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Homework Help: Am-241 radioactivity problem

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data
    A piece of Am-241 has a radioactivity of 10kBq. Determine how much Am-241 it contains.


    2. Relevant equations
    [tex]N(t)=N_0(\frac{1}{2})^{t/T_{1/2}} [/tex]


    3. The attempt at a solution
    Let A be the activity
    Let N be the number of atoms
    We know that [tex]A(t)=A_0(\frac{1}{2})^{t/T_{1/2}}[/tex] We can set our initial time to zero, which gives us [tex]A(t)=10kBq=A_0[/tex].
    Furthermore, we know that A(t)=-N(t). We also know that [tex]N(t)=N_0(\frac{1}{2})^{t/T_{1/2}} => N'(t)=-\frac{N_0ln2}{T_{1/2}}(\frac{1}{2})^{t/T_{1/2}}[/tex][tex]=>N'(0)=-\frac{N_0ln2}{T_{1/2}}=10kBq => N_0=\frac{10kBqT_{1/2}}{ln2}[/tex]. Now we use that one atom weights 241,0568229u and that the half-life is 432,2 y (I converted it to seconds). Then we get that the mass of our "piece" is approximately 78,7 micrograms. However, in the key it says 1,83 ng.
    What am I doing wrong here?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 15, 2010 #2

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Re: Radioactivity

    It's much simpler than that
    The number of decays in a second is just the number of atoms * the chance of a decay/second
    Which is just 1/mean lifetime - which you can easily get form the half life
     
  4. Apr 15, 2010 #3
    Re: Radioactivity

    Use this equation.
    [tex]
    A=A_0\exp\left(\frac{-t\ln 2}{T_{1/2}}\right),
    [/tex]
    [tex]A_0[/tex]=10000 Bq,
    [tex]T_{1/2}[/tex] is halflife
    PS: What is 't' (times passed from initial activity)?
     
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