Question

The half-life of carbon 14 is 5730 years. If a historical organic object has 63% of its Carbon 14 remaining today, how long ago did the object die? a. 2444.8 yearsb. 2424.8 yearsc. 4422 yearsd. 4222.8 yearse. 3819.5 years

Answer 1

The formula for finding the amount of carbon 14 remaining in time t is given by

`A=A_oa^t`

To get a, we use the information that the half-life is 5730 years. That is in 5730 years,

`A=(A_o)/(2)`

Therefore;

`\begin{gathered} (A_o)/(2)=A_oa^(5730)_{} \\ canceloutA_O \\ (1)/(2)=a^(5730) \\ \text{simplify to get} \\ a=0.999879039 \end{gathered}`

And;

`A=A_o(0.999879039)^t`

b. To find when the object died given that it has 63% of its Carbon 14 remaining today.

`\begin{gathered} 0.63A_0=A_0(0.999879039)^t \\ \text{cancel out A}_0 \\ 0.63=0.999879039^t \\ \text{take the natural log of both sides} \\ \ln 0.63=t(\ln 0.999879039) \\ t=(\ln 0.63)/(\ln 0.999879039) \\ t=3819.5 \end{gathered}`

Therefore, the correct answer is option e. 3819.5 years