When Does Alpha Emission Peak in the Decay of Bi-210 to Pb-208?

[ghetom]

Hi I'm trying to do a question on nuclear decay chains. The question is:

Bi-210 decays to Po-210 by beta decay (half life = 7.2 days), and this decays by alpha decay to Pb-208 (half life = 200 decays). If A substance is initially pure Bi-210, when does the alpha emmision peak?

So far I've got
Bi(0)=1, Po(0)=Pb(0)=o

Bi(t) = Bi(0)exp(-Lt)
(L being the decay constant, t_1/2 = ln(2)/L)

and
dPo(t)/dt = LBi(t) -L'Po(t)

so
dPo(t)/dt = L(Bi(0)exp(-Lt)) -L'Po(t)Obviously I just need to find when Po(t) is at a maximum, to find the corresponding maximum in alpha emmision, but I can't solve for Po(t). (I tryed dPo/dt = 0, but i still don't have an answer for Po(t) )can anyone help?

 

[NateD]

The solution is as follows:The rate of change of Po-210 can be written as:dPo(t)/dt = LBi(0)exp(-Lt) - L'Po(t) Setting dPo(t)/dt = 0 and solving for Po(t), we get:Po(t) = LBi(0)exp(-Lt)/L'This is the peak value of Po-210. This corresponds to the peak of alpha emission since it is the decay of Po-210 to Pb-208 which produces the alpha emission. Therefore, the peak of alpha emission occurs when Po(t) is at its maximum value, or LBi(0)exp(-Lt)/L'. To find the time at which this occurs, note that t = ln(2)/L. Substituting this into the equation for Po(t) gives:Po(t) = LBi(0)exp(-ln(2)/L)/L'Solving for t gives:t = ln(2L'/LBi(0))/LThis is the time at which the alpha emission peak occurs.