Question

Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h, respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)?

Short answer

The percent of the radioactivity would remain of the technetium-99 (Tc-99) and thallium-201 (Th-201) for each of the isotopes after 2 days are 0.39% and 63.4%.

Step-by-step solution

Reaction Rate

Reaction involved the effective collision of two reactants to produce the desired products. Reactions can be natural which are occurring in surrounding environment whereas it can be artificially done in the laboratory to form a desired required product.

The reaction rate can be defined as the speed of reaction to produce the products. The reaction rate can be slow, fast or moderate. The reaction can take a lesser than millisecond to produce products or it can take years to produce a desired product.

\({\bf{Rate = K }}{\left( {\bf{A}} \right)^{\bf{n}}}\)

A = concentration of reactant.

K = Rate constant.

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

Radioactivity can be defined as the radiation produced by the dis-integration of the ionizing unstable atomic nucleus.

Explanation                                                    

The half-life of the both technetium-99 and thallium-201

\(\begin{align}{t_{1/2}}\;for\;{T_c} - 99 &= 6hr\\{t_{1/2}}\;for\;{T_h} - 201 &= 73hr\end{align}\)

For two days of radioactivity of the compound,

\({\bf{2\;days\; = 2 \times 24\;hrs = 48\;hrs}}\) \(\)

The radioactivity percentage fortechnetium-99 (Tc-99):

\(\begin{align}{A_t} &= {A_o}{\left( {\frac{1}{2}} \right)^{t/{t_{1/2}}}}\\\frac{{{A_o}}}{{{\bf{ }}{A_t}}} &= {\left( {\frac{1}{2}} \right)^{48/6}}\\\frac{{{A_o}}}{{{\bf{ }}{A_t}}} &= {\left( {\frac{1}{2}} \right)^8}\\\frac{{{A_o}}}{{{\bf{ }}{A_t}}} &= \frac{1}{{256}}\end{align}\)

\(\begin{align}\% \;remained\; &= \frac{{{A_o}}}{{{\bf{ }}{A_t}}} \times 100\\\% \;remained &= {\bf{ }}\frac{1}{{256}} \times 100\\\% \;remained &= {\bf{ }}0.39\% \end{align}\)

The radioactivity percentage forthallium-201 (Th-201):

\(\begin{align}{A_t} &= {A_o}{\left( {\frac{1}{2}} \right)^{t/{t_{1/2}}}}\\\frac{{{A_o}}}{{{\bf{ }}{A_t}}} &= {\left( {\frac{1}{2}} \right)^{48/73}}\\\frac{{{A_o}}}{{{\bf{ }}{A_t}}} &= {\left( {\frac{1}{2}} \right)^{0.66}}\\\frac{{{A_o}}}{{{\bf{ }}{A_t}}} &= 0.634\end{align}\)

\(\begin{align}\% \;remained\; &= \frac{{{A_o}}}{{{\bf{ }}{A_t}}} \times 100\\\% \;remained &= {\bf{ }}0.634 \times 100\\\% \;remained &= {\bf{ }}63.4\% \end{align}\)

The percent of the radioactivity would remain of the technetium-99 (Tc-99) and the

allium-201 (Th-201) for each of the isotopes after 2 days are 0.39% and 63.4%.