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Share of Uranium isotopes

  1. Apr 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Current share of Uranium isotope on Earth is 99.28% (##^{238}U##) and 0.72% (##^{235}U##), half-life times are ##7.04\cdot 10^8 years## (##^{235}U##) and ##4.468\cdot 10^9 years## (##^{238}U##). Calculate the ratio between the isotopes ##4.5\cdot 10^9 years## ago.


    2. Relevant equations



    3. The attempt at a solution

    If hope it is ok to say that ##N(t)=N_0e^{-\frac{t}{\tau }}##.

    Let's now say that ##N_{238}(t)=Ae^{-\frac{t}{\lambda }}## where ##\lambda =\frac{t_{1/2}^{238}}{ln2}## and
    ##N_{235}(t)=Be^{-\frac{t}{\mu }}## where ##\mu =\frac{t_{1/2}^{235}}{ln2}##.

    Now we know that ##\frac{N_{238}(t)}{N_{238}(t)+N_{235}(t)}=0.9928## and also ##\frac{N_{235}(t)}{N_{238}(t)+N_{235}(t)}=0.0072##

    Knowing this, I can write:

    ##\frac{N_{238}(t)}{N_{238}(t)+N_{235}(t)}=\frac{Ae^{-\frac{t}{\lambda }}}{Ae^{-\frac{t}{\lambda }}+Be^{-\frac{t}{\mu }}}=0.9928## and

    ##\frac{Be^{-\frac{t}{\mu }}}{Ae^{-\frac{t}{\lambda }}+Be^{-\frac{t}{\mu }}}=0.0072##

    Dividing last two gives me:

    ##\frac{Ae^{-\frac{t}{\lambda }}}{Be^{-\frac{t}{\mu }}}=\frac{0.9928}{0.0072}## and

    ##A=Be^{-t/\mu +t/\lambda }\frac{0.9928}{0.0072}##

    Inserting this into ##\frac{Be^{-\frac{t}{\mu }}}{Ae^{-\frac{t}{\lambda }}+Be^{-\frac{t}{\mu }}}=0.0072## leaves me with the result that ##B=1##.

    Knowing this also gives me a result for ##A##, therefore ##A=459.8##.

    So.... If I am not mistaken, than following ratios should be the result I am searching:

    ##\frac{B}{B+A}=0.9978## for Uranium 235 and ##\frac{A}{B+A}=0.0022## for Uranium 238.

    Which to me is a bit confusing, so I would kindly ask somebody to tell me where I got it all wrong?
     
  2. jcsd
  3. Apr 17, 2014 #2

    BvU

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    Is not correct. It leaves you with the result that 0.0072 = 0.0072, which is not extremely useful.

    B = 1 is a strange answer anyway, right ?
    And: you should become suspicious when your answer stipulates that the fastest decaying isotope was less abundant in the past than the other one!

    And now for some more constructive stuff:
    First you want to check that 238U doesn't decay into 235U, which it doesn't.
    So both can be considered independent.

    You have expressions for the current ratio N238 / N235.
    With ##N(t)=N_0e^{-\frac{t}{\tau }}## you also have expressions for ##N(t)/N_0## for both.
    All you have to do is fill in the numbers !
     
  4. Apr 17, 2014 #3
    Fill in the numbers into what? o_O

    All I have is ##\frac{N_{238}(t)}{N_{238}(t)+N_{235}(t)}=0.9928## and ##\frac{N_{235}(t)}{N_{238}(t)+N_{235}(t)}=0.0072##, of course dividing those two brings me to current ratio, just like you said but... ?

    I don't get it what the next step is supposed to be..
     
  5. Apr 17, 2014 #4
    You reached the equation
    ##A=Be^{-t/\mu +t/\lambda }\frac{0.9928}{0.0072}##
    which is correct. You should now plug in the known values for t, λ, and μ. and solve for A/B.
     
  6. Apr 17, 2014 #5
    Ok...

    ##A=3.30B##.
     
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