To solve the given nuclear equation:
[tex]\[
{}^4_2\text{He} + {}^{23}_{11}\text{Na} \rightarrow {}^1_1\text{H} + {}^a_b\text{X}
\][/tex]
we need to balance both the mass numbers (the superscripts) and the atomic numbers (the subscripts) on both sides of the equation.
1. Balancing mass numbers (the superscripts):
The total mass number on the left-hand side is:
[tex]\[
4 \, (\text{He}) + 23 \, (\text{Na}) = 27
\][/tex]
On the right-hand side, the total mass number is:
[tex]\[
1 \, (\text{H}) + a \, (\text{X})
\][/tex]
Therefore, we have:
[tex]\[
27 = 1 + a
\][/tex]
Solving for [tex]\(a\)[/tex]:
[tex]\[
a = 27 - 1 = 26
\][/tex]
2. Balancing atomic numbers (the subscripts):
The total atomic number on the left-hand side is:
[tex]\[
2 \, (\text{He}) + 11 \, (\text{Na}) = 13
\][/tex]
On the right-hand side, the total atomic number is:
[tex]\[
1 \, (\text{H}) + b \, (\text{X})
\][/tex]
Therefore, we have:
[tex]\[
13 = 1 + b
\][/tex]
Solving for [tex]\(b\)[/tex]:
[tex]\[
b = 13 - 1 = 12
\][/tex]
With [tex]\(a = 26\)[/tex] and [tex]\(b = 12\)[/tex], we identify that element [tex]\(X\)[/tex] corresponds to the element with atomic number 12, which is aluminium ([tex]\( Al \)[/tex]).
Given these results, we must choose the correct row from the table:
\begin{tabular}{|c|c|c|}
\hline
Row & \multicolumn{1}{|c|}{} & [tex]$i i$[/tex] \\
\hline
A. & 27 and 13 & aluminium \\
\hline
B. & 27 and 13 & magnesium \\
\hline
C. & 26 and 12 & aluminium \\
\hline
D. & 26 and 12 & magnesium \\
\hline
\end{tabular}
The correct row is:
\begin{tabular}{|c|c|c|}
\hline
C. & 26 and 12 & aluminium \\
\hline
\end{tabular}
