Let's solve this step-by-step:
1. Understanding Alpha Decay:
- Alpha decay involves the emission of an alpha particle, which is a helium nucleus [tex]\(_{2}^{4}He\)[/tex]. This particle consists of 2 protons and 2 neutrons.
- When a nuclide undergoes alpha decay, its atomic number decreases by 2 (since it loses 2 protons), and its mass number decreases by 4 (since it loses 2 protons and 2 neutrons).
2. Isotope Produced After Decay:
- The problem states that the isotope produced after the alpha decay is [tex]\(_{92}^{235}\)[/tex]:
- Atomic number (Z) = 92
- Mass number (A) = 235
3. Determining the Original Nuclide [tex]\(X\)[/tex]:
- To determine the original nuclide [tex]\(X\)[/tex], we need to revert the changes that occurred during the alpha decay.
- Since the atomic number decreases by 2 during the decay, the original nuclide must have had an atomic number 2 units greater than the produced isotope. Therefore, the atomic number of [tex]\(X\)[/tex] before the decay was:
[tex]\[
Z_{X} = 92 + 2 = 94
\][/tex]
- Similarly, since the mass number decreases by 4 during the decay, the original nuclide must have had a mass number 4 units greater than the produced isotope. Therefore, the mass number of [tex]\(X\)[/tex] before the decay was:
[tex]\[
A_{X} = 235 + 4 = 239
\][/tex]
4. Identifying the Nuclide [tex]\(X\)[/tex]:
- The nuclide [tex]\(X\)[/tex] must have an atomic number of 94 and a mass number of 239.
- The element with atomic number 94 is Plutonium (Pu).
- Therefore, the original nuclide [tex]\(X\)[/tex] is [tex]\(_{94}^{239}\)[/tex] Pu.
5. Conclusion:
- The correct answer is [tex]\(_{94}^{239}\)[/tex] Pu.
Thus, the nuclide [tex]\(X\)[/tex] is [tex]\(\boxed{_{94}^{239} \text{Pu}}\)[/tex].
