In a living organism, a fixed fraction 1.30 × 10-12 of 12c is the radioactive isotope 14c, which has a half life of 5730 y.

A = 0.325 Bq

Explanation:

given,

half life of ¹⁴C = 5730 years

fixed fraction 1.30 × 10⁻¹² of ¹²C

half life = 5730 years

T_{1/2} = 5730 x 365 x 24 x 60 x 60

            = 1.807 x 10¹¹ s

radioactive decay constant

 λ =

 λ =

λ = 3.835 x 10⁻¹² /s

number of atom

mass m = 1.30 g

       n = 6.524 x 10²²

Number of ¹⁴C atoms in 1.3 g of sample N ' =

= 1.30 × 10⁻¹²  x 6.524 x 10²²

= 8.482 x 10¹⁰

Required activity, A =  λ N

A =   3.835 x 10⁻¹² x 8.482 x 10¹⁰

A = 0.325 Bq

the answer would be c