Question

⁷⁹As decays by emitting a β^- particle. This decay process has a half-life of 9.00 min. If the initial mass of a pure sample of this isotope is 4.55 µg, determine the number of ⁷⁹As nuclei remaining after 9.950×10¹ min. Report your answer to three significant figures in scientific notation.

Answer 1

initial mass = 4.55µg = 4.55 x 10^-6 g

half life = 9 min

rate constant = 0.693 / 9 = 0.077 min-1

after time = t = 9.950 x 10^1 min =99.5 min

for first order

k = 1/t * ln (Ao / At)

0.077 = 1/ 99.5 * ln (4.55 x 10^-6 / At)

At = 2.14 x 10^-9

remaining mass = 2.14 x 10^-9 g

moles of 79As = 2.14 x 10^-9 / 79 = 2.71 x 10^-11

1 mole ------------------------> 6.023 x 10^23 nuclie

2.71 x 10^-11 mol -------------> 2.71 x 10^-11   x 6.023 x 10^23 nuclie = 1.63 x 10^13

remaining nuclie = 1.63 x 10^13