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When was U-238 and U-235 50% at the same time?

  1. May 20, 2010 #1
    1. The problem statement, all variables and given/known data
    The uranium present in earth today is 99.28% for U-238 and 0.72% for U-235. The half-life of U-238 is 4.51E9 years and 7.0E8 years for U-235. How long ago was this uranium 50% U-238 and 50% U-235?


    2. Relevant equations
    A=Aoe-kt


    3. The attempt at a solution
    This is driving me nuts. I first thought of doing them both separately:

    Let A1 be U-238 and A2 be U-235
    A1=99.28
    A2=0.72

    k1= ln2/4.51E9=1.53E-10
    A1=Ao, 1e-k1t
    99.28=50e-1.53E-10t
    ln(99.28/50)=-1.53E-10t
    t=-4.4E9 years = 4E9 years

    k2= ln2/7.0E8=9.9E-10
    A2=Ao, 2e-k2t
    0.72=50e-9.9E-10t
    ln(0.72/50)=-9.9E-10t
    t=4.3E9 years = 4E9 years

    But then I tried solving them where Ao should be equal for them. But somehow that doesn't make sense unless Ao is the 50%??? Anyways, here's the other way I tried:

    Equation 1:
    99.28=Aoe-1.53E-10t

    Equation 2:
    0.72=Aoe-9.9E-10t

    Rearrange Equation 2:
    Ao=0.72/e-9.9E-10t

    Plug Equation 2 into Equation 1:
    99.28=(0.72e-1.53E-10t)/(e-9.9E-10t)
    99.28/0.72 = (e-1.53E-10t)/(e-9.9E-10t)
    137.9 = (e-1.53E-10t)/(e-9.9E-10t)
    ln(137.9) = -1.53E-10t + 9.9E-10t
    ln(137.9) = 8.36E-10t
    t = 5.9E9 years

    Arrrghh! This is so frustrating! I don't even know how to check which is the right answer. Can someone explain to me which method is correct and why? Am I putting the numbers in the correct A or Ao place??
     
  2. jcsd
  3. May 20, 2010 #2
    Your first method is incorrect. You cannot set 50% as your [itex]A_0[/itex], because [itex]A[/itex]'s in the exponential equations are absolute figures, not percentages. If you set your [itex]A_0[/itex] for U-238 to 50, and today's value to 99.28, then your equation says that the amount of U-238 has increased!

    Your second method yields the right answer (I think), but not really for the right reason. Again, you can't set your [itex]A[/itex]'s to percentages. Instead, work with absolute figures. Your equations should then look something like

    [tex]A_{235} = A_0 2^{-t/t_{235}};
    A_{238} = A_0 2^{-t/t_{238}}.[/tex]

    Here, [itex]A_0[/itex] is the (absolute!) amount of U-238 and U-235 at the time when there was an equal amount of both, and the left hand sides represent the amounts today. Now, what can you say about the relationship between [itex]A_{235}[/itex] and [itex]A_{238}[/itex]?
     
  4. May 20, 2010 #3
    Thanks for replying :) but please bear with me - I learn pretty slow. :blushing:

    When you say absolute figures, what do you mean? I tried googling. Are they the same as absolute values? Baaarrghh....what? Do I need to look up the exact amount today? Wait - is that why you made the base a "2" instead of "e"?

    And, again I apologize if I'm slow, U-238 is not supposed to increase because of the decay chain, right? (I just now looked it up.) And U-235 is on a totally different decay chain, right? (I just looked that one up too.) So, they should definitely be unrelated except for the fact that one point in time they were both at 50%.

    Then, does that mean you could set the equation to be:

    Ao,235 = Ao,238

    I rearranged the equations and got t=5.9x109 years like I did earlier, however, I didn't recognize that type of equation you gave me from anything I've read in my PChem book. I was wondering if you could explain each parts (particularly the "2" in there).

    Ahhhh, thank you so much for pointing me in a different direction. I was getting confused by the wording of the homework question.
     
  5. May 21, 2010 #4
    Don't worry about 'learning slow', it's hard to judge in what stage of education you are, so I may not have provided enough details.

    I do not mean absolute in the sense of an absolute value. I mean absolute as opposed to relative. If last year my apple tree produced ten apples, and this year it produced eleven, I can put this in two ways. I can say that the production of apples increased by 1, or by 10%. The first number is absolute, the second is relative.

    I used the 2 instead of [itex]e[/itex] because you gave half-lives. Look at the equation

    [tex]A = A_0 2^{-t/T}.[/itex]

    The half-life [itex]T[/itex] is defined as the time it takes to cut the original amount in half. Looking at the above equation, you see that indeed, when [itex]t=T[/itex], you get

    [tex]A = A_0 2^{-1} = A_0/2.[/tex]

    There's no difference between my way of writing it and your way of writing it. You write

    [tex]A = A_0 e^{-k t}[/tex]

    and you define [itex]k = \ln{2}/T[/itex], so that is

    [tex]A = A_0 e^{-(\ln{2}/T) t} = A_0 2^{-t/T}.[/tex]

    No, neither of the two amounts are supposed to increase, since you're dealing with decay.

    Though it is true that [itex]A_{0,235} = A_{0,238}[/itex], this is already incorporated in the equations I mentioned by the fact that I simply called both of these numbers [itex]A_0[/itex]. Instead, look at what you can say about the left-hand sides of the equations, i.e. the present-day values. You do not need to know the actual amounts to find a relationship between them. Try to find their ratio.
     
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