Sure, let’s solve the problem step-by-step to determine the identity of the parent atom that undergoes radioactive decay.
1. Understand the Decay Process:
- The given equation indicates the product of the decay: [tex]\(_{54}^{129} \text{Xe}\)[/tex] (Xenon-129) and [tex]\(_{-1}^{0} e\)[/tex] (an electron).
- This type of decay is known as beta decay. In beta decay, a neutron is converted into a proton, and an electron (beta particle) is emitted. This increases the atomic number by 1 while the mass number remains the same.
2. Determine the Characteristics of the Parent Atom:
- Since the atomic number (Z) of Xenon (Xe) is 54 after the decay, the atomic number of the parent atom must have been [tex]\(54 - 1 = 53\)[/tex] before the decay.
- The mass number (A) remains the same during beta decay. Therefore, the mass number of the parent atom must have been 129.
3. Identify the Parent Atom:
- We then look for an element with an atomic number of 53 and a mass number of 129. This corresponds to Iodine (I), which has an atomic number of 53.
4. Verify against the Choices:
- Choice A: [tex]\(_{55}^{133} \text{Cs}\)[/tex]
- This element has an atomic number of 55 and a mass number of 133. Clearly, it cannot be the parent atom since both the atomic number and mass number do not match our requirements.
- Choice B: [tex]\(_{53}^{127} \text{I}\)[/tex]
- This element has an atomic number of 53 (correct) but a mass number of 127 (incorrect). Thus, this cannot be the parent atom.
- Choice C: [tex]\(_{55}^{129} \text{Cs}\)[/tex]
- This element has an atomic number of 55 (incorrect) and a mass number of 129 (correct). Therefore, this cannot be the parent atom considering the atomic number.
- Choice D: [tex]\(_{53}^{129} \text{I}\)[/tex]
- This element has an atomic number of 53 (correct) and a mass number of 129 (correct). This matches our requirements perfectly, making it the correct choice.
Therefore, the identity of the parent atom is:
D. [tex]\(_{53}^{129} \text{I}\)[/tex]
