Respuesta :
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 [tex]0.5^n[/tex]
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 = [tex]330/110=3[/tex]
Hence the amount remaining would be = [tex]0.5 ^3 \times 125 =15.625[/tex]
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
[tex]$k=\frac{0.693}{t_{\frac{1}{2}}}=0.00632$[/tex]
t = time taken = 5 hours 30 minutes = 330 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 (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}$[/tex]
[A] = 17.4mg is the Answer
