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Half-life and decay

  1. Aug 27, 2006 #1
    (a) The half-life of radium-226 is 1620 years. Write a formula for the quantity, Q, of radium left after t years, if the initial quantity is Q0.

    Check me on this one:
    Q = (Q0 / 2(1620/t))

    (b) What percentage of the original amount of radium is left after 500 years?

    Check this one as well:
    Q = (Q0 / 21620/500)
    Q = (Q0 / 23.24)
    Q = Q0 * 2-3.24
    Q = Q0(0.1058)
    10.6%
     
  2. jcsd
  3. Aug 27, 2006 #2

    Astronuc

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    In part 'a', the exponent on the 2 should reflect the number of half-lives, which is the ratio of t/T1/2, where T1/2 is the half-life.

    So after 1 half-life, Q/Qo= 1/2, and after two half-lives, Q/Qo= (1/2)2 = 1/4, . . .

    In the part 'b', the half-life of Rn-226 is 1620 years, the point at which 50% would be remaining, and 500 years is less than 1/3 of the half-life, so does 10.6% look right?
     
    Last edited: Aug 27, 2006
  4. Aug 27, 2006 #3

    HallsofIvy

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    In other words, you have the exponent "upside down". It should be
    t/1620, not 1620/t.
     
  5. Aug 27, 2006 #4
    how's this?

    (a) Q = (Q0 / 2t/1620)

    (b) Q = (Q0 / 2500/1620)
    Q = (Q0 / 20.3086...)
    Q = Q0 * 2-0.3086...
    Q = Q0(0.8074...)
    80.7%
     
  6. Aug 27, 2006 #5

    Astronuc

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    Better. :approve:
     
  7. Aug 28, 2006 #6

    HallsofIvy

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    Do you understand why it is t/1620 rather than 1620/t? The "half life" of a substance is the time it takes to degrade to half its original value. Every time one "half life" passes, the amount is multiplied by 1/2: if the original amount is M, after one "half life" the amount is (1/2)M. After a second "half life", it is (1/2)((1/2)M)= (1/2)2M. After a third "half life" we multiply by 1/2 again: (1/2)((1/2)2M)= (1/2)3M. That is, the exponent just counts the number of "half lives" in the t years. If the "half life" is 1620, that "number of half lives" is t/1620 so the amount will be (1/2)t/1620M= M/2t/1620.
     
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