Question

If 140. mg of fluorine-18 is shipped at 8:00 A.M., how many milligrams of the radioisotope are still active when the sample arrives at the radiology laboratory at 1:30 P.M.

Answer 1

Answer:

To answer the question we need to know the half-life of fluorine for reference we have chlorine half-life as 110. The fraction of chlorine remaining would be given by $0.5^n$

Explanation:

Here we know and is known as the number of Half-life that have elapsed during this process  Now we are giving the time of start as 8 a.m. and time of finishing at 1:30 p.m., so the time between is 5 ½  hours. Which on converting in minutes will give 330 minutes.

So Half Life elapsed would be given by = $330/110=3$

Hence the amount remaining would be = $0.5 ^3 \times 125 =15.625$

Answer 2

Answer:

17.4 mg of the radioisotope are still active when the sample arrives at the radiology laboratory at 1:30 P.M.

Explanation:

Half-life of F-18 is found to be 109.7 minutes

Rate constant      

$k=\frac{0.693}{t_{\frac{1}{2}}}=0.00632$

t = time taken = 5 hours 30 minutes = 330 minutes

$\ln [A]=\ln [A]_{0}-k t$

[A] is the final quantity  

$[A]_0$  is the initial quantity

Plugging the values and solving for [A]

$\\ \ln [A]=\ln (140 m g)-\left(0.00632 \min ^{-1} \times 330 \min \right) \\\\ \ln [A]=4.942-2.085 \\\\ \ln [A]=2.857 \\\\ [A]=e^{2.857}$

[A] = 17.4mg is the Answer