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Deriving the Bateman equation of Nuclear Decay Chains

by Elariel
Tags: bateman, chains, decay, deriving, equation, nuclear
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Elariel
#1
Mar13-07, 08:31 PM
P: 1
1. The problem statement, all variables and given/known data

Derive Bateman equation for a decay chain
a->b->c->d where each decays with a given mean life let decay constant be L, where L=1/mean life
Na(0)=No, Nb(0)=Nc(0)=Nd(0)=0

2. Relevant equations

Want to derive Nb(t)={(No)(La)/(Lb-La)}*{exp[-La*t]-exp[-Lb*t]}
extend for Nc(t)

3. The attempt at a solution

dNa(t)/dt=-La*Na(t)
Na(t)=No*exp[-La*t]

dNb(t)/dt=-Lb*Nb(t)+LaNa(t)
dNb(t)/dt=-Lb*Nb(t)+La{No*exp[-La*t]}

this is a none homogenous differential equation. I can't find a way to solve it.

dNc(t)/dt=-Lc*Nc(t)+LbNb(t)

I'm really not sure where to go from here.

If anyone could lend a hand it would be greatly appreciated.
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Dick
#2
Mar14-07, 11:09 AM
Sci Advisor
HW Helper
Thanks
P: 25,235
Ok, so what you are really asking is how to solve y'(t)+L*y(t)=f(t). You know the homogenous solution is exp(-L*t). Guess the solution will be of the form y(t)=g(t)*exp(-L*t). Put this guess into your original equation and get:

g'(t)*exp(-L*t)-L*g(t)*exp(-L*t)+L*g(t)*exp(-L*t)=f(t).

So g'(t)=exp(L*t)*f(t) and you can just integrate to get g(t).


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