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Equilibrium between release and decay Kr-85

  1. Aug 29, 2007 #1
    1. The problem statement, all variables and given/known data

    A power plant releases 2,0 grams of Kr-85 in to the atmosphere every day. At some point there's an equilibrium between what the power plant releases into the atmosphere and the decay in the atmosphere -> The decay in atmosphere is equal to 2,0 grams per day

    Calculate the mass of Kr-85 in the atmosphere for this to be possible


    2. Relevant equations

    I do not have any. Maybe: N = N0*e^-k*t but I am not sure

    3. The attempt at a solution

    I can't find a solution to this problem at all.

    Thank you very much on beforehand!

    /Thomas
     
  2. jcsd
  3. Aug 29, 2007 #2

    Astronuc

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    Staff: Mentor

    k is the decay constant, but [itex]\lambda[/itex] is conventionally used.

    So in equilibrium, the production rate matches the decay rate.


    The activity (A) is proportional to the number of atoms (N) present by k.

    One is actually looking for the mean activity and mean mass, since the problem doesn't state if the release is instantaneous (i.e. a puff) or if it is continuous.
     
  4. Aug 30, 2007 #3
    It is a continuous stream/release

    I thought this: I know that the half-life is 10,8 years. I need to find a mass that enables the Kr-85 decay to release 2 grams/day. There must be an equation since it's impossible for me to calculate it since the half-life is an eks. function.

    I know the formula and what it means, but I am not sure wheter it is the right one to use, and if I've got the rigt infos. to just plot them in to the equation.?

    Thanks alot for for your help.
    best regards
    /Thomas
     
  5. Aug 30, 2007 #4

    Astronuc

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    Staff: Mentor

    That is an average activity, so convert 2.0 grams to number of atoms Nd decaying, and the mean activity A (decay rate) is simply Nd/time.

    Then A/k = N, where k is the decay constant and N is the number of atoms present for that decay rate.

    See where that takes one.
     
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