Question

An ancient club is found that contains 130 g of pure carbon and has an activity of 6.5 decays per second. Determine its age assuming that in living trees the ratio of (14 C /12 C ) atoms is about 1.20 × 10^ − 12 . Note that the half life of carbon-14 is 5700 years and the Avogadro number is 6.02 × 10^23 .

Answer 1

Given,

mass of carbon, m = 130 g

Ratio of (C14/C12) = 1.20 * 10^-12

Half life of C-14 , t_1/2 = 5700 years

Now,

Activity, A = A₀*e^(t/k)

=> (A/A₀) = e^-t/k

at t = 5700 years, (A/A₀) = 0.5

=> 0.5 = e^-t/k

Taking natural log both sides

=> ln(0.5) = -5700/k

=> -0.693 = -5700/k

=> k = 5700/0.693

         = 8225.11 years

1 year = (365*24*60*60) seconds

=> k = 8225.11*365*24*60*60

         = 2.59 * 10¹¹ s

Now,

Number of moles of C-12, n = 130/12

                                              = 10.83 moles

Number of atoms of C-12, N = 10.83*6.02*10²³

                                                = 6.52*10²⁴ atoms

Now,

Number of C-14 atoms, N' = 1.20 * 10^-12 * 6.52*10²⁴

                                            = 7.824 * 10¹²

Now,

Initial decay, d = (Number of C-14 atoms)/k

                         = (7.824 * 10¹² )/(2.59 * 10¹¹)

                         = 30.21 decays/sec

Given, decay per second is 6.5 decays per second

So,

fractional of original activity remaining = (6.5/30.21)

                                                                 = 0.215

Now,

We know that,

=> A = A₀*e^-(t/k)

=> 0.215 = 1*e^-(t/8225.11)

Taking natural log,

=> ln(0.215) = -t/8225.11

=> -1.54 = -t/8225.11

=> t = 8225.11*1.54

         = 12666.67 years