Re: chains in equillibrium

[BERNARD L COHEN <blc+@PITT.EDU>]

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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