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Radioactive Decay Differential Equation Question

  1. Nov 9, 2008 #1

    TFM

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    1. The problem statement, all variables and given/known data

    Unobtainium is a radioactive (and fictional) element, with its rate of decay being proportional to the amount of Unobtainium, [tex] u' = −k u [/tex]. The half-life of Unobtainium, in other words the time it takes for half the initial amount to decay, is [tex] 2.3 × 10^9 [/tex]yr.

    (a) Find the value of k.

    (b) How many decay events will there be per second in a 1 kg block of Unobtainium, given
    that it has an atomic mass of 415?

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I think I am doing the right thing, but I can't seem to work out k. I have a feeling I need limits to me integrals. so far I have:

    [tex] \frac{du}{dt} = -ku [/tex]

    [tex] \frac{1}{u} dt = -k dt [/tex]

    [tex] ln u = -kt + C [/tex]

    Where C is a constant of integration

    taking exponentials of both sides gives:

    [tex] u = e^{-kt} + C [/tex]

    If we had a [tex] u_{0} [/tex] somewhere, we know the half life, and thus could say that [tex]u = \frac{1}{2} u_o [/tex], stick some values for u and u0 in, and thus could probably work out k.

    Any ideas where I am going wrong?

    TFM
     
  2. jcsd
  3. Nov 9, 2008 #2

    tiny-tim

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    Hi TFM! :smile:

    Nooo … u = Ce-kt :wink:
     
  4. Nov 9, 2008 #3

    TFM

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    That was rather silly of me, thanks.

    I still need a [tex] u_0 [/tex] in there somewhere though to calculate k, don't I...?

    I also need to work out the constant C as well.

    TFM
     
  5. Nov 9, 2008 #4

    tiny-tim

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    Are you tired? :redface:

    How about u0 = C?
     
  6. Nov 9, 2008 #5

    TFM

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    How did you get C to equal [tex] u_0 [/tex]

    ???

    TFM
     
  7. Nov 9, 2008 #6

    tiny-tim

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    If u = Ce-kt, what is u when t = 0? :smile:
     
  8. Nov 9, 2008 #7

    TFM

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    I see, so:

    [tex] u = Ce^{-k*0} [/tex]

    [tex] u = Ce^0 [/tex]

    and:

    [tex] e^0 = 1 [/tex]

    Thus:

    [tex] u = C [/tex] at t = 0

    thus C = [tex] u_0 [/tex]

    Thanks,

    So now:

    [tex] u = u_0e^{-kt} [/tex]

    Insert values for u and u_0 that follow half-life, eg u_0 = 10, u = 5:

    [tex] 5 = 10e^{-k*2.3 * 10^9} [/tex]

    Divide both sides by 10:

    [tex] \frac{5}{10} = e^{-k*2.3 * 10^9} [/tex]

    Take logs of both sides:

    [tex] ln(\frac{1}{2}) = -k*2.3 * 10^9 [/tex]

    divide by half life:

    [tex] \frac{ln\frac{1}{2}}{2.3 * 10^9} = -k [/tex]

    thus

    [tex] k = -\frac{ln\frac{1}{2}}{2.3 * 10^9} [/tex]

    Does this look right?

    TFM
     
  9. Nov 9, 2008 #8

    tiny-tim

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    Yes (but a bit long-winded :wink:).

    Now simplify -ln(1/2). :smile:
     
  10. Nov 9, 2008 #9

    TFM

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    So:

    [tex] k = -\frac{ln\frac{1}{2}}{2.3 * 10^9} [/tex]

    [tex] -ln\frac{1}{2} [/tex]

    Thi simplifies to:

    [tex] ln(1) - ln(2) [/tex]

    ln(1) = 0

    thus:

    [tex] k = -\frac{-ln2}{2.3 * 10^9} = \frac{ln2}{2.3 * 10^9} [/tex]

    TFM
     
  11. Nov 9, 2008 #10

    tiny-tim

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  12. Nov 9, 2008 #11

    TFM

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    Excellent. Thanks.

