To identify the product nuclide for the given nuclear decay reaction, it is important to understand the process of electron capture. Electron capture is a process in which an inner shell electron is captured by the nucleus of its own atom. This process typically decreases the atomic number (Z) of the element by 1 while keeping the mass number (A) the same.
Here is a step-by-step solution for the given reaction:
1. Identify the initial nuclide and the particle involved:
- The initial nuclide is iodine ([tex]\({ }_{53}^{133} I\)[/tex]).
- The particle involved in the reaction is an electron ([tex]\({ }_{-1}^0 e\)[/tex]).
2. Understand the electron capture process:
- During electron capture, an electron combines with a proton in the nucleus to form a neutron. This process decreases the atomic number by 1 but the mass number remains unchanged.
3. Determine the atomic number and mass number of the product nuclide:
- The initial atomic number (Z) of iodine is 53.
- The mass number (A) remains the same, which is 133.
- The atomic number of the product nuclide after losing one proton (due to electron capture) is [tex]\( 53 - 1 = 52 \)[/tex].
4. Identify the element with atomic number 52:
- The element with atomic number 52 is tellurium (Te).
5. Combine the mass number and the symbol for the nuclide:
- The mass number remains 133 and the symbol for tellurium is Te.
Therefore, the product nuclide is [tex]\({ }_{52}^{133} Te\)[/tex].
Given the options:
- [tex]\({ }_{54}^{133} Xe\)[/tex]
- [tex]\({ }_{52}^{133} Te\)[/tex]
- [tex]\({ }_{52}^{130} Te\)[/tex]
- [tex]\({ }_{54}^{132} Xe\)[/tex]
The correct answer is:
[tex]\[
{ }_{52}^{133} Te
\][/tex]
