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Calculating natural abundance.

  1. Feb 28, 2013 #1
    I thought this area was appropriate for this question from a previous exam paper which I just need checking. So:

    2013_02_26_22_28_17.jpg

    Working out:

    The half-life of 244Hb is 10 million years.
    The half-life of 244Hb is 5.2595×10^12 minutes.
    The mean life of 244Hb is 7.58783677×10^12 minutes.
    So if you had a sample of 7.58783677×10^12 atoms of 244Hb, you could expect about 1 decay a minute.

    You have x grams of 244Hb and 1-x grams of 240Hb.
    Thus you have about x/244 moles of 244Hb.
    Thus you have about 2.468×10^21×x atoms of 244Hb.
    Since this sample decays one atom per minute, we know 2.468×10^21×x = 7.58783677×10^12.
    Or x = (1/ln 2)(10 million years / 1 minute) (244 grams/mole / ( N_A per mole * 1 gram ) = 3.07×10^-9

    Natural abundance of 240Hb on a per atom basis is ( x / 244)/ ( ( x / 244) + ( (1-x) / 240) ) = 60 x/(61 - x). Why?
    So the natural abundance of 240Hb is about 3.02×10^-9.

    If there is no natural source of 244Hb, this implies that the sample of Hibernium is no more than about 283 million years old.
     
  2. jcsd
  3. Feb 28, 2013 #2
    Feel free to move if its in the wrong place (which I think it is!)
     
  4. Feb 28, 2013 #3
    anyone?
     
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