Model for series radioactive decay

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SUMMARY

This discussion focuses on modeling radioactive decay in a series involving three nuclei: A, B, and C. The differential equations governing the decay rates are established as follows: for nucleus A, the equation is dA/dt = -a A(t); for nucleus B, dB/dt = a A(t) - b B(t); and for nucleus C, the correct formulation is dC/dt = b B(t) - c C(t), without including a term for A(t). The importance of methodical step-by-step formulation of each differential equation is emphasized to avoid confusion.

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If nucleus A decays to nucleus B in rate a, and B decays to C in rate b, and C is decaying at the rate c. To setup a model for that process, we start from A

<br /> \frac{dA}{dt} = - a A(t)<br />

and for B, part for B is dying out from it's own decaying process but some amount of A will decay into B, so

<br /> \frac{dB}{dt} = a A(t) - bB(t)<br />

But for C, it is quite confusing, the total number of C is proportional to the number of B directly but also to that of A indirectly, of course it will dying out due to it's own decaying process. My question is should I include a term for A(t) in the different equation for C? That is, should I write

<br /> \frac{dC}{dt} = a A(t) + bB(t) - cC(t)<br />

or

<br /> \frac{dC}{dt} = bB(t) - cC(t)<br />
 
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The reason you are having trouble is because you are trying to do too many steps at once. Write each variable's differential equation first, e.g.

\frac{dB}{dt} = - \frac{dA}{dt} - bB(t)

i.e. the amount added minus the amount lost. Then and only then start plugging in known quantities.

I know it's tempting to take short cuts, but you are really much better off being methodical.
 
Vanadium 50 said:
The reason you are having trouble is because you are trying to do too many steps at once. Write each variable's differential equation first, e.g.

\frac{dB}{dt} = - \frac{dA}{dt} - bB(t)

i.e. the amount added minus the amount lost. Then and only then start plugging in known quantities.

I know it's tempting to take short cuts, but you are really much better off being methodical.

Got it. Thanks
 

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