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Number of decayed atoms in a simultaneous decay process

  1. Jun 24, 2011 #1
    A certain amount [itex]N_0[/itex] of a radioactive isotope with decay constant [itex]\lambda_1[/itex] is injected into a pacient. Besides that isotope's natural decay process, there's also a biological elimination process, with decay constant [itex]\lambda_2[/itex].

    Now, at time t, the number of remaining isotope atoms is given by [itex]N(t) = N_0 e^{-(\lambda_1 + \lambda_2) t} [/itex]. 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 [itex]\lambda_1[/itex] 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 [itex]\lambda[/itex].

    How to proceed then?
    Thanks in advance for your help.
     
  2. jcsd
  3. Jun 24, 2011 #2
    Try working with your equation such that you can separate the contribution of each process.

    Remember that you can freely separately the exponentials.
     
  4. Jun 24, 2011 #3

    Dick

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    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.
     
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