Question

I need help with this equation! Carbon-14 has a half-life of 5730 years. Derek has a sample of a fossil that has 33mg of its original 100 mg of carbon-14. Find the decay constant, k, and use it to answer the questions that follow.
First use the formula P(t)=Ae ^{kt} , to find the decay constant, k

Answer 1

First all, the decay formula is $P(t)=Ae^{-kt}$
where:
$P(t)$ is the remaining quantity after $t$ years
$A$ is the initial sample
$t$ is the time in years 
$k$ is the decay constant 

From the problem we know that $A=33$ and $P=100$, but we don't have the time $t$; to find it we will take advantage of the half-life of the Carbon-14. If you have a sample of 100 mg and Carbon-14 has a half-life of 5730, after 5730 years you will have half of your original sample i.e. 50 mg. We also know that after $t$ years we have a remaining sample of 33mg, so the amount of the sample that decayed is $100mg-33mg=67mg$. Knowing all of this we can set up a rule 3 and solve it to find $t$:
$\frac{100mg--\ \textgreater \ 5730years--\ \textgreater \ 50mg}{100mg--\ \textgreater \ (t)years---\ \textgreater \ 67mg}$
$\frac{5730years--\ \textgreater \ 50mg}{(t)years---\ \textgreater \ 67mg}$
$\frac{5730}{t} = \frac{50}{67}$
$t= \frac{(5730)(67)}{50}$
$t=7678.2$

Now that we know our time $t$ lets replace all the values into our decay formula:
$33=100e^{-7678.2k}$
Notice that the constant $k$ we need to find is the exponent; we must use logarithms to bring it down, but first lets isolate the exponential expression:
$\frac{33}{100} = e^{-7678.2k}$
$e^{-7678.2k} = \frac{33}{100}$
$ln(e^{-7678.2k} )=ln( \frac{33}{100} )$
$-7678.2k=ln( \frac{33}{100} )$
$k= \frac{ln( \frac{33}{100}) }{-7678.2}$
$k=-0.000144$

We can conclude that the decay constant $k$ is approximately  -0.000144