Deriving the Bateman equation of Nuclear Decay Chains

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SUMMARY

The discussion focuses on deriving the Bateman equation for a nuclear decay chain represented as a series of decays from species a to b to c to d, each characterized by a decay constant L, defined as L = 1/mean life. The derived equation for Nb(t) is Nb(t)={(No)(La)/(Lb-La)}*{exp[-La*t]-exp[-Lb*t]}. The participants also explore the non-homogeneous differential equations governing the decay processes, specifically dNb(t)/dt and dNc(t)/dt, and provide insights into solving these equations using the method of integrating factors.

PREREQUISITES
  • Understanding of differential equations, particularly non-homogeneous types.
  • Familiarity with decay constants and mean life in nuclear physics.
  • Knowledge of exponential functions and their properties.
  • Basic grasp of the Bateman equation and its applications in decay chains.
NEXT STEPS
  • Study the method of integrating factors for solving non-homogeneous differential equations.
  • Learn about the Bateman equation and its derivation in detail.
  • Explore applications of decay chains in nuclear physics and radiochemistry.
  • Investigate numerical methods for solving differential equations when analytical solutions are complex.
USEFUL FOR

Students and professionals in nuclear physics, mathematicians focusing on differential equations, and anyone involved in modeling decay processes in various scientific fields.

Elariel
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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.
 
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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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