An archaeologist finds the ¹⁴C in a sample of $2.10 \mathrm {~g}$ of material to be decaying at 107 Bq. A modern $1.00 - g$ sample of the same material decays at 151 Bq,and assume this rate has also held in the past. The half-life of ¹⁴C is 5730 years. How old is the sample?

Question

Quizplus · Accepted Answer

First, we need to find the initial activity of the ancient sample using the modern sample as a reference:
\begin{align*}
	ext{Activity}_	ext{ancient} &= 	ext{Activity}_	ext{modern} \cdot \frac{	ext{Mass}_	ext{ancient}}{	ext{Mass}_	ext{modern}}\
&= 151~\mathrm{Bq} \cdot \frac{2.10~\mathrm{g}}{1.00~\mathrm{g}}\
&= 317~\mathrm{Bq}
\end{align*}

Next, we can use the half-life of the sample to find the decay constant:
\begin{align*}
\lambda &= \frac{\ln(2)}{t_{1/2}}\
&= \frac{\ln(2)}{5730~\mathrm{yr}}\
&= 0.000120968~\mathrm{yr^{-1}}
\end{align*}

Using the decay equation $N(t) = N_0e^{-\lambda t}$, we can solve for the age of the ancient sample:
\begin{align*}
N_0 &= \frac{	ext{Activity}}{\lambda}\
&= \frac{317~\mathrm{Bq}}{0.000120968~\mathrm{yr^{-1}}}\
&= 2.618 	imes 10^6\
t &= -\frac{\ln(N/N_0)}{\lambda}\
&= -\frac{\ln(0.5)}{\lambda}\
&= 5730~\mathrm{yr} \cdot \frac{\ln(2)}{\lambda}\
&= 8,980~\mathrm{yr}
\end{align*}

So the answer is A) 8980 years.

An Archaeologist Finds the 14C in a Sample Of 2.10 g2.10 \mathrm {~g}2.10 g

An Archaeologist Finds the ¹⁴C in a Sample Of $2.10 \mathrm {~g}$