Charcoal prepared from a living tree exhibits a disintegration rate of 15.3 disintegrations of ¹⁴ C per gram per minute. An ancient charcoal sample has a ¹⁴ C activity of 12.3 disintegrations per gram per minute. Determine the age of this sample given that the half-life of ¹⁴ C is 5730 y .

Step-by-step solution

The radioactive decay law is given by: t = 2.303 ( A₀ A ) The decay constant is: = 0.693 t_ 1/2 = 0.693 5730 y ⁻¹ Given: A₀ = 15.3 dis min ⁻¹ g ⁻¹ A = 12.3 dis min ⁻¹ g ⁻¹ Substituting the values into the decay equation: t = 2.303 5730 0.693 ( 15.3 12.3 ) t = 13196.19 0.693 (1.2439) t = 19042.1 0.0948 t 1805 y