# Number of decayed atoms in a simultaneous decay process

1. Jun 24, 2011

### 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 occuring (the one characterized by the used value of $\lambda$.

How to proceed then?

2. Jun 24, 2011

### rickz02

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

Remember that you can freely separately the exponentials.

3. Jun 24, 2011

### 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.