How Do You Calculate the Amounts of B and C in Radioactive Decay Over Time?

elsen_678
Messages
3
Reaction score
0
hi I'm new here hope you can help me with this problem
nuclei A decays to B and then to C(stable)
A and B have the same decay constant, n
Initially the amount of A is N(0)>0 while B and C are zero
then the amount of A at any time is :
N(t)=N(0)exp(-nt)
can somebody derive the equation for the amount of sample B and C as a function of time ?
thanks
 
Physics news on Phys.org
If you have any exposure to elementary diff. eq., the set up is as follows:

B'(t)=n(A(t)-B(t))
C'(t)=nB(t)
where A(t)=A(0)exp(-nt)
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top