Answer:
[tex]\large \boxed{\text{19.9 mg}}[/tex]
Explanation:
Two important equations in radioactive decay are
[tex](1) \,t_{\frac{1}{2}} = \dfrac{\ln2}{k }\\\\(2) \,\ln \dfrac{N_{0} }{N_{t}} = kt[/tex]
We can use them to solve this problem.
1. Calculate the radioactive decay constant
The half-life of ¹⁸F is 109.8 min.
[tex]\begin{array}{rcl}t_{\frac{1}{2}}& = &\dfrac{\ln2}{k }\\\\k& = &\dfrac{\ln2}{t_{\frac{1}{2}}}\\\\ & = & \dfrac{\ln2}{\text{109.8 min}}\\\\ & = & 6.313 \times 10^{-3}\text{ min}^{-1}\\\end{array}[/tex]
(ii) Calculate the mass of ¹⁸F remaining
Time in transit = 13:30 - 08:00 = 05:30 h = 5.50 h = 330 min
[tex]\begin{array}{rcl}\ln \dfrac{N_{0} }{N_{t}} & = & kt\\\\\ln \dfrac{\text{160. mg}}{ N_{t}} & = & 6.313 \times 10^{-3}\text{ min}^{-1} \times \text{330 min}\\\\& = & 2.083\\\dfrac{\text{160. mg}}{ N_{t}} & = & e^{2.083}\\\\& = & 8.030\\\text{160. mg} & = & 8.030N_{t}\\N_{t} & = & \dfrac{\text{160. mg}}{8.030}\\\\& = & \textbf{19.9 mg}\\\end{array}\\\text{The mass of the sample is $\large \boxed{\textbf{19.9 mg}}$}[/tex]
