Question

An ancient club is found that contains 100 g of pure carbon and has an activity of 6 decays per second. Determine its age assuming that in living trees the ratio of ¹⁴C/¹²C atoms is about 1.10 × 10^-12. Note that the half life of carbon-14 is 5700 years and the Avogadro number is 6.02 × 10²³.

Answer 1

First, calculate the decay constant for 14C, which has a half-life of 5700 years.
decay constant=In2/ T1/2
=In2/ (5700 years)
=1.2160x10^-4 y^-1

The number of 14C nuclei can be calculated in two steps.
First, the number of 12C nuclei in 100 g of carbon is
N(12C)= (6.02 x 10^23 nuclei/mol) x (100g)/ 12.0 g/mol
=5.0166x10^24 nuclei


Ratio of 14C to 12C is 1.2 x 10^-12

N<original>(14C) = (1.2 x 10^-12) (5.01666x10^24)
= 6.02x 10^12 nuclei

Therefore, the initial activity of the sample is
R<o>= decay constant x N<o>
=(1.216x10^-4 y^-1)(6.02x 10^12 nuclei)
= 9.5164x10^8 decays/y

Now calculate the age of the ancient club using Equation R=(R<o>)(e^(-decay constant x time), which relates the activity R at any time t to the initial activity R<o>:
R = R<o>e^(-decay constant x t)
R/ R<o> =e^(-decay constant x t)

Given that R=6(3.1526x10^7) decays/y.
We found that R<o> = 9.5164x10^8 decays/y.
Calculate t by taking the natural logarithm of both sides of the last equation:

ln(R/R<o>)=In e^(-decay constant x t)
-decay constant x t = ln(R/R<o>)
-1.216x10^-4 t= -1.6156
t=13286.18 years