Write the nuclear equation for the release of a positron by \(_{54}^{117} \mathrm{Xe} .\)

Short Answer

Expert verified
\(_{54}^{117} \mathrm{Xe} \rightarrow _{1}^{0}\mathrm{e} + _{53}^{117}\mathrm{I}\)

Step by step solution

01

Identify the initial atom

We have an atom of Xenon (\(_{54}^{117} \mathrm{Xe}\)). Xenon-117 means that the atomic number (number of protons) is 54 (this is the atomic number of Xenon) and the nucleon number (sum of protons and neutrons) is 117.
02

Understand the decay process

In positron emission, a proton in the atom's nucleus is changed into a neutron, with the emission of a positron (\(_{1}^{0}\mathrm{e}\)) and a neutrino (not included in the nuclear equation as it has negligible mass and no charge). This will result in change of atomic and nucleon number of the atom.
03

Apply Conservation Principle

In a nuclear reaction, both nucleon number and atomic number are conserved. That is the sum of atomic numbers (bottom number) and sum of nucleon numbers (top number) on the left side of the equation must equate to their sums on the right side of the equation. Our original atom has 54 protons, and after emitting a positron, it will have 53 protons. The nucleon number doesn't change because a neutron replaces the lost proton. This means the resulting atom will be Iodine (\(_{53}^{117}\mathrm{I}\)). So our nuclear equation becomes :\(_{54}^{117} \mathrm{Xe} \rightarrow _{1}^{0}\mathrm{e} + _{53}^{117}\mathrm{I}\)

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