Question 595563
The half life of carbon14 is 5730 yrs.
 How long does it take for 2.8 grams of carbon 14 to be reduced to 1.1 grams ofcarbon-14 by radioactive decay?
 (Don't round to the final answer.)
:
The radioactive decay formula: A = Ao*2^(-t/h), where
A = resulting amt after t time
Ao = initial amt
t = decay time
h = half-life of substance
:
2.8*2^(-t/5730) = 1.1
Divide both sides by 2.8
2^(-t/5730) = {{{1.1/2.8}}}
2^(-t/5730) = .392857
Using nat logs
ln[2^(-t/5730)] = ln(.392857)
log equiv of exponents
{{{-t/5730}}}*ln(2) = ln(.392857)
{{{-t/5730}}} = {{{ln(.392857)/ln(2)}}}
find the ln
{{{-t/5730}}} = -1.3479
Multiply both sides by -5730
t = -1.3479 * -5730
t = 7,723.6 yrs to decay to 1.1 grams