Question

A sample of ${}^{18}F$ is used internally as a medical diagnostic tool to look for the effects of the positron decay $(T_{\frac{1}{2}}=110\text{min})$ . 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
$-ln100=-\frac{ln2\times t}{T_{\frac{1}{2}}}$
Which on solving for $\text{t}$ yields
$\therefore$ $t=\frac{ln100}{ln2}\times T_{\frac{1}{2}}$
$=\frac{log100}{log2}\times T_{\frac{1}{2}}$
$=\frac{2}{0.3010}\times 110$
$=731min=12.2h$