anonymous
  • anonymous
a student in greece discovers a pottery bowl that contains 62% of it's original amount of c-14 find the age of the bowl to the nearest year
Mathematics
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SOLVED
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katieb
  • katieb
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IrishBoy123
  • IrishBoy123
you'll need the half life if C-14 to start with
anonymous
  • anonymous
5730 is the half life of c-14
anonymous
  • anonymous
u have any options

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anonymous
  • anonymous
no, i have the formula N=N0e^kt
anonymous
  • anonymous
k= 0.0001
IrishBoy123
  • IrishBoy123
adding some colour to the decay equation: \(N(t)=N_oe^{kt}\) or \(N(t)/N_o= e^{kt}\) it is decay so k= minus number the half life tells you when \(N(t)/N_o= 0.5\) but \(e^{-0.0001 * 5730} \ne 0.5\) you need \(k = -1.21 \times 10^{-4}\) so find t for \(N(t)/N_o= 0.62\) lots of clues in there :p
anonymous
  • anonymous
I really don't understand it at all would you mind going step by step to solve it?
IrishBoy123
  • IrishBoy123
OK lets say a long long time ago you started with \(N_o\) atoms of C-14. [C_14 atoms are unstable and undergo \(\beta \) decay. \(\mathrm{~^{14}_{6}C}\rightarrow\mathrm{~^{14}_{7}N}+ e^- + \bar{\nu}_e\)] so the amount N you have at any time t is given by the equation you have \(N(t) = N_o e^{-kt}\), an exponential decay equation. if at time t you have 62% left, then \(N(t)/N_o = 0.62\) your job is to find t, knowing that \(k=−1.21×10^{−4} \) and \(0.62 = e^{-kt}\)
anonymous
  • anonymous
how would I go about solving that? My calculator won't do it
IrishBoy123
  • IrishBoy123
use your brain, its much more powerful :p thought about take logs of both sides?
anonymous
  • anonymous
logs?
IrishBoy123
  • IrishBoy123
natural logarithms....
anonymous
  • anonymous
okay i have no clue what that is.
anonymous
  • anonymous
I'm pretty sure k=-0.000121
anonymous
  • anonymous
and you told me .62 is e^-kt
IrishBoy123
  • IrishBoy123
well that's very surprising. your teacher should not be asking you to do stuff like this unless they want you to spend all night guessing the answer. all i can say is that to calc, say, natural log 5, you find the \(ln\) button on your calculator and go for it using the number 5 and if you have something like \(e^z\), then \(ln \ ( e^z) = z\) so you can take log on both sides to unravel this....... and that's a close as i can get to actually doing it for you, in fact i may well be too close already, and that will get me in trouble in these parts :-(

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