• Dec 5th 2008, 11:50 AM
fobos3
Hi!

I understand the concept of radioactive decay when you have only one element.
$\displaystyle \frac{dy}{dt}=-k\times y$

But what happens when one element decays into a second element and the second decays into a third. What is the quantity of not decayed material of the second element? And what if you have a longer chain? For example:
Element A has a half-life $\displaystyle T_1=2 s.$
Element B has a half-life $\displaystyle T_2=5 s.$
Element C is stable.
A decays to B which decays to C.

A(t), B(t), C(t) - is the quantity of A, B, C at time t
$\displaystyle A(t)=2^{-t/2}\times A(0)$
B(t)=?
C(t)=?
• Dec 5th 2008, 12:39 PM
running-gag
Hi

Between t and t+dt
A goes from A(t) to A(t+dt) = A(t) - a A(t) dt
B goes from B(t) to B(t+dt) = B(t) - b B(t) dt + a A(t) dt
C goes from C(t) to C(t+dt) = C(t) + b B(t) dt

It is clear that for every t, A(t) + B(t) + C(t) is constant

A(t+dt) = A(t) - a A(t) dt

$\displaystyle \frac{A(t+dt) - A(t)}{dt} = -a A(t)$

$\displaystyle \frac{dA}{dt} = -a A(t)$

$\displaystyle A(t) = A_0 e^{-at}$

B(t+dt) = B(t) - b B(t) dt + a A(t) dt

$\displaystyle \frac{B(t+dt) - B(t)}{dt} = -b B(t) + a A(t)$

$\displaystyle \frac{dB}{dt} = -b B(t) + a A_0 e^{-at}$

General solution of equation $\displaystyle \frac{dB}{dt} = -b B(t)$ is
$\displaystyle B(t) = \lambda e^{-bt}$

Particular solution of $\displaystyle \frac{dB}{dt} = -b B(t) + a A_0 e^{-at}$ is
$\displaystyle \frac{a A_0}{b-a} e^{-at}$ if a and b are different

$\displaystyle B(t) = \lambda e^{-bt} + \frac{a A_0}{b-a} e^{-at}$
$\displaystyle B(t) = B_0$

$\displaystyle B(t) = (B_0 - \frac{a A_0}{b-a})e^{-bt} + \frac{a A_0}{b-a} e^{-at}$

To be checked !!
But I think that the general idea is not so far !