Answer:
54.9 mg of the radioisotope are still active after 110 min.
Explanation:
Half-life of F-18 is found to be 109.7 minutes
Rate constant
[tex]$k=\frac{0.693}{t_{\frac{1}{2}}}=0.00632$[/tex]
t = time taken = 110 minutes
[tex]$\ln [A]=\ln [A]_{0}-k t$[/tex]
[A] is the final quantity
[tex][A]_0[/tex] is the initial quantity
Plugging the values and solving for [A]
[tex]\\$\ln [A]=\ln (110 m g)-\left(0.00632 \min ^{-1} \times 110 \min \right)$\\\\$\ln [A]=4.700-0.6952$\\\\$\ln [A]=4.0048$\\\\$[A]=e^{4.0048}$[/tex]
[A] = 54.9 mg is the Answer
