To make things easier, I already know about exponential decay, half lives, etc. What I am unsure of is doing calculations when the parent nuclide decays into a daughter nuclide that is also radioactive.
For example:
Nuclide A has a half-life of 300 days
Nuclide B 100 days
Nuclide C stable
Let’s suppose you are starting out with 100 grams of element A. How much of A, B and C will be present after 300 days?
After 300 days, element A has undergone 1 half life and so there’s 50 grams remaining.
Element A created 50 grams of element B which has undergone 3 half lives in 300 days. Does this mean what remains of element B is an eighth of the 50 grams or 6.25 grams?
Is the calculation that simple ?
Nope.
Ignore A for the moment, and try starting out with 50 grams of B.
At day 300, you’ll have gone through 3 half lives and have 6.25 grams of B left. That’s the same answer as you proposed for your A->B->C reaction, but in that case you started out with less than 50 grams of B. You started out with no B at all!
You’re looking at two sequential reactions here, so the abundance curves will look something like this.
Certainly not that simple. That 50 grams of B was produced over a period of time, so most of it hasn’t been around for a full three B-half-lifes, and you’ll still have more B than that left.
The proper way to approach this problem is with a system of coupled differential equations. Probably, though, someone else has already solved those differential equations and come up with a simple formula, since this is a fairly common situation. Unfortunately, I don’t know what it is off the top of my head.
Though I was hoping otherwise, I had a feeling that the calculations had to be a little more complex than my example.
Before asking a question in GQ, you can rest assured I have done a somewhat thorough job of web surfing. So, Chronos, any idea where I might locate those coupled differential equations?
Suppose A decays at the rate a and B decays at the rate b.
Then at time t the amount left is A(t) = A_0*exp(-at).
At any time t, A is producing B at the rate aA(t), but B is decaying at the rate bB(t) so the differential equation for the amount of B is:
dB/dt = aA(t) - bB(t) = aA_0*exp(-at) - bB.
A particular solution to this equation is
B(t) = aA_0/(b-a)*exp(-at)
The homogeneous solution to the equation is B(t) = K*exp(-bt). So the entire solution is
B(t) = aA_0/(b-a)exp(-at) + Kexp(-bt).
We must set K so that B(0) = 0. That is, K = -aA_0/(b-a). Thus the solution is:
B(t) = aA_0/(b-a)*[exp(-at) - exp(-bt)].
We could get the amount of C in a similar fashion, but it’s easier to note that
The formula for the amount of the second element in a radioactive decay chain is:
λ[sub]a[/sub] * (Beginning amount of A) * (e[sup]-λ[sub]a[/sub]*t[/sup] - e[sup]-λ[sub]b[/sub]*t[/sup])
(λ[sub]b[/sub] - λ[sub]a[/sub])
Where λ[sub]a[/sub] = the decay rate of A or (ln(2) / half-life) = (0.6931471805599 / 300 days) =
0.00231049060187
The decay rate of B (or λ[sub]b[/sub]) = 0.00693147180560
I tried writing the formula in Excel and got the following data:
Starting with 128 grams of A and 0 grams of B
After 0 days we have:
128 A and 0 grams of B (pretty good so far)
100 days 101.59 A 18.797 B
200 days 80.635 A 24.317 B
300 days 64 A 24 B
400 days 50.80 A 21.398 B
The amount of B seems to peak at about 250 days which fits in with the type of “rate curve” in the chart that squink had referenced.
Chronos
Yes, I’ve tried looking there but those calculators are only applicable for a “mere” single decay reaction. (Still, a fine site nonetheless).
If you are wondering if this information will end up as a calculator on my site - no. Why? If for no other reason, I’ve found that the complexity of a calculator is inversely proportional to its popularity. A calculator for determining data for circles, spheres and cylinders? Wildly popular !!! An advanced calculator for Kepler’s 3rd Law? [crickets chirping]
Anyway, this all came about because I was thinking about adding more information to the half-life calculator (explaining λ, τ, etc) but like most “Dopers” I got a little carried away and wondered about, daughter nuclides, etc. So, if a “Doper” wishes to offer a solution beyond a 2 decay chain that would be great but I appreciate the help already given.
