91Ó°ÊÓ

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.

The characteristics of isotopes of carbon are examined in this paper.

\(^{{\rm{131}}}{\rm{I}}\)and \(^{{\rm{123}}}{\rm{I}}\)used in thyroid scans and we want to determine their masses.

We are provided all of the information we require in the problem statement.

isotopes, \(^{{\rm{131}}}{\rm{I}}\) and \(^{{\rm{123}}}{\rm{I}}\), their activity of \({\rm{50}}\,{\rm{\mu Ci}}\)and \({\rm{70}}\,{\rm{\mu Ci}}\), respectively and their half-lives of \({\rm{8}}{\rm{.04\;d}}\) \(\left( {{\rm{694656 s}}} \right)\)and \(13.2\;{\rm{h}}\) \(\left( {{\rm{1140480 s}}} \right){\rm{.}}\)

To compute the masses, we must first recall the activity's expression:

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

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

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

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

So we can now convert expressed activity from Curie to Becquerel for both isotopes.

\(^{{\rm{131}}}\)I we have:

\(\frac{{50 \times {{10}^{ - 6}}Ci}}{{2.70 \times {{10}^{ - 11}}}} = 1.85 \times {10^6}\;Bq\)And for \(^{{\rm{123}}}\)I we have:

\(\frac{{70 \times {{10}^{ - 6}}\,Ci}}{{2.70 \times {{10}^{ - 11}}}} = 2.59 \times {10^6}\;\,Bq\)

And putting in the numbers for both isotopes we get:

\(\begin{aligned} {N_{{}^{131}I}} &= \frac{{R\left( {{t_{1/2}}} \right)}}{{0.693}}\\ &= 1.85 \times {10^{12}}\end{aligned}\)

and for the other one we have:

\(\begin{aligned} {N_{{}^{123}I}} &= \frac{{R\left( {{t_{1/2}}} \right)}}{{0.693}}\\ &= 4.26 \times {10^{12}}\end{aligned}\)

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:

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

\(\begin{aligned} {m_{{}^{131}I}} &= \left( {1.85 \times {{10}^{12}}} \right)\left( {\frac{{mol}}{{6.02 \times {{10}^{23}}}}} \right)\left( {131\frac{g}{{mol}}} \right)\\ &= 4.03 \times {10^{ - 10}}\;g\end{aligned}\)

And for the other isotope:

\(\begin{aligned} {m_{{}^{123}I}} &= \left( {4.26 \times {{10}^{12}}} \right)\left( {\frac{{mol}}{{6.02 \times {{10}^{23}}}}} \right)\left( {123\frac{g}{{mol}}} \right)\\ &= 8.70 \times {10^{ - 10}}\;g\end{aligned}\)

Hence, the masses are so minute, on the scale of a few tens of Nanograms, it must be combined with other stable iodine to guarantee proper dispersion throughout the body.