To determine the nuclide symbol of [tex]\( X \)[/tex] in the nuclear equation:
[tex]\[ X \longrightarrow { }_{89}^{228} \text{AC} + { }_{-1}^0 \beta \][/tex]
we will use the principles of conservation of mass number and atomic number.
### Step-by-Step Solution:
1. Conservation of Mass Number:
- The mass number (A) is the total number of protons and neutrons in the nucleus of an atom.
- In a nuclear reaction, the mass numbers on both sides of the equation must balance.
- Let [tex]\( A_X \)[/tex] be the mass number of [tex]\( X \)[/tex].
- The mass number of [tex]\( \text{Ac} \)[/tex] is 228.
- The mass number of [tex]\( \beta \)[/tex] (a beta particle) is 0.
According to the conservation of mass number:
[tex]\[
A_X = 228 + 0
\][/tex]
[tex]\[
A_X = 228
\][/tex]
2. Conservation of Atomic Number:
- The atomic number (Z) is the number of protons in the nucleus of an atom, which determines the element.
- In a nuclear reaction, the atomic numbers on both sides of the equation must balance.
- Let [tex]\( Z_X \)[/tex] be the atomic number of [tex]\( X \)[/tex].
- The atomic number of [tex]\( \text{Ac} \)[/tex] (Actinium) is 89.
- The atomic number of [tex]\( \beta \)[/tex] (a beta particle) is -1.
According to the conservation of atomic number:
[tex]\[
Z_X = 89 + (-1)
\][/tex]
[tex]\[
Z_X = 88
\][/tex]
3. Determine the Nuclide Symbol of [tex]\( X \)[/tex]:
- The mass number [tex]\( A_X \)[/tex] of [tex]\( X \)[/tex] is 228.
- The atomic number [tex]\( Z_X \)[/tex] of [tex]\( X \)[/tex] is 88.
Looking up the periodic table, the element with atomic number 88 is Radium (Ra).
Therefore, the nuclide symbol of [tex]\( X \)[/tex] is:
[tex]\[
{ }_{88}^{228} \text{Ra}
\][/tex]
Given the options:
- [tex]\({ }_{90}^{230} \text{Th}\)[/tex]
- [tex]\({ }_{89}^{229} \text{Ac}\)[/tex]
- [tex]\({ }_{90}^{228} \text{Th}\)[/tex]
- [tex]\({ }_{88}^{228} \text{Ra}\)[/tex]
The correct answer is:
[tex]\[
{ }_{88}^{228} \text{Ra}
\][/tex]
