Question

Radioactive Phosphorus-32 has a half-life of 14.3 days. If a dose with an activity of 126μCi is given to a patient, how much radioactivity is μCi will remain after 57.2 days?

Answer 1

Given:

Half life = 14.3 days

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

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(14.3)

= 4.846*10^-2 days-1

we have:

[A]o = 126 uCi

t = 57.2 days

k = 4.846*10^-2 days-1

use integrated rate law for 1st order reaction

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

ln[A] = ln(126) - 4.846*10^-2*57.2

ln[A] = 4.836 - 4.846*10^-2*57.2

ln[A] = 2.064

[A] = e^(2.064)

[A] = 7.88 uCi

Answer: 7.88 uCi