T_B})({(\frac{1}{2})}^{\frac{t}{T_A}} - {(\frac{1}{2})}^{\frac{t}{T_B}})$$

in terms of the half-lives.

The instant $$\tau$$ at which $$N_B$$ is maximised is easily computed by setting $$\frac{dN_B}{dt} = 0$$

giving $$\tau = \frac{1}{\lambda_B - \lambda_A}ln(\frac{\lambda_B}{\lambda_A}) = \frac{ln(\frac{T_A}{T_B})(T_A)(T_B)}{(ln 2)(T_A - T_B)}$$

Last edited: May 3, 2006
5. May 3, 2006

### Curious3141

I'm bumping this up, because I would really appreciate someone commenting on my method and checking the math if possible. Actually, sequential decay is a problem I've wanted to solve for a long time, but it always slipped my mind or I never had the motivation. So this was a good excuse.

After we sort it out on the forum, I'd like the OP to take the solution back to the teacher and get his opinion on whether that's the simplest way to do it, and whether that's what's expected.

6. May 6, 2006

### Curious3141

Raul, have you got an answer to this from your instructor ? I'm interested to know, too.

7. May 6, 2006

### raul_l

I can get you the answer on monday. I'm interested myself as well :)

8. May 6, 2006

### Curious3141

Thank you. If it's not too much trouble, show your instructor my work and see if he agrees. Find out if there's an easier way.

9. May 6, 2006

### Staff: Mentor

Solution is correct, and the method is the most direct, i.e. simplest method.

10. May 6, 2006

### Curious3141

Thank you!!!

11. May 8, 2006

### raul_l

I consulted my teacher and your method is indeed the simplest one. It's weird that I was given a problem with that difficulty, because it's way beyond my level (it's taken from some kind of international olympiad).

Also, T1 is not greater than T2 as I originally posted. Sorry about that. Otherwise the problem is not solvable.

12. May 8, 2006

### Curious3141

That doesn't surprise me, since this is roughly the level of problem I had to solve when training for my Physics Olympiad, back in the day. (In fact, those problems were usually a lot harder).

I don't see that this should matter. The equation works whichever half-life is greater. Can you explain why this shouldn't be so ?

13. May 8, 2006

### raul_l

Come to think of it, I'm not sure anymore. It seemed logical to me at first, because if the mass of element B is initially 0 and if it decays faster than the increase of its mass by the decaying of A, then the mass of B should always remain close to 0. But I guess you're right, there should still be a point at which the mass is maximised.
However, if I set T[a]=T I get t=0/0. Where's the logic in that?

14. May 8, 2006

### Curious3141

Decay is fundamentally a random process. From a micro perspective, there is no statistical reason why B cannot accumulate in mass over a few "ticks" even if its half-life is shorter than A's.

Ah, try doing the limit, setting $$T_A = kT_B$$ and letting k tend to unity (where the half-life becomes the same, let's call that T). It's a good exercise (you will find it easier with L'Hopital's Rule). You will find that $$\tau = \frac{T}{\ln 2}$$ and that $$max(N_B) = \frac{N_0}{e}$$. For interest's sake, in this case, the time when the mass of B is maximised is actually the mean life of a single particle of B (which is the same as the mean life of a particle of A, since the half-lives are the same). Of course, this is the reciprocal of the common decay constant. The value of the maximal mass has a nice relationship to e, the base of the natural log.

Fascinating stuff, isn't it ?

Last edited: May 8, 2006
15. May 8, 2006

### raul_l

Thing were so much more simple before this sequential decay problem... :)
Anyway, I was thinking the same thing (that I should probably take the limit to get rid of the 0/0 problem). However, the equations are above my level and I still can't fully understand all of them, but I'm woking on it. :)

16. May 8, 2006

### Curious3141

Give it a little time (and effort). They're not that complicated, really.