    So for part b)

    How many decay events will there be per second in a 1 kg block of Unobtainium, given
    that it has an atomic mass of 415?


    We know now that:

    [tex] u = u_0e^{\frac{ln\frac{1}{2}}{2.3 * 10^9}t} [/tex]

    Do we insert t = 1 second, and u_0 = 1?

    TFM
     
  13. Nov 9, 2008 #12

    tiny-tim

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    t = 1, and u0 is the number of atoms of Unobtainium in 1kg :smile:
     
  14. Nov 9, 2008 #13

    TFM

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    The question states that it has an atomic mass of 415. This means, if I remember correctly, it has 415 atoms per kilogram, thus 1 kg = 415...?

    TFM
     
  15. Nov 9, 2008 #14

    tiny-tim

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    oooh, that's chemistry … i've no idea! :redface:

    though that does see a rather heavy atom …

    (and, come to think of it :rolleyes:, doesn't hydrogen have an atomic mass of 1?)
     
  16. Nov 9, 2008 #15

    TFM

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    Well on wikipedia:

    and gives an equation:

    [tex] m_{\rm{u}}={m_{\rm{grams}} \over N_{A}} [/tex]

    Is this useful?

    TFM
     
  17. Nov 10, 2008 #16

    tiny-tim

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    If you can use it, it is useful.

    If it is useful, use it. :wink:
     
  18. Nov 10, 2008 #17
    Hey,

    It should be [tex]415 g*mol^{-1}[/tex]. To find the number of mols divide the total mass by the molar mass. To find the number of atoms multiply the number of mols by Avogadro's number. Then put the number of atoms as your value of u to find the number of atoms decaying per second.
     
  19. Nov 10, 2008 #18

    TFM

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    Thanks, so its

    [tex] 415 grams m^{-1} [/tex]

    we have 1000 grams

    so:

    [tex] \frac{1000}{415} = 2.4 mols [/tex]

    So now we times this number by Avagaro's Constant, [tex] 6.022 * 10^{23} [/tex]

    This gives the number of particles to be:

    [tex] 1.45 * 10^{24} [/tex]

    So the equation is:

    [tex] u = u_0e^{\frac{ln\frac{1}{2}}{2.3 * 10^9}t} [/tex]

    t = 1
    [tex] u_0 = 1.45 * 10^{24} [/tex]

    [tex] u = (1.45 * 10^{24})e^{\frac{ln\frac{1}{2}}{2.3 * 10^9}1} [/tex]

    This gives the answer to be

    U = 1.45e^24

    :mad:

    But thats because the answer is so small.

    we need to find the number of decay events, so I assume we take away u from u_o,

    so this gives:

    [tex] 4.27*10^{14} [/tex] events per second. Seem right?

    TFM
     
  20. Nov 10, 2008 #19
    Your amount for [tex]u_{0}[/tex] is correct.

    The rest is incorrect. [tex]u \&\ u_{0}[/tex] are simply the number of atoms in your sample and as such have no units. You want the number of atoms decaying per second in a 1 kg sample. You have the total number of atoms in 1kg, [tex]u_{0}[/tex]. You have already found the decay constant, in this case k. You need only to multiply these two values to find your answer. Can you see why? (Consider the units of k.)

    Also don't forget to change k from per year to per second.
     
  21. Nov 10, 2008 #20

    TFM

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    Well the decay constant is 1/mean lifetime

    So the k at the moment is:

    [tex] k = -\frac{ln\frac{1}{2}}{2.3 * 10^9} [/tex]

    Converting into seconds gives a value of [tex] -9.4923*10^{-3} [/tex]

    So multiplying u0 with k(secs) gives: [tex] 1.376*10^22 [/tex]

    ???

    TFM
     
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