Question

Follow the decay of parent 87Rb and growth of daughter 87Sr in a granite sample over the course of six half-lives. Assume that the granite initially contains 1.2 x 10^20 atoms of 87Rb and 0.3 x 10^20 atoms of 87Sr. The half-life of 87Rb is 48.8x10^9 years. Show the results graphically on the graph paper below.

Answer 1

Ans:

We know that all radioactive decay follows first order kinetics. Accordingly, we have

N_t = N_o . e^(-λt)

where, Nt = current no. of radioactive isotopes (⁸⁷Rb)

               N_o = no of radioactive isotopes present initially    (⁸⁷Rb)

               t = time of decay for (⁸⁷Rb)

t½ = Half-life of (⁸⁷Rb) = 48.8 x 10⁹ years

λ = decay constant of (⁸⁷Rb) = 0.693/t½ = 0.693/(48.8 x 10⁹) = 1.42 x 10^-11 years^-1

1. We know that in a first order reaction after every half-life, the no. of radioactive isotopes reduces to half of the initial value.
2. Also, we know that one ⁸⁷Rb after decay gives one ⁸⁷Sr.

	87Rb atoms	87Sr atoms
Initial	1.2 x 1020	0.3 x 1020
After 1 half life	0.6 x 1020	(0.3 x 1020) + (0.6 x 1020) = 0.9 x 1020
After 2 half life	0.3 x 1020	(0.3 x 1020) + (0.6 x 1020) + (0.3 x 1020) = 1.2 x 1020
After 3 half life	0.15 x 1020	(0.3 x 1020) + (0.6 x 1020) + (0.3 x 1020) + (0.15 x 1020) = 1.35 x 1020
After 4 half life	0.075 x 1020	(0.3 x 1020) + (0.6 x 1020) + (0.3 x 1020) + (0.15 x 1020) + (0.075 x 1020) = 1.425 x 1020
After 5 half life	0.0375 x 1020	(0.3 x 1020) + (0.6 x 1020) + (0.3 x 1020) + (0.15 x 1020) + (0.075 x 1020) + (0.0375 x 1020) = 1.4625 x 1020
After 6 half life	0.01875 x 1020	(0.3 x 1020) + (0.6 x 1020) + (0.3 x 1020) + (0.15 x 1020) + (0.075 x 1020) + (0.0375 x 1020) + (0.01875 x 1020) = 1.48125 x 1020