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Decay of an isotope

  1. Apr 25, 2013 #1
    1. The problem statement, all variables and given/known data

    The half life of a particular isotope is 10.84 days. Find the number of atoms of this isotope that would be necessary to produce a sample with an activity of 1.79 micro-Ci.

    2. Relevant equations

    [itex] R=N_{0} \lambda e^{-\lambda t} [/itex]

    where R is the decay rate
    lambda is the decay constant
    N is the number of atoms
    and t is the half life



    3. The attempt at a solution

    First I converted the half life to seconds: 936576 s

    than I solved for the decay rate constant: lambda= 7.401*10^-7 1/s

    and R is 1.79*10^-6 Ci = 1.79*10^-6 decays/second

    When I plugged these values into the equation though, I get 4.84, which is wrong.
     
  2. jcsd
  3. Apr 25, 2013 #2

    SammyS

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    Which equation did you plug these values into?

    What did you use for t ?
     
  4. Apr 25, 2013 #3
    I'm not sure how I messed that up, but I totally didn't use that equation.

    What I actually did was solve for lambda and then used the equation

    [tex]N_{0} = \frac{R_0}{\lambda}[/tex]

    and got N to be 2.418, but that's wrong
     
  5. Apr 25, 2013 #4
    According to Wikipedia,

    "The Curie (symbol Ci) is a non-SI unit of radioactivity, named after Marie and Pierre Curie.[1][2] It is defined as

    1 Ci = 3.7 × 10^10 decays per second. "

    1 micro-Ci = 3.7E4 decays /sec so R should be 1.79 X 3.7E4 = 6.62 E4.

    Maybe this is the problem
     
  6. Apr 25, 2013 #5
    Deleted post.

     
    Last edited by a moderator: Apr 25, 2013
  7. Apr 25, 2013 #6
    That is definitely the problem. Thanks a ton!
     
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