## anonymous one year ago 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

1. IrishBoy123

2. anonymous

5730 is the half life of c-14

3. anonymous

u have any options

4. anonymous

no, i have the formula N=N0e^kt

5. anonymous

k= 0.0001

6. 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

7. anonymous

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

8. 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 $$0.62 = e^{-kt}$$

9. anonymous

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

10. IrishBoy123

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

11. anonymous

logs?

12. IrishBoy123

natural logarithms....

13. anonymous

okay i have no clue what that is.

14. anonymous

I'm pretty sure k=-0.000121

15. anonymous

and you told me .62 is e^-kt

16. 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 :-(