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Re: chains in equillibrium
Solve the differential equations (not difficult) or consult the Skrable reference
I cited in a previous message.
The opinions expressed are strictly mine.
It's not about dose, it's about trust.
Curies forever.
Bill Lipton
liptonw@dteenergy.com
BERNARD L COHEN wrote:
> On Thu, 31 Oct 2002, William V Lipton wrote:
> > >
> > > The decay rate for all members of the chain is equal to the decay
> > > rate of the parent nucleus for the chain.
> > >
> > >
> >
> > NOT NECESSARILY
> >
> > There are 3 general cases:
> > let K1 = decay constant of parent, A1e = equilibrium activity of parent
> > K2 = decay constant of daughter, A2e = equilibrium activity of daughter
> > (Assume no branching.)
> >
> > (1) transient equilibrium: K2 > K1, i.e. the parent is longer lived than the
> > daughter: Then: A2e = [K2/(K2 - K1)]*A1e
> > i.e., A2e > A1e
>
> --How is this possible? The rate at which the daughter is formed
> is A1e. How can it decay at a greater rate than the rate at which it is
> formed and still be in an equilibrium situation?
>
> > (2) secular equilibrium: K2 >> K1, really a special case of transient
> > equilibrium:
> > Then: A2e = A1e
>
> --I don't know what you mean by "transient equilibrium". I assume
> that equilibrium means what you call secular equilibrium.
>
> >
> > (3) K2 < K1: no equilibrium
> >
> > It gets more complicated if there's branching.
>
> --Not really. In equilibrium, each nucleus decays at the same rate
> as the parent, regardless of branching.
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