Question

A fossil is discovered that has only 12.5% of the carbon-14 that it would have had originally (when the animal was alive). If the half-life of carbon-14 is 5700 years, approximately how old is the fossil?

Answer 1

Answer:

The fossil is 17,100 years old.

Explanation:

The decay equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt}= -\lambda N$

$\Rightarrow \frac{dN}{N}= -\lambda dt$

Integrating both sides

$\Rightarrow \int\frac{dN}{N}= \int-\lambda dt$

$\Rightarrow ln |N|=-\lambda t+c$

When t=0, N=$N_0$ = initial amount

$ln |N_0|=-\lambda .0+c$

$\Rightarrow ln |N_0|=c$

$\therefore ln |N|=-\lambda t+ln|N_0|$

$\Rightarrow ln |N|-ln|N_0|=-\lambda t$

$\Rightarrow ln |\frac {N}{N_0}|=-\lambda t$

$\Rightarrow \frac {N}{N_0}=e^{-\lambda t}$

$\Rightarrow N=N_0e^{-\lambda t}$

The decay equation is

$N=N_0e^{-\lambda t}$

Given that,

The half life of carbon - 14 is 5700 years.

For half life, $N=\frac{1}{2} N_0$

To find the value of  $\lambda$, we need to put the value of N and t in the decay equation.

$\frac12N_0=N_0e^{-\lambda \times 5700}$

$\Rightarrow \frac12=e^{-\lambda \times 5700}$             [ Divided $N_0$ both sides]

Taking ln both sides

$\Rightarrow ln| \frac12|=ln|e^{-\lambda \times 5700}|$

$\Rightarrow ln| \frac12|={-\lambda \times 5700}$

$\Rightarrow \lambda= \frac{ln| \frac12|}{-5700}$

$\Rightarrow \lambda= \frac{ln|1|-ln|2|}{-5700}$             [ $ln|\frac mn|= ln |m|-ln |n|$]

$\Rightarrow \lambda= \frac{ln|2|}{5700}$                      [ln 1= 0]

The fossil has only 12.5% of the carbon carbon-14 that it would have had originally.

So, $N=\frac{12.5}{100} N_0$

Then,

$\frac{12.5}{100} N_0=N_0e^{-\frac{ln|2|}{5700}t$

$\Rightarrow \frac{12.5}{100} =e^{-\frac{ln|2|}{5700}t$

Taking ln both sides

$\Rightarrow ln|\frac{12.5}{100} |=ln|e^{-\frac{ln|2|}{5700}t}|$

$\Rightarrow ln|\frac{12.5}{100} |={-\frac{ln|2|}{5700}t}$

$\Rightarrow t=\frac{ ln|\frac{12.5}{100}|} {-\frac{ln|2|}{5700}}$

$\Rightarrow t=\frac{ ln|\frac{12.5}{100}|\times 5700} {-{ln|2|}}$

$\Rightarrow t=17,100$

The fossil is 17,100 years old.