Table 32.1 indicates that \(7.50\,mCi\)of \({}^{99m}{\rm{Tc}}\) is used in a brain scan. What is the mass of technetium?

The spontaneous emission of radiation in the form of particles or high-energy photons as a result of a nuclear process is known as radioactivity.

We are given the activity of technetium utilised for the brain scan in this issue, and we must estimate the mass of the technetium.

Where \({\rm{N}}\) is the exact number of atoms that will allow us to compute the element's mass.

\(R = \frac{{0.693N}}{{{t_{1/2}}}}\)

First, we must remember the relationship between Curie and Becquerel radioactive units, which is:

\(1\;\,Bq = 2.70 \times {10^{ - 11}}\,Ci\)

Following this, now we can convert \({\rm{7}}{\rm{.50mCi}}\)to Becquerel’s, so we have:

\(\frac{{7.50 \times {{10}^{ - 3}}Ci}}{{2.70 \times {{10}^{ - 11}}}} = 2.778 \times {10^8}\,\;Bq\)

Now, using the first equation to represent the number of atoms, we get:

\(N = \frac{{R\left( {{t_{1/2}}} \right)}}{{0.693}}\)

Now, we can observe from the book's Appendix b that technetium has a half-life of \({\rm{6}}{\rm{.02}}\)hours.

\(N = \frac{{R\left( {{t_{1/2}}} \right)}}{{0.693}}\)

Therefore, the result is:

\(N = 8.66 \times {10^{12}}\)

Now, we need to know Avogadro number and molar mass in order to appropriately link the number of atoms to the amount of the material, thus we have the following relationship:

\(N = \frac{m}{M}{N_A}\)

or expressing only the mass we got:

\(m = \frac{{MN}}{{{N_A}}}\)

Now, let's put all of the numbers together (remember, molar mass is mass per mole, so 99 gram per mol):

\(m = \left( {8.696 \times {{10}^{12}}} \right)\left( {\frac{{mol}}{{6.02 \times {{10}^{23}}}}} \right)\left( {99\frac{g}{{mol}}} \right)\)

Therefore, the mass of technetium is \(m = 1.43 \times {10^{ - 9}}\;g\).