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anonymous

  • 5 years ago

How old is a bone that has lost 25% of its carbon 14? (hint:the half life of Carbon 14 is is 5,770yrs)

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  1. anonymous
    • 5 years ago
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    2374.4 yrs

  2. anonymous
    • 5 years ago
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    t = 1/k x ln[100/(100-25)] k = ln 2/ t(1/2)

  3. anonymous
    • 5 years ago
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    Are you sure? That's not one of the possible answers.

  4. anonymous
    • 5 years ago
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    Answeres can be: 21,106 yrs 2,390 yrs 3,120 yrs 5,770 yrs

  5. anonymous
    • 5 years ago
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    Answers*

  6. anonymous
    • 5 years ago
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    2390.thats the closest

  7. anonymous
    • 5 years ago
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    approximations in the logarithms lead to deviations

  8. anonymous
    • 5 years ago
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    True. Thank you.

  9. anonymous
    • 5 years ago
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    Another way is: \[A=A_0e^{rt}\] Work out what the rate of change is, in other words r for the C14 \[1/2 = e^{r5770}\] take ln of both sides \[\ln(1/2)=r\times5770\] \[r=\frac{\ln(1/2)}{5770} \approx 0.00012013\] substitute that back into the equatio nfor 25% gives \[0.75 = e^{-0.00012013t}\] finally \[\frac{\ln(0.75)}{-0.0002013}=t \approx2394.75\]

  10. anonymous
    • 5 years ago
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    yet another way is \[.75=(\frac{1}{2})^\frac{t}{5700}\] \[\frac{ln(.75)}{ln(.5)}=\frac{t}{5700}\] \[t=5700\times \frac{ln(.75)}{ln(.5)}\]

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