Considering a 2-time decay

2011-03-16 1:28 am

Some radioactive decaying processes do not end in only one step. For example, Uranium-238 will not stop decaying after decaying once to Th-234, and it will continue to decay to Pb-206 (Uranium Series). If we do not assume the intermediate nuclei have negligible half lives, how can I interpret the situation? Please use the following example to explain

Example:

A sample of Ba-140 of initial activity of 3.2 x 10^17 Bq decays to La-140, with a half life of 12.8 days. Then La-140 (half life = 1.7 days) continues to decay to Ce-140. What will be the equation in finding the total activity at time t? (Without assumption: A = Ao e^-kt is not applicable.)

回答 (1)

天同

2011-03-16 6:18 am

✔ 最佳答案

The equation describing the number of atoms in the intermediate daughter La-140 us,

d(N2)/dt = -(k2)(N2) + (k1).(N1)
where N1 and N2 are the number of Ba-140 and La-140 atoms at time t respectively
k1 and k2 are the corresponding decay constants

Since N1 = (N1o)e^(-(K1).t)
where (N1o) is the initial number of atoms of Ba-140
hence, d(N2)/dt = -(k2)(N2) + (k1).(N1o)e^(-(K1).t)
i.e. d(N2)/dt + (k2)(N2) = (k1).(N1o)e^(-(K1).t)

It is then only a matter of mathematics to solve this equation for N2, the solution is,
N2 = (N1o).e^[-(k2)t].[1 - e^(-(k1)t]

Since activity A2 is proportional to the number of atoms N2
A2 = (A1o)e^[-(k2)t].[1 - e^(-(k1)t]

The total activity is thus equal to (A1 + A2), where A1 = (A1o)e^[-(k1)t]

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