Sure! Let's work through this problem step-by-step.
1. Understanding Alpha Decay:
- In alpha decay, a nucleus emits an alpha particle, which is a helium-4 nucleus [tex]\(({}_{2}^{4}He)\)[/tex].
- This means that the nucleus loses 2 protons and 2 neutrons.
- Therefore, the atomic number of the nucleus decreases by 2, and the mass number decreases by 4.
2. Given Information:
- The product of the alpha decay is an isotope of Uranium: [tex]\({}_{92}^{235}U\)[/tex].
- We need to determine the original nuclide [tex]\(X\)[/tex].
3. Determine the Atomic and Mass Numbers of [tex]\(X\)[/tex]:
- Let the atomic number of [tex]\(X\)[/tex] be [tex]\(Z_X\)[/tex] and the mass number be [tex]\(A_X\)[/tex].
- Upon emitting an alpha particle ([tex]\({}_{2}^{4}He\)[/tex]), the resulting nucleus is [tex]\({}_{92}^{235}U\)[/tex].
Using the properties of alpha decay:
[tex]\[
\text{Atomic number of } X = \text{Atomic number of } U + 2 = 92 + 2 = 94
\][/tex]
[tex]\[
\text{Mass number of } X = \text{Mass number of } U + 4 = 235 + 4 = 239
\][/tex]
4. Identifying the Original Nuclide [tex]\(X\)[/tex]:
- [tex]\(X\)[/tex] must have an atomic number of 94 and a mass number of 239.
5. Matching with the Options:
1. [tex]\({}_{91}^{235}Fa\)[/tex] - Atomic number 91, Mass number 235 (This is incorrect as 91 ≠ 94 and 235 ≠ 239)
2. [tex]\({}_{93}^{235}Np\)[/tex] - Atomic number 93, Mass number 235 (This is incorrect as 93 ≠ 94 and 235 ≠ 239)
3. [tex]\({}_{92}^{236}\cup\)[/tex] - Atomic number 92, Mass number 236 (This is incorrect as 92 ≠ 94 and 236 ≠ 239)
4. [tex]\({}_{94}^{239}Pu\)[/tex] - Atomic number 94, Mass number 239 (This is correct as 94 = 94 and 239 = 239)
Therefore, the original nuclide [tex]\(X\)[/tex] is [tex]\({}_{94}^{239}Pu\)[/tex].
So, the correct answer is:
[tex]\[
{}_{94}^{239}Pu \quad \text{(Option 4)}
\][/tex]
