A half-life question, involving 2 elements

In summary, the conversation discusses the decay of two elements, A and B, with half-lives of T1 and T2 respectively. The question is when is the mass of element B the greatest, which can be calculated by solving two first-order differential equations. After multiple attempts, the participants agree that one of the solutions is the simplest method and can be used to solve the problem, regardless of T1 and T2 being in a specific order.
  • #1
raul_l
105
0
Element A has a half-life of [tex] T_{1} [/tex] and decays into element B, which has a half-life of [tex] T_{2} [/tex], while [tex] T_{2}<T_{1} [/tex]. 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.
 
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  • #2
Can we see one of your attempts?
 
  • #3
Ok.
Lets say elements A and B have masses [tex] m_{1} [/tex] and [tex] m_{2} [/tex]. The equation to calculate the mass of a decaying element is [tex] m=m_{0}\times2^{-\frac{t}{T/2}} [/tex], right?
For element A it should be [tex] m=m_{1}\times2^{-\frac{t}{T_{1}}} [/tex] and for element B [tex] m=m_{2}\times2^{-\frac{t-T_{1}}{T_{2}}} [/tex]. I think it should be [tex] t-T_{1} [/tex], 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:
[tex] m=(m_{1}\times2^{-\frac{t}{T_{1}}})\times2^{-\frac{t-T_{1}}{T_{2}}} [/tex]
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 [tex] T_{1}>T_{2} [/tex]. I'm still not sure, what difference does it make :).
 
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  • #4
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) occurring simultaneously.

Let [tex]N_0, N_A, N_B, T_A, T_B, \lambda_A, \lambda_B[/tex] 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 [tex]\lambda = \frac{ln 2}{T}[/tex].

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

[tex]\frac{dN_A}{dt} = -\lambda_AN_A[/tex] ---(1)
[tex]\frac{dN_B}{dt} = -\frac{dN_A}{dt} - \lambda_BN_B = \lambda_AN_A - \lambda_BN_B[/tex]---(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 [tex]N_B[/tex] as a function of t is :

[tex]N_B = (N_0)(\frac{\lambda_A}{\lambda_B - \lambda_A})(e^{-\lambda_At} - e^{-\lambda_Bt})[/tex]

or equivalently,

[tex]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}})[/tex]

in terms of the half-lives.

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

giving [tex]\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)}[/tex]
 
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  • #5
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
Raul, have you got an answer to this from your instructor ? I'm interested to know, too.
 
  • #7
I can get you the answer on monday. I'm interested myself as well :)
 
  • #8
raul_l said:
I can get you the answer on monday. I'm interested myself as well :)

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
Curious3141 said:
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.
Solution is correct, and the method is the most direct, i.e. simplest method. :smile:
 
  • #10
Astronuc said:
Solution is correct, and the method is the most direct, i.e. simplest method. :smile:

Thank you! :approve: :smile:
 
  • #11
Curious3141 said:
Raul, have you got an answer to this from your instructor ? I'm interested to know, too.

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
raul_l said:
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).

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). :wink:

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

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
Curious3141 said:
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 ?

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
raul_l said:
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.

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.

However, if I set T[a]=T I get t=0/0. Where's the logic in that?


Ah, try doing the limit, setting [tex]T_A = kT_B[/tex] 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 [tex]\tau = \frac{T}{\ln 2}[/tex] and that [tex]max(N_B) = \frac{N_0}{e}[/tex]. 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 ? :wink:
 
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  • #15
Curious3141 said:
Fascinating stuff, isn't it ? :wink:

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
raul_l said:
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. :)

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

1. What is a half-life?

A half-life is the amount of time it takes for half of a given sample of a radioactive substance to decay into a more stable form.

2. How is the half-life of an element determined?

The half-life of an element can be determined through experimentation and observation. Scientists measure the rate at which the element decays and calculate the time it takes for half of the sample to decay.

3. What factors can affect the half-life of an element?

The half-life of an element can be affected by the temperature, pressure, and chemical environment of the sample. It can also vary depending on the isotope of the element.

4. How can the half-life of an element be used in practical applications?

The half-life of an element is commonly used in radiometric dating to determine the age of rocks and fossils. It is also used in medical imaging techniques, such as PET scans, to track the movement of radioactive substances in the body.

5. Can the half-life of an element change?

No, the half-life of an element is a constant value that does not change. However, the amount of a radioactive substance can decrease over time, making it appear as though the half-life has changed.

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