Question

what is the activity in curies of a 2.05 g sample of nickel-69. it has a half life of 127 minutes?

Answer 1

First, we need to convert the 2.05 g sample of nickel-69 into numbers of atoms.

= (2.05 g / 69 g mol^-1) (6.022 x 10²³ atoms/mol) = 1.79 x 10²² atoms

Second, we use the rate equation to calculate the activity in atoms/second and then convert the answer into curies

r = k[N]

for that k = 0.693/t_1/2 = 0.693 / 127 min = 0.693 / (127 x 60 s) = 9.09 x 10 ^-5 s^-1

r = (9.09 x 10 -5 s-1)( 1.79 x 10²² atoms) = 16.27 x 10¹⁷ atoms decay/s

but, one curie = 3.700 x 10¹⁰ atoms that decay/second.

(16.27 x 10¹⁷ atoms/s)(1 Ci/3.700 x 10¹⁰ atoms/s) = 4.4 x 10 ⁷ Ci