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Radioactive decay question, how to do it?

  1. Feb 1, 2015 #1
    1. The half-life of 238 92U is 4.5 × 109 years, which is much larger than all the other half-lives of the decays in the series. A rock sample when formed originally contained 3.0 × 1022 atoms of 238 92U and no 206 82Pb atoms. At any given time most of the atoms are either 238 92U or 206 82Pb with a negligible number of atoms in other forms in the decay series.

    (ii) A certain time, t, after its formation the sample contained twice as many 238 92U atoms as 206, 82 Pb atoms. Show that the number of 238, 92 U atoms in the rock sample at time t was 2.0 × 1022.

    The answer shows an equation u/(3 x1022 - u) = 2
    (u represents the number of uranium atoms then)

    How did they get this equation? What kind of way of thinking got them to this point... don't get it. If someone could explain please. Also, if it helps, there was also part (i) where the task was to show how the number of 238 92U atoms and the number of 206 82Pb atoms in the rock sample vary over a period of 1.0 × 1010 years from its formation using a graph of N against time t.

    Maybe I should add more to be up with the forum rules... I want to use N=N0e-yt and worked out the decay constant y to be ln(2)/Half life of U = ln(2)/4.5x109 = 1.54x10-10.

    Question 2 in the paper: http://filestore.aqa.org.uk/subjects/AQA-PHYA5-1-QP-JUN12.PDF
    Mark scheme: http://filestore.aqa.org.uk/subjects/AQA-PHYA52A-W-MS-JUN12.PDF
     
    Last edited: Feb 1, 2015
  2. jcsd
  3. Feb 1, 2015 #2

    SteamKing

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    It's pretty clear that a manipulation of the half-life equation is in order. Here is another discussion of this equation:

    http://lpc1.clpccd.cc.ca.us/lpc/hanna/HistoricalGeology/HalfLifeEquations.pdf
     
  4. Feb 1, 2015 #3
    Thank you my teacher didn't even give me that equation but it was very useful. This is my working...
    t = Half life * log2(1/y) = 4.5 x109log2(3/2) = 2632331253
    so now N = N0e-1.54x10-10x 2632331253 = 2.0 x1022 (2sf)

    Very useful and definitely better than the answer they given but still don't know how they got that equation... anyone?
     
  5. Feb 1, 2015 #4
    Hi. The numbers involved and the description makes all of this look more complicated than the question really is. So, to simplify:
    You have a number N of particles A to begin; these can either stay as A or change into B.
    Question: if at time t you have twice as many A than you have B, how many A do you have?
    n(A) + n(B) = u + Pb = N = 3.0 × 1022, at any time.
    at t: u + u/2 = N, so u = 2N/3.
    Also from the equation above you can take u to the right of the equal sign and obtain:
    u/2 = N–u
    -> u/(N–u) = 2.
     
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