Number of decayed atoms in a simultaneous decay process

[mike_M]

A certain amount N_0 of a radioactive isotope with decay constant \lambda_1 is injected into a pacient. Besides that isotope's natural decay process, there's also a biological elimination process, with decay constant \lambda_2.

Now, at time t, the number of remaining isotope atoms is given by N(t) = N_0 e^{-(\lambda_1 + \lambda_2) t}. My question is, how do I calculate the number of atoms that, at time t, have decayed by *one* specific process (e.g. by radioactive decay alone)? I don't think I can use the above equation, with \lambda_1 instead of the sum, because such an equation would describe the number atoms assuming that that only one decay process is occurring (the one characterized by the used value of \lambda.

How to proceed then?
Thanks in advance for your help.

 

[rickz02]

Try working with your equation such that you can separate the contribution of each process.

Remember that you can freely separately the exponentials.

 

[Dick]

Your equation comes from solving N'=-lamba1*N-lambda2*N. Once you've found the total number of atoms as a function of time N(t) (as you've done) then the number of atoms going by one process or another is given by N'_rad(t)=-lambda1*N(t) and N'_elim(t)=-lambda2*N(t). Solve those.