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 %

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