Question

predict the missing component in the nuclear equation.
_{12}^{23}mg \to x + _{1}^{0}e
_{12}^{23}mg
_{13}^{23}al
_{11}^{23}na

Explanation:

Step1: Apply conservation of atomic number

In a nuclear reaction, the sum of atomic numbers on both sides must be equal.
The atomic number of \(^{23}_{12}\text{Mg}\) is \(12\), and for \(^{0}_{1}\text{e}\) it is \(1\). Let the atomic number of \(X\) be \(Z\). Then \(12=Z + 1\), so \(Z=12 - 1=11\).

Step2: Apply conservation of mass number

The mass number of \(^{23}_{12}\text{Mg}\) is \(23\), and for \(^{0}_{1}\text{e}\) it is \(0\). Let the mass number of \(X\) be \(A\). Then \(23=A+0\), so \(A = 23\).

Answer:

\(^{23}_{11}\text{Na}\)