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

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- IrishBoy123

you'll need the half life if C-14 to start with

- anonymous

5730 is the half life of c-14

- anonymous

u have any options

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- anonymous

no, i have the formula N=N0e^kt

- anonymous

k= 0.0001

- 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

I really don't understand it at all would you mind going step by step to solve it?

- 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

how would I go about solving that? My calculator won't do it

- IrishBoy123

use your brain, its much more powerful :p
thought about take logs of both sides?

- anonymous

logs?

- IrishBoy123

natural logarithms....

- anonymous

okay i have no clue what that is.

- anonymous

I'm pretty sure k=-0.000121

- anonymous

and you told me .62 is e^-kt

- 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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