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Iodine in the body is preferentially taken up by the thyroid gland. Therefore, radioactive iodine in small doses is used to image the thyroid and in large doses is used to kill thyroid cells to treat some types of cancer or thyroid disease. The iodine isotopes used have relatively short half-lives, so they must be produced in a nuclear reactor or accelerator. One isotope frequently used for imaging is ${}^{\mathbf{123}}\mathit{I}$; it has a half-life of 13.2 h and emits a 0.16-MeV gamma-ray photon. One method of producing ${}^{\mathbf{123}}\mathit{I}$is in the nuclear reaction ${}^{\mathbf{123}}\mathit{T}\mathit{e}\mathbf{+}\mathit{p}\mathbf{→}{}^{\mathbf{123}}\mathit{I}\mathbf{+}\mathit{n}$. The atomic masses relevant to this reaction are ${}^{\mathbf{123}}\mathit{T}\mathit{e}$, 122.904270 u; ${}^{\mathbf{123}}\mathit{I}$, 122.905589 u; n, 1.008665 u; and ${}^{\mathbf{1}}\mathit{H}$, 1.007825 u. The iodine isotope commonly used for treatment of disease is ${}^{\mathbf{131}}\mathit{I}$, which is produced by irradiating ${}^{\mathbf{130}}\mathit{T}\mathit{e}$in a nuclear reactor to form ${}^{\mathbf{131}}\mathit{T}\mathit{e}$. The ${}^{\mathbf{131}}\mathit{T}\mathit{e}$then decays to ${}^{\mathbf{131}}\mathit{I}$. ${}^{\mathbf{131}}\mathit{I}$undergoes beta- decay with a half-life of 8.04 d, emitting electrons with energies up to 0.61 MeV and gamma-ray photons of energy 0.36 MeV. A typical thyroid cancer treatment might involve administration of 3.7 GBq of ${}^{\mathbf{131}}\mathit{I}$.

Which reaction produces ${}^{\mathbf{131}}\mathit{T}\mathit{e}$ in the nuclear reactor?

(a) ${}^{\mathbf{130}}\mathit{T}\mathit{e}\mathbf{+}\mathit{n}\mathbf{→}{}^{\mathbf{131}}\mathit{T}\mathit{e}$;

(b) ${}^{\mathbf{130}}\mathit{I}\mathbf{+}\mathit{n}\mathbf{→}{}^{\mathbf{131}}\mathit{T}\mathit{e}$;

(c) ${}^{\mathbf{132}}\mathit{T}\mathit{e}\mathbf{+}\mathit{n}\mathbf{→}{}^{\mathbf{131}}\mathit{T}\mathit{e}$;

(d) ${}^{\mathbf{132}}I\mathbf{+}\mathit{n}\mathbf{→}{}^{\mathbf{131}}\mathit{T}\mathit{e}$.

Step by step solution

01

Conservation of atomic number

First, we write the reaction equation with the unknown particle ${}_Z{}^{A}X$follows

${}_{52}{}^{130}Te+{}_Z{}^{A}X→{}_{52}{}^{131}Te$

From the conservation of atomic number and the number of nucleons, we get

$$\begin{gathered} 52+Z→52 \\ Z=0 \\ 130+A→131 \\ A=1 \end{gathered}$$

The particle with Z = 0 and A = 1 is the neutron.

02

Conclusion

Therefore, the answer is : ${}_{52}{}^{130}Te+n→{}_{52}{}^{131}Te$.

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