Question

The following isotope has a half-life of 5,730 years. If we have 8.7×10-4 μμg of sample, how many years will pass until only 29% of the original sample remains. The decay follows first-order kinetics.

Answer 1

Sol :- Given initial concentration of sample say [A]₀ = 8.7 x 10^-4 μμg = 8.7 x 10^-16 g

because 1 μμg = 10^-12 g = 1 pm

Concentration after time "t" say [A]_t = 29 % of 8.7 x 10^-16 g = 2.523 x 10^-16 g

Half life period = t_1/2 = 5730 years

For first order kinetics

t_1/2 = 0.693 / k

and

k = 0.693 / 5730 years = 1.209 x 10^-4 years^-1

also

From Integrated rate equation for first order reaction , we have

t = 2.303/k log [A]₀ / [A]_t

t = 2.303 / 1.209 x 10^-4 years^-1 log 8.7 x 10^-16 g / 2.523 x 10^-16 g

t = 19048.8 years ( log 3.448275862)

t = 19048.8 years ( 0.5376)

t = 10240.6 years