Certainly! Let's solve this problem step by step to determine the nuclide symbol of [tex]\( X \)[/tex] in the given nuclear equation.
First, consider the nuclear equation given:
[tex]\[
X \longrightarrow \, _{89}^{228}\text{Ac} + \, _{-1}^0\beta
\][/tex]
1. Understanding Beta Decay:
- In beta decay ([tex]\( \beta^- \)[/tex] decay), a neutron in the nucleus is converted into a proton. This conversion results in the emission of a beta particle ([tex]\( \beta^- \)[/tex]).
- The atomic number of the new element increases by 1, while the mass number remains unchanged.
2. Analyzing the Product:
- The product in this decay process is [tex]\( _{89}^{228}\text{Ac} \)[/tex]. This means the product nuclide has an atomic number of 89 and a mass number of 228.
3. Determining the Original Nuclide [tex]\( X \)[/tex]:
- Since the atomic number increases by 1 during beta decay, the original nuclide [tex]\( X \)[/tex] must have an atomic number that is 1 less than the product nuclide.
- The atomic number of the product nuclide (Ac) is 89.
- Subtracting 1 from the atomic number of Ac gives us the atomic number of [tex]\( X \)[/tex]:
[tex]\[
89 - 1 = 88
\][/tex]
- The mass number remains the same during beta decay, so the mass number of [tex]\( X \)[/tex] is also 228.
4. Forming the Nuclide Symbol for [tex]\( X \)[/tex]:
- Combining the atomic number (88) and the mass number (228), we find that the nuclide symbol [tex]\( X \)[/tex] corresponds to:
[tex]\[
{ }_{88}^{228}\text{Ra}
\][/tex]
5. Matching with Given Choices:
- From the choices provided:
[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 nuclide symbol for [tex]\( X \)[/tex] is:
[tex]\[
{ }_{88}^{228}\text{Ra}
\][/tex]
So, the nuclide symbol of [tex]\( X \)[/tex] is [tex]\( \boxed{_{88}^{228}\text{Ra}} \)[/tex].
