Chemical Forums
Specialty Chemistry Forums => Nuclear Chemistry and Radiochemistry Forum => Topic started by: Ann on October 10, 2005, 12:43:51 AM
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Suppose "A" is decaying to "B" and "B" to "C". "B" is an intermediate that coverts to "C" immediately. We know landa(A) and landa(B) as decay constants of A to B and B to C, respectively. Also, we know that at t=0, the initial value of A is A(0) and the initial values of B and C is both equal to zero. What decaying equation can I get for B and C? Any suggestion would be helpful.
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The first thing you should do is look into the different types of equilibrium that exist. What you have stated above is one of the “3" standard equilibrium cases.
If you want to try to derive the equations yourself, then start at the beginning. You know that at the beginning, you only have A available, so any amount of B is formed from the decay of A. You can then write out the amount of B formed at time t is given by the amount of A that decays. So the amount of B formed is, –dN(A)/dt = N(t)*lamda(A). But then you have the added complication that B is also decaying. So the rate of decay of B is dN(B)/dt = N(t)*lamda(A) - N(t)*lamda(B), or the production of B, minus the amount lost through decay.
That is not the full answer, but it is a start.
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For the future reference, it's not "landa", or "lamda", it's lambda ... Greeks would be pissed if they saw this ;D
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although this question is on radioactive decay, it's pretty much the same math stuff we do for reaction kinetics. The nice part about radioactive decay is that it always involves first order reaction.
;)