The half-life of ^(14)C (Carbon-14) is 5730 years. That is, it takes this many years for half of a sample of ^(14)C to decay. If the decay of ^(14)C is modeled by x' = -rx,…

Question

The half-life of
`^(14)C` (Carbon-14) is 5730 years. That is, it takes this many years for half of a sample of
`^(14)C` to decay. If the decay of
`^(14)C` is modeled by x' = -rx, where x is the amount of
`^(14)C`, find the decay constant r. (Answer: r = 0.000121 yr⁻¹). In an artifact the percentage of the original
`^(14)C` remaining at the present day was measured to be 20 %. How old is the artifact?

Answer 1

a)
`(1)/(2) x_o = x_o e^(-r(5730)`

We can cancel
`x_o` and we got:

`(1)/(2)= e^(-r(5730))`

We apply natural log and we got:

`ln((1)/(2)) = -5730 r`

And
`r =(ln((1)/(2)))/(-5730)=0.000121`

b)
`0.2 x_o = x_o e^(-0.000121 t)`

We can cancel
`x_o` in both sides and we got:

`0.2 = e^(-0.000121 t)`

Now we can apply natural log on both sides and we got:

`ln(0.2) = -0.00121 t`

And if we solve for t we got:

`t =(ln(0.2))/(-0.000121)=13301.139 yars`

So in order to have 20% of the original amount
`x_o` the total time is 13301.14 years approximately.

Explanation:

Part a

For this case we have the following model given by the differential equation:

`x' = (dx)/(dt)= -rt`

The solution for this model is given by:

`X(t) = x_o e^(-rt)`

Using the half life we know that for 5730 years we need to have 1/2 of the initial amount
`x_o` so if we replace we have this:

`(1)/(2) x_o = x_o e^(-r(5730))`

We can cancel
`x_o` and we got:

`(1)/(2)= e^(-r(5730))`

We apply natural log and we got:

`ln((1)/(2)) = -5730 r`

And
`r =(ln((1)/(2)))/(-5730)=0.000121`

Where
`x_o` represent the initial amount. For this case we know the value for the rate of decay
`r = 0.000121 (1)/(year)`

Part b

And the half like is 5730 years. And we want to find the time in years in order to have 20% of the original amount. So we can write the following expression:

`0.2 x_o = x_o e^(-0.000121 t)`

We can cancel
`x_o` in both sides and we got:

`0.2 = e^(-0.000121 t)`

Now we can apply natural log on both sides and we got:

`ln(0.2) = -0.00121 t`

And if we solve for t we got:

`t =(ln(0.2))/(-0.000121)=13301.139 yars`

So in order to have 20% of the original amount
`x_o` the total time is 13301.14 years approximately.