# Radioactive Decay: A Problem in Differential Equations

1. Apr 24, 2010

### Bachelier

I've been working with this problem for almost two weeks trying to find a good equation for the decay of Bismuth to no avail.

Can someone give me insightful comments:

Here's the problem:

1. The problem statement, all variables and given/known data

In the radioactive decay series of Uranium (238, 92), isotopes of lead, bismuth occur as products of two successive Beta decays with half -lives of 19.7 minutes and 26.8 minutes, respectively.

Decays are each proportional to the amount of isotope present.
Assume initially at time =0, we have 100 mg of lead and 150 mg of bismuth?

We are asked to find the amount of lead and bismuth at any time?

2. Relevant equations

So first we have to formulate a DE for the decay.

The one equation for Lead is simple.

Let L(t) be the amount of lead at any time, then the DE model is:

dL/dt = -a.L ​
(a: constant of decay)

after integration and get L(t) = 100.e^(-a.t)

Using the Initial Value Problem and half-life value we get an equation:
L(t)= 100.e^(-.035185.t)

Now the Differential Equation for Bismuth is: (B(t): the amount of Bismuth at any time)

dB/dt= -b.B + a.L​
(a.L: quantity of lead decayed added to the Bismuth which equals 100.e^(-.035185.t)) and b: constant of decay for Bismuth

Solving this 1st Order DE we get:

[100.e^(-.035185.t) + C.100.e^(-b.t)]​
B(t)= _____________________________________
[b -.035]​

C is a constant of integration.

First: Is my B(t) equation correct based on the problem we have?

And second, how can I solve for b and C based on the IVP I have?

Can you guys help?

thx :)

2. Apr 24, 2010

### xaos

shouldn't we need to know how much uranium there is to know how much lead is created from the decay of uranium? i don't think there is loss of mass in beta decay, so the conversion is 1:1.

3. Apr 25, 2010

### Bachelier

Actually what we are studying here is the decay of both Lead and Bismuth.

The Uranium is mentioned here to give an idea that both isotopes are derived from Ur. :)

4. Apr 25, 2010

### Bachelier

O Come on guys, is this such a difficult problem?