Question

In a living organism, a fixed fraction 1.30 x 10-12 of 12C is the radioactive isotope 14C, which has a half life of 5730 y. What is the activity of the 14C in .95g of 12C found in living tissue in Bq?

Answer 1

Half life T_1/2 = 5730 years = 5730 x 365 x24x60 x60 = 1.8*10^11

Decay constant = ln2/T_1/2 = 0.693 / T_{1/2 =} 3.85*10^-12 s^-1

Mass m = 0.95 g

Molecular weight of 12C is M = 12 g

Number of moles n = m/M = 0.95 /12 = 0.079 mol

Number of atoms in n moles N= 0.079* (6.023 x10^23) = 4.75 x10 ²²

Number of 14C atoms in 0.95 g of sample N ' = (0.95 x10 ^-12 )(4.75 x10^22)= 4.51x10^10

Required activity A = N '

                            =(3.85 x10^-12)* (4.51x10^10 )

                            = 0.173 Bq