SOLUTION: This is a carbon decay question that I haven't encounted yet. Can anyone give me a hand here? Thanks!
Question: Only 25% of the carbon-14 in a wooden bowl remains. How old i
Algebra ->
Customizable Word Problem Solvers
-> Misc
-> SOLUTION: This is a carbon decay question that I haven't encounted yet. Can anyone give me a hand here? Thanks!
Question: Only 25% of the carbon-14 in a wooden bowl remains. How old i
Log On
Question 62855This question is from textbook college algebra
: This is a carbon decay question that I haven't encounted yet. Can anyone give me a hand here? Thanks!
Question: Only 25% of the carbon-14 in a wooden bowl remains. How old is the bowl?
Thanks for your help! This question is from textbook college algebra
You can put this solution on YOUR website! This is a carbon decay question that I haven't encounted yet. Can anyone give me a hand here? Thanks!
:
Question: Only 25% of the carbon-14 in a wooden bowl remains. How old is the bowl?
:
The Half-life formula: A = Ao*.5^(t/h)
Where:
A = resulting amt; 25% here
Ao= Initial amt; 1 here
t = time (in years here)
h = half life, (carbon 14 half-life given as 5730 yrs)
:
.25 = 1 * .5^(t/5730)
:
Use the log equiv of exponent:
ln(.25) = (t/5730) * ln(.5)
-1.38629 = (t/5730) * -.693147
-1.38629 = -.693147t/5730
:
Get rid of denominator, mult equation by 5730
-1.3829*5730 = -.693147t
-7943.466689 = -.693147t
t = -7943.466689/-.693147
t = 11,460 yrs
:
Check:
A = 1*.5^(11460/5740); using a good calc
A = .25
