Question

Radioisotopes are often used in diagnostic imaging for detecting disease. The isotope F-18, which has a half-life of 110 min, is used in medical imaging. What percentage of the original acitivity in the sample remains after 300 min? Remember that half life is given by k=0.693/t1/2 ? Show your work.

Answer 1

Given:
Half life = 110 min
use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k
= 0.693/(half life)
= 0.693/(110)
= 6.3*10^-3 min-1
we have:
[F-18]o = 100 (Let initial activity be 100)
t = 300.0 min
k = 6.3*10^-3 min-1

use integrated rate law for 1st order reaction
ln[F-18] = ln[F-18]o - k*t
ln[F-18] = ln(100) - 6.3*10^-3*300
ln[F-18] = 4.6052 - 6.3*10^-3*300
ln[F-18] = 2.7152
[F-18] = 15.1

final activity is 15.1
This is out of 100
So, this is 15.1 %

Answer: 15.1 %