Deciphering Decay Chains: Understanding the Complexity of Radioactive Decay

[nmsurobert]

I've been reading about decay chains and I have confused myself on how decay works.

Take a sample of 238U for example.. its half-life is 4.5 billion years. At that point half of the sample is now 238U and the other half is 206Pb.
But looking at a decay chain, (and maybe I am reading it wrong) it shows that the change from 238U to 234Th is 4.5 billion years. Are the other half lives in the chain are so short relative to first step that we are quickly left with half 238U and half 206Pb?

Anytime I see figures of this half life, only first first and last isotopes are shown. There is never a discussion about the series. I started reading about this because I wanted a physical answer to how do we know the half life of 232Th without using the decay formula.

This is the chain I've been looking at. http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/radser.html#c3

 

[Vanadium 50]

As your chart shows, the next longest isotope is U-234 at 246,000 years. Long for us, but practically instantaneous with respect to a few billion.

how do we know the half life of 232Th without using the decay formula.

We look it up. It's 14 billion years.

 

[etotheipi]

Also, for series like this, eventually the activities $A_i = \lambda_i N_i$ of all of the unstable daughter isotopes in the chain equalise to that of the mother isotope ($^{238}\text{U}$). Once this equilibrium has been reached we can effectively "ignore" all of the isotopes in the middle, since the $^{206}\text{Pb}$ will be produced at exactly the same rate that the $^{238}\text{U}$ is used up.

 

[nmsurobert]

Also, for series like this, eventually the activities $A_i = \lambda_i N_i$ of all of the unstable daughter isotopes in the chain equalise to that of the mother isotope ($^{238}\text{U}$). Once this equilibrium has been reached we can effectively "ignore" all of the isotopes in the middle, since the $^{206}\text{Pb}$ will be produced at exactly the same rate that the $^{238}\text{U}$ is used up.

I thought that would be the case. I tried to scribble some math and show that, but I was just frustrating myself even more. Thought I'd come here and ask the pros. Thanks.

 

[sophiecentaur]

I tried to scribble some math and show that, but I was just frustrating myself even more.

Me too!
If a chain consists of a set of different reactions with different rates r for the different steps then the decay will not necessarily be a simple exponential. Thats just a bit of Maths but I think it has to apply.
A(t) = A₀ Exp(r₁/t) where r₁ is the rate of one reaction but then there will be A₀ - A(t) of product of that reaction which will be changing and decaying at its own rate. Overall that doesn't look like a simple exponential decay so it won't have a simple half life. But the r for one very long r step would dominate so you would often get near enough a half life.

If I really wanted to know the answer then I would probably reach for Excel. That would give a fair answer for a two step chain. Trouble is that every time I do reach for Excel, I find the basic skills are getting more and more tired and forgotten. Sad thing is that there's probably a bit of long forgotten book work that would allow one to do it analytically.

 

[etotheipi]

For the first two in the chain, e.g. $N_1$ and $N_2$, then we could say something like$$dN_{2} = A_{1}dt - A_{2}dt = \lambda_{1}N_{1}dt - \lambda_2 N_2 dt$$ $$\frac{dN_{2}}{dt} = \lambda_{1}N_{1} - \lambda_2 N_2= \lambda_{1}N_{1}(0)e^{-\lambda_{1}t} - \lambda_2 N_2$$We can multiply through by $e^{\lambda_2 t}$,$$e^{\lambda_2 t}\frac{dN_{2}}{dt} + e^{\lambda_2 t} \lambda_2 N_2 = \lambda_1 N_1(0)e^{(\lambda_2-\lambda_1) t}$$ $$N_2(t) e^{\lambda_2 t} = \lambda_1 N_1(0) \int e^{(\lambda_2 - \lambda_1)t} dt = \lambda_1 N_1(0) \frac{1}{\lambda_2 - \lambda_1} e^{(\lambda_2 - \lambda_1)t} + c$$Now if we know $N_2(t)$ maybe it is possible to obtain the form for $N_3(t)$, and so on... but it looks like it might get a little tricky!

 

[snorkack]

We actually have a quite notable decay chain where the simplifications do NOT hold.
²⁴⁴Pu->²⁴⁰U+α 80 million years
²⁴⁰U->²⁴⁰Np+β+νˇ_e 14 h
²⁴⁰Np->²⁴⁰Pu+β+νˇ_e 62 min
²⁴⁰Pu->²³⁶U+α 6 560 years
²³⁶U->²³²Th+α 23,4 million years
²³²Th->²²⁸Ra+α 14 050 million years

Note that since the half-life of ²³⁶U is shorter than, but a substantial fraction of, the half-life of ²⁴⁴Pu, the activity of ²³⁶U relative to that of ²⁴⁴Pu will stabilize at a level substantially higher than that of unity.
The half-life of ²³²Th is longer than that of ²⁴⁴Pu and therefore the ratio of ²³²Th activity to ²⁴⁴Pu grows without bound.

 

[sophiecentaur]

but it looks like it might get a little tricky!

. . . so I don't need to feel bad about it!