Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min. (A positron is a particle with the mass of an electron and a single unit of positive charge; the equation is (\({_{{\bf{518}}}^{\bf{9}}}{\bf{F}}\)⟶\({_{{\bf{18}}}^{\bf{8}}}{\bf{O + e - }}\).) Physicians use\(^{{\bf{18}}}{\bf{F}}\)to study the brain by injecting a quantity of fluoro-substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment.

(a) What is the rate constant for the decomposition of fluorine-18?

(b) If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h?

(c) How long does it take for 99.99% of the\(^{{\bf{18}}}{\bf{F}}\)to decay?

Intermolecular forces are the interaction which are formed by the attraction of the two having opposite charge (partial positive and partial negative charge). The opposite charge is formed by the presence of the electron-negative atom in the molecule. Due to the presence of electron-negative atom, there is and induced partial positive charge is generated on the electron-positive charge (or less electron-negative charge) atom. There will be an attraction between the both oppositely charges to form a bond.

Half-life period can be defined as the time period at which half concentration of the reactants gets converted into product.

The half-life of the first-order reaction is:

\(Half - life{\bf{ }}period{\bf{ }} = {\bf{ }}\frac{{\ln {\bf{ }}\left( 2 \right)}}{k}\)

The half-life period of the reaction does not depend upon the concentration of the reactant.

Half-life period of the \(^{{\bf{18}}}{\bf{F}}\) decay\({\bf{ = 109}}{\bf{.7min}}\)

\(\begin{align}Half - life{\bf{ }}period{\bf{ }} &= {\bf{ }}\frac{{0.693}}{k}\\109.7\min {\bf{ }} &= {\bf{ }}\frac{{0.693}}{k}\\k &= \frac{{0.693}}{{109.7\min }}\\k &= 0.0063{\min ^{ - 1}}\end{align}\)

\(\begin{align}Time{\bf{ }} &= {\bf{ }}5.59{\bf{ }}hour{\bf{ }}\\ &= {\bf{ }}335.4{\bf{ }}\ minutes\end{align}\)

\(\begin{align}k &= {\bf{ }}\frac{{2.303}}{t}Log{\bf{ }}\frac{{\left( A \right)}}{{{{\left( A \right)}_0}}}\\0.0063 &= {\bf{ }}\frac{{2.303}}{{335.4}}Log{\bf{ }}\frac{1}{{{{\left( A \right)}_0}}}\\0.0063 \times \frac{{335.4}}{{2.303}} &= {\bf{ }}Log{\bf{ }}\frac{1}{{{{\left( A \right)}_0}}}\\Log{\bf{ }}\frac{1}{{{{\left( A \right)}_0}}} &= 0.92\end{align}\)

\(\begin{align}{}\frac{1}{{{{\left( A \right)}_0}}} &= {\bf{ }}AntiLog(0.92)\\\frac{1}{{{{\left( A \right)}_0}}} &= 8.32\\{\left( A \right)_0} &= \frac{1}{{8.32}}\\{\left( A \right)_0} &= 0.12\end{align}\)

Radioactivity may be defined as the dis-integration of radioactive atom.

\(\begin{align}{\bf{The percentage of the radioactivity = 0}}{\bf{.12 \times 100 }}\\{\bf{ = 12\% }}\end{align}\)

The rate constant of the decay of \({\bf{ = 0}}{\bf{.0063mi}}{{\bf{n}}^{{\bf{ - 1}}}}\)

\(\begin{align}kt &= {\bf{ }}Ln{\bf{ }}\frac{{\left( A \right)}}{{{{\left( A \right)}_0}}}\\0.0063 \times t &= Ln{\bf{ }}\frac{{{{\left( A \right)}_0}}}{{0.0001{{\left( A \right)}_0}}}\\0.0063 \times t &= {\bf{ }}Log{\bf{ }}\frac{1}{{0.0001}}\\0.0063 \times t &= 9.210\end{align}\)

\(\begin{align}t &= \frac{{0.0063}}{{9.210}}\\t &= 1458\min \\t &= \frac{{1458}}{{3600}}\\t &= 24.29hour\end{align}\)

Therefore, it takes 24.29hour for the 99.99% of the \(^{{\bf{18}}}{\bf{F}}\) to decay.