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Age of a meteorite

  1. Sep 1, 2015 #1

    Rectifier

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    The problem statement
    The ratio between stable Argon atoms(##^{40}Ar##) and radioactive Potassium atoms(##
    ^{40}K##)in a meteorite is 10.3. Assume that these Ar atoms are produced by decay of Potassium-atoms, whose half-life is ## 1.25 \cdot 10^9 ## years. How old is the meteorite?
    Translated from Swedish.

    The attempt at a solution
    Radiactive decay can be calculated with ## N(t) = N_0e^{- \lambda t}## where t is time and lambda is constant. Half-life gives us ## \frac{1}{2} = N_0e^{- \lambda t}##

    And we know that
    ## \frac{N_{r}}{N_s}=10.3 ## and ##N_{r} + N_s = N_0## thus ##N_r = \frac{N_0}{11.3} ##
    ## N(t) = N_0e^{- \lambda t}## and ##N_r = \frac{N_0}{11.3} ## give us ##\frac{N_0}{11.3} = N_0e^{- \lambda t}## and ## t=4.3 \cdot 10^9 ## which is wrong :/.

    Can someone help please?
     
    Last edited: Sep 1, 2015
  2. jcsd
  3. Sep 1, 2015 #2

    DrClaude

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    The way I read it, there is 10.3 times more argon than potassium, not the other way around.
     
  4. Sep 1, 2015 #3

    Rectifier

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    Tanks for the reply!

    So it should be:
    ## \frac{N_{s}}{N_r}=10.3 ## and ##N_{s} + N_r = N_0## thus ##N_r = \frac{N_0}{11.3} ##
    ## N(t) = N_0e^{- \lambda t}## and ##N_r = \frac{N_0}{11.3} ## give us ##\frac{N_0}{11.3} = N_0e^{- \lambda t}## and ## t=4.3 \cdot 10^9 ##

    and I get the same answer :D
     
  5. Sep 1, 2015 #4

    DrClaude

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    I didn't notice that you had made a second error cancelling the first one!

    I notice now that
    is not correct.

    What is the value of ##\lambda##?
     
  6. Sep 1, 2015 #5

    Rectifier

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    Is it ## \lambda = \frac{ln2}{T_{1/2}}## ?

    EDIT:
    Thus ## \lambda = \frac{ln2}{T_{1/2}} =\frac{ln2}{1.25 \cdot 10^9 } ##
     
  7. Sep 1, 2015 #6

    DrClaude

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    That's correct.

    I made the calculation myself, and find the same value as you. Why do say the answer is wrong?
     
  8. Sep 1, 2015 #7

    Rectifier

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    My teacher told me so. Well I guess that he is wrong then.
     
  9. Sep 1, 2015 #8

    DrClaude

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    Didn't he give any hint as to what was wrong?

    The only thing I can see is that your answer does not have the correct number of significant digits.
     
  10. Sep 1, 2015 #9

    Rectifier

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    Ah sh*t :D
    "Svara med två decimaler"

    Then its 4.37 :D
     
  11. Sep 1, 2015 #10

    DrClaude

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    Precis!

    You could also infer it from the fact that both the ratio and the half-life are given with three significant digits.
     
  12. Sep 1, 2015 #11

    Rectifier

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    Thank you for your help!
     
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