The half-life for the radioactive decay of calcium-47 is 4.5 d. If a sample has an activity of 1.9 μCi after 27 d, what was the initial activity, in microcuries, of the sample?

Question

The half-life for the radioactive decay of calcium-47 is 4.5 d. If a sample has an activity of 1.9 μCi after 27 d, what was the initial activity, in microcuries, of the sample?

Answer 1

The easiest way to do this is to use the simplified equation for a first order decay. That is...

Fraction remaining = 0.5ⁿ where n is the number of half lives that have elapsed.

Fraction remaining = ?

n = 27 days x 1 half life/4.5 days = 6 half lives have elapsed. Now we find fraction remaining

Fraction remaining = 0.5⁶

Fraction remaining = 0.015625

This equates to 1.9 uCi so we can now solve for the initial amount present (in micro curies).

0.015625 x = 1.9 uCi

x = 121.6 uCi (or 120 uCi to 2 significant figures)