Question

A sample of $^{18} F$ is used internally as a medical diagnostic tool to look for the effects of the positron decay $\left(\right. T_{\frac{1}{2}} = 110 \, \text{min} \left.\right)$ . How long does it take for $\text{99\%}$ of the $^{18}F$ to decay?

• A. $\text{12} . \text{4 h}$
• B. $\text{12} .0 \text{ h}$
• C. $\text{12} . \text{2 h}$
• D. $\text{12} . \text{5 h}$

Answer 1

Answer: (C)

Radioactive decay equation is
$N=N_{0}e^{- \lambda t}=N_{0}e^{- ln \left(2\right) \frac{t}{T_{\frac{1}{2}}}}$
After decay of 99% of the initial sample only 1% will be left, and $\frac{N}{N_{0}}=1\%$ . We then have
$\frac{N}{N_{0}}=\frac{1}{100}=e^{- ln \left(2\right) \frac{t}{T_{\frac{1}{2}}}}$
If we take the natural logarithm, we have
$-ln 100=-\frac{ln ⁡ 2 \times t}{T_{\frac{1}{2}}}$
Which on solving for $\text{t}$ yields
$\therefore $ $t=\frac{ln 100}{ln ⁡ 2}\times T_{\frac{1}{2}}$
$=\frac{log 100}{log ⁡ 2}\times T_{\frac{1}{2}}$
$=\frac{2}{0.3010}\times 110$
$=731min = 12.2 \, h$