Question

Question:Technetium has not been found in nature. It can be readily obtained readily as a product of uranium fission in nuclear power plants, however, and is now produced in quantities of many kilograms per year. One medical use relies on the tendency of ${}^{\mathbf{99}\mathbf{m}}\mathbf{Tc}$(an excited state of ${}^{\mathbf{99}}\mathbf{Tc}$) to concentrate in abnormal heart tissue. Calculate the total activity (in disintegrations per second) caused by the decay of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu g}$of role="math" localid="1660796739343" ${}^{\mathbf{99}\mathbf{m}}\mathbf{Tc}$, which has a half-life of $\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{hour}$.

Short answer

The total activity caused by decay of${}^{99\mathrm{m}}\mathrm{Tc}$ is $1.92\times {10}^{11}$ disintegrations per second.

Step-by-step solution

Decay constant

The half-life of a radioactive substance is the time required by the substance to decay to half its initial amount. We can calculate the decay constant from half-life with the following expression:

$\begin{array}{l}{\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{k}}\\ \mathrm{k}=\frac{0.693}{6.0}{\mathrm{hr}}^{-1}\\ =0.1155{\mathrm{hr}}^{-1}\end{array}$

 Total activity of Tc99m

Number of atoms in $1.0\mathrm{\mu g}{}^{99\mathrm{m}}\mathrm{Tc}$$=\frac{1\times {10}^{-6}\mathrm{g}}{99\mathrm{g}/\mathrm{mol}}\times \frac{6.023\times {10}^{23}\mathrm{atoms}}{1\mathrm{mol}}=0.06\times {10}^{17}\mathrm{atoms}$

The total activity can be calculated as:

$$\begin{gathered} \mathrm{A}=\mathrm{kN} \\ =\frac{0.1155}{1\mathrm{hr}}\times \frac{1\mathrm{hr}}{3600\mathrm{s}}\times 0.06\times {10}^{17}\mathrm{atoms} \\ =1.92\times {10}^{11}\mathrm{disintegrations}\mathrm{per}\mathrm{second} \end{gathered}$$