Question

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

Answer 1

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

Answer 2

5730 is the half life of c-14

Answer 3

u have any options

Answer 4

no, i have the formula N=N0e^kt

Answer 5

k= 0.0001

Answer 6

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

Answer 7

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

Answer 8

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 \(0.62 = e^{-kt}\)

Answer 9

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

Answer 10

use your brain, its much more powerful :p

thought about take logs of both sides?

Answer 11

logs?

Answer 12

natural logarithms....

Answer 13

okay i have no clue what that is.

Answer 14

I'm pretty sure k=-0.000121

Answer 15

and you told me .62 is e^-kt

Answer 16

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 :-(