Question

The half-life for the radioactive decay of calcium-47 is 4.5 d. if a sample has an activity of 4.5 Ci after 9.0 d, what was the initial activity of the sample?

Answer 1

Given:

Half life = 4.5 days

use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(4.5)

= 0.154 days-1

we have:

[A] = 4.5 Ci

t = 9.0 days

k = 0.154 days-1

use integrated rate law for 1st order reaction

ln[A] = ln[A]o - k*t

ln(4.5) = ln[A]o - 0.154*9

ln[A]o = 1.504 + 0.154*9

ln[A]o = 2.89

[A]o = e^(2.89)

[A]o = 18.0 Ci

Answer: 18 Ci