Deriving the Bateman equation of Nuclear Decay Chains

  • Thread starter Elariel
  • Start date
  • #1
1
0

Homework Statement



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

Homework Equations



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

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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619
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).
 

Related Threads on Deriving the Bateman equation of Nuclear Decay Chains

Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
815
  • Last Post
Replies
0
Views
1K
Replies
23
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
9
Views
768
Top