A piece of burnt wood of mass 20 g is found to have a 14 C activity of 4 decay s - 1 . How long has the tree that this wood belonged to be dead? Given T 1 2 o f 14 C = 5730 year.

Step-by-step solution

The decay constant of 14 C is λ = 0.693 T 1 2 = 0.693 5730   × 3.16   × 10 7 = 3.83   × 10 - 12   s - 1 ∵       1   year = 3 .17   × 10 7 s To find the number of 14 C nuclei in 20 g of burnt wood, we first calculate the number of 12 C nuclei in 20 g of carbon (burnt wood). Thus N 12 C = 6.02   × 10 23 12   × 20   ≃ 10 24 Now, assuming that the ratio of 14 C   t o   12 C   i s   1.3   &