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Half-life homework problem

  1. Dec 4, 2009 #1
    1. The problem statement, all variables and given/known data

    At present, the concentration of uranium 235 in naturally occurring uranium deposits is approximately 0.77%. What will the concentration be one billion years from now?


    2. Relevant equations

    N(t) = No e^ -(ln(2)/half-life)t

    3. The attempt at a solution

    i attempted this problem and got an answer of .69%.. What i did was put XNo= .77No e^ -(ln(2)/half-life)t .. that was the No can cancel out.. and then I just solved for X.. I feel like I am close but just making a slight mistake.
     
    Last edited by a moderator: Mar 7, 2013
  2. jcsd
  3. Dec 4, 2009 #2

    Redbelly98

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    Re: Half-life

    You need to know the half-life of uranium to solve this problem.
     
  4. Dec 4, 2009 #3
    Re: Half-life

    well i looked it up and found that it was equal to 7.04E8.. sorry that i did not include this information.. it wasn't included in the original question but i looked it up in the textbook.. but with that information, am i going about solving the problem correctly?
     
  5. Dec 4, 2009 #4
    Re: Half-life

    Your equation is right, although that .77 should either be .77% or .0077. I get something around half the value you reported when I plug the half-life and 1 Billion years into my calculator.

    One thing you're not accounting for is the decay of the U-238 though. Does the problem say to ignore that? About 15% of it would decay away in a billion years and you would have to account for that to get the new concentration of U-235.
     
  6. Dec 4, 2009 #5

    Redbelly98

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    Re: Half-life

    U-235 is not part of the decay chain of U-238, so the U-238 decay does not matter here.

    I agree with Grogs, the % U-235 is less than the .69% calculated by JDioro. JDioro, what do you calculate for the quantity:

    (ln(2)/half-life)t​

    It's probably a simple arithmetic error somewhere.
     
  7. Dec 4, 2009 #6
    Re: Half-life

    It's not in the decay chain, but it is in the denominator of the equation used to find the fraction of U-235 in natural uranium:

    frac(U-235) = N(235) / [N(235) + N(238)] (Ignoring the really small amount of U-234)

    That's why most of the times I saw this type of question in basic physics courses the question stated to ignore U-238 decay. Otherwise you have to solve for both and find the new fraction.

    If the question is asking what's the concentration in the rock (it seems a little ambiguous the way it's worded - "uranium deposit" sounds like we're just talking just uranium enrichment) then the U-238 just converts to Th-234 and so on so you don't have to account for it. If you're just doing basic radioactive decay I suspect that's what the instructor is looking for, but it's good to state that you're making that assumption.
     
  8. Dec 4, 2009 #7
    Re: Half-life

    my calculations were as follows:

    (ln(2)/half-life)t = .0985

    and then i just calculate e^ -.0985 and now i got .906.. and then when i multiply by the .77 I get .697
     
  9. Dec 4, 2009 #8

    Redbelly98

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    Re: Half-life

    If the half-life is 7.04e8 years, and t=1e9 years, then that expression should be 0.985, not 0.0985.

    Yes, good point. But I agree, we're probably supposed to ignore that, and that should have been stated explicitly.
     
  10. Dec 4, 2009 #9
    Re: Half-life

    UGH!... Im using the E button on my calculator and didn't account for an extra.. i was doing 1E8 instead of 10E8.. thanks for the help.. really appreciate it..
     
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