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Homework Help: Uranium decay

  1. May 10, 2007 #1


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    1. The problem statement, all variables and given/known data
    Hi, I've worked this out (I think) but would appreciate someone xhecking over my method.

    U 235 and U 238 found in ratio of 7.3E-3 to 1, but thought to have been produced equally (ie equal quantities)

    half lives are 1.03E9 and 6.49E9 respectively.

    Estimate time since formation

    2. Relevant equations

    Decay equation: N(t) = No e^(-wt) (used w for lambda here)

    3. The attempt at a solution

    As No is the same for both

    Ln (N(t) / e^(-w1.t)) = Ln No

    Ln N(t) + w1.t = Ln N(t) + w2.t

    w1 = ln2 / 1.03E-9 = 6.729E-10
    w2 = ln2 / 6.49E9 = 1.068E-10

    Put these into above equ and re-arranging

    ln (7.3E-3) - ln(1) = (t (6.729E-10)-(1.068E-10))


    t = (ln (7.3E-3) - ln(1)) / ((6.729E-10)-(1.068E-10))

    which I work out to be 8.69E9 years.

    I'm not quite sure if this is correct, but any advice would be appreciated.

  2. jcsd
  3. May 10, 2007 #2
    Hi, your method is correct except a missed symbol in equation:
    ln (7.3E-3) - ln(1) = t((6.729E-10)-(1.068E-10))
    but it doesn't matter. If no calculation mistake, the answer is a correct one.
  4. May 10, 2007 #3


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    If you mean that I have missed 't' out, then that is because both w1 and w2 are being multipled by t, so I factored it out
  5. May 10, 2007 #4


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    Yes, it's right.

    Notice you can always double check by plugging in the initial equation!
  6. May 11, 2007 #5

    Andrew Mason

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    I am not clear on your units. The half life of U238 is 4.51x10^9 years. The half life of U235 is 7.04x10^8 years.

    [tex]N/N0 = 1/2 =e^{\omega_{238}(4.51 x 10^9)}[/tex] so

    [tex]\omega_{238} = \ln{.5}/(4.51 x 10^9) = -1.5x10^{-10}[/tex]


    [tex]\omega_{235} = \ln{.5}/(7.04 x 10^8) = -9.8x10^{-10}[/tex]

    So if they started out in the same proportion after a SuperNova, the SuperNova occurred about:

    [tex].0073 = e^{-\omega_{235}t}/e^{-\omega_{238}t} = e^{(\omega_{238}-\omega_{235})t} [/tex]

    [tex]t = \ln{.0073}/(\omega_{238}-\omega_{235}) = 4.92/(7.3x10^{-10}) = 6.74 x 10^9 \text{years}[/tex]

  7. May 13, 2007 #6


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    Good observation. It looks like the OP is being given inverse decay rates rather than half-lives. kel would be advised to carefully check the wording of the question.
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