Question

the half-life of carbon-14 is 5700 years. find the age of a sample at which 9% of the radioactive nuclei originally present have decayed? a. 1776 years b. 676 years c. 1226 years d. 776 years

Answer 1

This can be solved as a ratio, can't it?
\[{50/100 \over 5700} = {9/100 \over x}\]

Answer 2

yes

Answer 3

It's a differential equation problem. This is kind of a classic differential equation which has the solution x(t) = e^(-k * t) * e^c

x(t) is the amount of carbon 14 left after t years.

e^c is the initial amount of carbon 14 (put in t=0 to see this).

x(5700) = 1/2 * e^c.
Some algebra gives you k = ln(2^(1/5700))

now to find the time that gives you 9% loss of carbon:
x(t) = 0.91 * e^c
some algebra gives you t = ln(1/(0.91^(1/ln(2^(1/5700)))))

This comes out as 775.551...
So after 776 years 9% will have decayed.