Question

A sample has a 14 6 C activity of 0.0072 Bq per gram of carbon.

(a) Find the age of the sample, assuming that the activity per gram of carbon in a living organism has been constant at a value of 0.23 Bq.

(b) Evidence suggests that the value of 0.23 Bq might have been as much as 40% larger. Repeat part (a), taking into account this 40% increase.

Answer 1

Half life of carbon 14 is T = 5730 years

Decay constant = ln2/T

= 0.6931/ 5730

= 1.209 x10 ^-4 yr ^-1

Initial activity Ao = 0.23 Bq

present activity A = 0.0072 Bq

We know A = Ao e ^-t

0.0072 = 0.23 e ^-t

e ^-t =0.0072 / 0.23

= 0.0313

-t = ln0.0313

= -3.463

t = 3.463

t = 3.463 /(1.209 x10 ^-4 yr ^-1 )

= 28651.76 yrs

(d).Ao = 0.23Bq +40% of 0.23 Bq

= 0.322 Bq

A = Ao e ^-t

0.0072 = 0.322 e ^-t

   e ^-t =0.0072 / 0.332

= 0.0223

-t = ln0.0223

= -3.8

t = 3.8

t = 3.8 /(1.209 x10 ^-4 yr ^-1 )

= 31434.82 yrs