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

The fossil is 17,100 years old.

Step-by-step explanation:

The decay equation:

`(dN)/(dt)\propto -N`

`\Rightarrow (dN)/(dt)= -\lambda N`

`\Rightarrow (dN)/(N)= -\lambda dt`

Integrating both sides

`\Rightarrow \int(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=(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 * 5700)`

`\Rightarrow \frac12=e^(-\lambda * 5700)` [ Divided
`N_0` both sides]

Taking ln both sides

`\Rightarrow ln| \frac12|=ln|e^(-\lambda * 5700)|`

`\Rightarrow ln| \frac12|={-\lambda * 5700}`

`\Rightarrow \lambda= (ln| \frac12|)/(-5700)`

`\Rightarrow \lambda= (ln|1|-ln|2|)/(-5700)` [
`ln|\frac mn|= ln |m|-ln |n|`]

`\Rightarrow \lambda= (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=(12.5)/(100) N_0`

Then,

`(12.5)/(100) N_0=N_0e^{-(ln|2|)/(5700)t`

`\Rightarrow (12.5)/(100) =e^{-(ln|2|)/(5700)t`

Taking ln both sides

`\Rightarrow ln|(12.5)/(100) |=ln|e^)/(5700)t|`

`\Rightarrow ln|(12.5)/(100) |=-(ln`

`\Rightarrow t=\frac(12.5)/(100) 2`

`\Rightarrow t=\frac* 5700 {-2}`

`\Rightarrow t=17,100`

The fossil is 17,100 years old.