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

[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.

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.

 

[skeeter]

let A_o = amount of carbon 14 in the leaf at t = 0

so then, 0.29A_o would be the amount of carbon 14 left at the time t you are looking for, correct?

 

[soroban]

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.

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: $\displaystyle \:t \:\approx\:9983\text{ years}$

 

[ochocki]

I didn't know of the half life formula, thanks for showing that to me.

 

[soroban]

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.

 

[ochocki]

Thank you guys so much, you really helped me out.