Question

1. An ancient club is found that contains 200 g of pure carbon and has an activity of 7.5 decays per second. Determine its age assuming that in living trees the ratio of (14C12C)(14C12C) atoms is about 1.30×10−121.30×10−12. Note that the half life of carbon-14 is 5700 years and the Avogadro number is 6.02×10236.02×1023.

Answer 1

Given data from the question t_1/2 = 5700 years Activity of 7.5 decays per second we have decay constant rightdoublearrow lambda = 0.693/t_1/2 = 0.693/5700 = 1.215789 times 10^- 4/years And Number of carbon atoms in 200 g N = [200/12 g/mol] (6.02 times 10^23) N = 1.0033 times 10^25 For 14 c nuclei in a sample form a living tree is N_14 = (1.30 times 10^- 12) (1.0033 times 10^25) N_14 = 1.30429 times 10^13 nuclei To calculate its age, we have XN = lambda N_14 e^- lambda t 7.5 decays = [1.215789 times 10^- 4/y/3.10 times 10^7 s/y] (1 - 30429 times 10^13) e^- lambda t 0.14945 = e^- 1.215789 times 10^- 4 t ln(0.14945) = -1.215789 times 10^- 4 xt t = ln (0.14945)/1.215789 times 10^- 4 t = 1.56 times 10^4 years