Given the half life of C-14 is 5,700 how do you find the percentage of original C-14 after 2,588 years?

Question

The answer is supposed to be 73% but I dont know how they got that.

Answer 1

The equation is x = (1/2)^(t/5700)

This here is x = (1/2)^(2588/5700)
= .72999825

or about 73 %

Answer 2

y = y0*e^(kt) where y0 is the initial amount and y is the amount left after t.

First you need to find k

.5 = 1e^(k*5700) after 5700 years you will have half as much, so plug those into the formula

ln(.5) = ln e^(k*5700)

ln(.5) = (k*5700)

k = ln(.5)/5700 = -.0001216 approximately

Now plug this in the formula again using the numbers that you want to find. In this case you start with 100% (y0) and t=2588, and you are looking for y

y = 100e^(-.0001216*2588)

y = 73, which is 73% of the original 100

Answer 3

amount left is original times (1/2)^(2588/5700), since you have a fraction of the half life.

(1/2)^(2588/5700) = 0.7299982509

which is 73% of the original.

Answer 4

You need to use the formula for the amount of C-14 as a function of time:

C14(t) = c14(0)* 2^(-t/halflife)

To test this, if halflife is 1, and t is 1, then you should have half left after 1 year. c14(1) = c14(0)*2^(-1) = c14/2.

so, c14(2588) = c14(0)*2^(2588/5700)
which, using my calculator, I compute to be .7299 - pretty close