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Determine the amount of americium-241 remaining after 15 years.

  1. Jun 27, 2013 #1
    1. The problem statement, all variables and given/known data
    The half life of americium-241 is 457.699 years. A typical smoke detector contains 33.1 kBq of americium-241. Determine the amount of americium-242 remaining after 15 years.


    2. Relevant equations
    A=Aoe - - 0.693t/T
    1/2


    3. The attempt at a solution
    =33.1xe - (0.693x15/457.699)
    A=32.1kBq
     
  2. jcsd
  3. Jun 27, 2013 #2

    rude man

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    About right, although you should use more significant digits than ln(0.5) = 0.693. I got 32.4 for my answer.
     
  4. Jun 27, 2013 #3

    HallsofIvy

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    Rather than use that formula, you could simply use the fact that "The half life of americium-241 is 457.699 years" means that Americium reduces by 1/2 every 457.699 years. 15 years is 15/457.699= 0.033 "half lives" so the sample is reduced by [tex](1/2)^{0.033}= 0.98[/tex] times: 0.98(33.1)= 32 mg.

    I am using two significant figures for my answer because "15 years", given to 2 significant figures, is the least accurate figure.
     
  5. Jun 27, 2013 #4

    rude man

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    "15 years" could have meant 15.000 years,; it probably did.

    Not extending log(0.5) to more than 3 sig. digits means an error of 0.4/32.357 = 1.24%.
     
  6. Jun 27, 2013 #5

    mfb

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    You cannot compare the digits like that.
    A more extreme example: if the initial radiation is 33.10000 kBq, what is the radiation after 1 second? Sure, the time has just one digit, but the activity will be 33.10000 kBq after 0 seconds and 2 seconds, too. 33kBq would be a wrong answer.
     
  7. Jun 27, 2013 #6

    rude man

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    Yes indeed.
    Assume it was for 30 years instead of 15. The answer to 2 sig. digits would still have been 32 kBq!

    Good point, enforcing precision on the 15 yrs was pointless; but log(0.5) should have been extended to more than 3 sig. digits. Agreed?
     
  8. Jun 27, 2013 #7

    haruspex

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    Good point - this is exponentiation, not multiplication or division. A 6 month error in the 15 years (though I agree with rude man that the question probably does not intend you to assume that) would be ±3.3%, but the resulting error in 2-n is only 0.08%.
    OTOH, if the question had asked how much had decayed, there would be a subtraction involved, and the error would now be back to ±3.3% (not a coincidence).
     
  9. Jun 27, 2013 #8

    rude man

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    A 6 month error in the 15 years would have given an error of 0.077%:

    let τ = -457.699/ln(0.5) = 660.32

    A = 33.1 exp(-t/τ)
    A = 33.1 exp(-15/660.32) = 32.357
    A' = 33.1 exp(-15.5/660.32) = 32.332
    (A - A')/A = 0.077%.
     
  10. Jun 27, 2013 #9

    haruspex

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    Wasn't 0.08% near enough?
     
  11. Jun 27, 2013 #10

    rude man

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    You said 3.3%, maybe I misinterpreted?
     
  12. Jun 27, 2013 #11

    haruspex

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    I wrote that ±6 months on 15 years was a 3.3% error, producing a 0.08% error in the answer.
     
  13. Jun 28, 2013 #12

    mfb

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    I would use more digits (or 2^...), but that is a small correction. The difference between your answers is just a calculation error in the first post. I get 32.35671 with the exact value and 32.35655 with the ln(2)-approximation.
    The relative error on ln(2) can be compared to the relative error of those 15 years - the answer is not very sensitive to it.
     
  14. Jun 28, 2013 #13

    rude man

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    You're right, the OP's first (& only so far) post did not do the arithmetic correctly & I hadn't checked it.
     
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