Question

in the above chemical equation solve for a, z, and x.

_{9}^{12}b \to _{z}^{a}x + _{-1}^{0}e

Explanation:

Step1: Apply mass - number conservation

In a nuclear reaction, the sum of mass numbers on the left - hand side is equal to the sum of mass numbers on the right - hand side.
The mass number of \(_{9}^{12}B\) is \(A_{B}=12\), and for \(\beta^{-}\) particle (\(_{-1}^{0}e\)) the mass number is \(A_{e}=0\). So, \(A + 0=12\), which gives \(A = 12\).

Step2: Apply atomic - number conservation

The sum of atomic numbers on the left - hand side is equal to the sum of atomic numbers on the right - hand side.
The atomic number of \(_{9}^{12}B\) is \(Z_{B}=9\), and for \(\beta^{-}\) particle (\(_{-1}^{0}e\)) the atomic number is \(Z_{e}=- 1\). So, \(Z+( - 1)=9\), which gives \(Z = 10\).

Step3: Identify the element

The element with atomic number \(Z = 10\) is Neon (\(Ne\)).

Answer:

\(A = 12\), \(Z = 10\), \(X=Ne\)