A fossilized leaf contains 29% of it's normal amount of.....

O

ochocki

New member
Joined
Feb 7, 2005
Messages
46
A fossilized leaf contains 29% of it's normal amount of carbon 14. How old is the fossil (to the nearest year)? Use 5600 years as the half-life of carbon 14.

I'm usually pretty good at decay problems but this problem seems to be missing some info.

Can anyone please just give me a small bump in the right direction? I know the decay formula well, im just having a hard time geting this going.
 
let A<sub>o</sub> = amount of carbon 14 in the leaf at t = 0

so then, 0.29A<sub>o</sub> would be the amount of carbon 14 left at the time t you are looking for, correct?
 
Re: A fossilized leaf contains 29% of it's normal amount of.

Hello, ochocki!

A fossilized leaf contains 29% of it's normal amount of carbon 14.
How old is the fossil to the nearest year?
Use 5600 years as the half-life of carbon 14.
Click to expand...

There is enough information, but you should be aware of the half-life formula.

. . \(\displaystyle \L A\;=\;A_oe^{-kt}\;\;\;\)where \(\displaystyle \,k \:=\:\frac{\ln 2}{\text{half-life}}\)


In this problem: \(\displaystyle \,k\:=\:\frac{\ln 2}{5600} \:=\:0.000123776\)

Hence: \(\displaystyle \L\:A \;=\;A_oe^{-0.000124t}\)


Since \(\displaystyle A\:=\:0.29A_o\), we have: \(\displaystyle \:0.29A_o \:=\:A_oe^{-0.000124t}\)

Then: \(\displaystyle \:e^{-0.000124t} \:=\:0.29\;\;\Rightarrow\;\;0.000124t \:=\:\ln(0.29)\)

. . \(\displaystyle t \:=\:\frac{\ln(0.29}{-0.000124} \:=\:9982.85771\)


Therefore: \(\displaystyle \:t \:\approx\:9983\text{ years}\)

 
Hello again, ochocki!

If you ever forget the half-life formula (I do it all the time),
. . you can derive it yourself.

We have: \(\displaystyle \L\:A\:=\:A_oe^{-kt}\)

How long does it take for the substance be one-half the original amount?
. . That is, when does \(\displaystyle A\:=\:\frac{1}{2}A_o\) ?

. . So we have: \(\displaystyle \:A_oe^{-kt} \:=\:\frac{1}{2}A_o\;\;\Rightarrow\;\;e^{-kt}\:=\:\frac{1}{2}\)

Then: \(\displaystyle \:-kt \:=\:\ln\left(\frac{1}{2}\right)\:=\:\ln\left(2^{-1}\right) \;=\;-\ln(2)\)

Hence: \(\displaystyle \:t \:=\:\frac{-\ln(2)}{-k} \:=\:\frac{\ln(2)}{k}\)


Recall that \(\displaystyle t\) is the number of years in the half-life of the substance.

So we have: \(\displaystyle \:\text{half-life} \:=\:\frac{\ln(2)}{k}\;\;\Rightarrow\;\;\fbox{k \:=\:\frac{\ln(2)}{\text{half-life}}}\)

And that is how we determine the constant \(\displaystyle k\) is the formula.

 
You must log in or register to reply here.
Top