Question

The half-life of a radioactive substance is the time required for half of a sample to undergo radioactive decay, or for the quantity to fall to half its original amount. Carbon 14 has a half-life of 5,730 years. Suppose given samples of carbon 14 weigh (fraction 5/8) of a pound and (fraction 7/8) of a pound. What was the total weight of the samples 11460 years ago

(show work please)

Answer 1

Amount of C-14 taken were 2.5 pounds and 3.5 pounds respectively.

Explanation:

Radioactive decay is an exponential process represented by

`A_(t)=A_(0)e^(-kt)`

where
`A_(t)` = Amount of the radioactive element after t years

`A_(0)` = Initial amount

k = Decay constant

t = time in years

Half life period of Carbon-14 is 5730 years.

`(A_(0) )/(2)=A_(0)e^(-5730k)`

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

Now we take ln (Natural log) on both the sides

`ln((1)/(2))=ln[e^(-5730k)]`

-ln(2) = -5730kln(e)

0.69315 = 5730k

`k=(0.69315)/(5730)`

`k=1.21* 10^(-4)`

Now we have to calculate the weight of samples of C-14 taken for the remaining quantities
`(5)/(8)` and
`(7)/(8)` of a pound.

`(5)/(8)=A_(0)e^{(-1.21* 10^(-4)* 11460)}`

`(5)/(8)=A_(0)e^{(-1.21* 10^(-4)* 11460)}`

`(5)/(8)=A_(0)e^((-1.3863))`

`A_(0)=(5)/(8)* e^(1.3863)`

`A_(0)=(5)/(8)* 4`

`A_(0)=(5)/(2)`

`A_(0)=2.5` pounds

Similarly for
`(7)/(8)` pounds

`(7)/(8)=A_(0)e^{(-1.21* 10^(-4)* 11460)}`

`(7)/(8)=A_(0)e^{(-1.21* 10^(-4)* 11460)}`

`(7)/(8)=A_(0)e^((-1.3863))`

`A_(0)=(7)/(8)* e^((1.3863))`

`A_(0)=(7)/(8)* 4`

`A_(0)=(7)/(2)`

`A_(0)=3.5` pounds

Answer 2

That's two half lives. 5/8 * 2 * 2 = 20/8 = 5/2 pounds7/8 * 2 * 2 = 28/8 = 7/2 pounds