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Radioactive chain decay

  1. Jun 18, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm looking at a problem from MIT's Open Courseware on radioactive chain decay, i.e. one element decays into another decays into another, finding the quantity at time t.


    2. Relevant equations
    The standard linear differential equation governing exponential decay.


    3. The attempt at a solution
    I'd just like to make sure I'm going about this the right way - to calculate say the amount of substance 2 present at time T you'd solve the differential equation for the first substance finding [tex]N(t) = N_o {e}^-{\lambda t}[/tex], then take that and plug it back in to the standard exponential differential equation as the quantity, solve that differential equation etc. God knows I'm not going to try to code it up in Latex, but I'd just like to know that I'm on the right track.:biggrin:
     
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  3. Jun 18, 2009 #2

    HallsofIvy

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    What are the differential equations you refer to?
     
  4. Jun 18, 2009 #3
    The differential equation governing exponential decay, i.e. [tex]\frac{dN}{dT} = -\lambda N[/tex]. Solving this through separation of variables should give me the amount of the first substance remaining at time T, that is [tex] N(t) = N_o {e}^{-\lambda t} [/tex]. If I want to find the amount of the next product of the chain decay at time T, I assume I would substitute the second equation for the amount of the first substance at time T back into the original differential equation as N and solve that to get [tex]N_2(t)[/tex], the amount of the second substance...
     
  5. Jun 18, 2009 #4
    Actually, I think I might have to substitute [tex](N_o - N_o e^{-\lambda t})[/tex] since at time T = 0 there isn't any of the second substance yet produced to begin decaying!
     
  6. Jun 19, 2009 #5

    HallsofIvy

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    The "second product" would be the result of the decay of the first product and that will have a different "[/itex]\lambda[/itex]".
    In the formula you give,
    [tex]\frac{dN}{dt}= -\lambda N[/tex],
    N is the amount of the original substance at time t. The "product" N1 is
    [tex]N_0- N_1(t)= N_0(1- e^{-\lambda t}[/tex]
    as you say.
     
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