91Ó°ÊÓ

- A: mass number

- Z: atomic number

X: element produced

A balanced nuclear equation is one where the sum of the mass numbers (the top number in notation) and the sum of the atomic numbers balance on either side of an equation.

- Alpha particle:${}_2^4\text{He}$

${}_{92}^{234}\text{U}{→}_2^4\text{He}+\underset{\text{Z}}{\text{A}}\text{X}$

First, let us calculate the values of A and Z

$\begin{array}{c}234=4+\text{A}\\ \text{A}=234−4\\ =230\\ 92=2+\text{Z}\end{array}$

Solve further

$\begin{array}{c}\text{Z}=92−2\\ =90\end{array}$

The element with atomic number 90 is thorium, so the product is thorium-230

${}_{92}^{234}\text{U}{→}_2^4\text{He}{+}_{90}^{230}\text{Th}$

- Electron:- ${}_{−1}^0\text{e}$Atomic number of neptunium is 93

${}_{93}^{232}\text{Np}{+}_{−1}^0\text{e}{→}_{\text{Z}}^{\text{A}}\text{X}$

First, let us calculate the values of A and Z

$\begin{array}{c}232+0=\text{A}\\ \text{A}=232\\ 93−1=\text{Z}\\ \text{Z}=92\end{array}$

The element with atomic number 92 is uranium, so the product is uranium-232

${}_{93}^{232}\text{Np}{+}_{−1}^0\text{e}{→}_{92}^{232}\text{U}$

Positron emission by$\frac{12}{7}{\text{N}}_{+1}^0\text{e}$

${}_7^{12}\text{N}{→}_{+1}^0\text{e}{+}_{\text{Z}}^{\text{A}}\text{X}$

First, let us calculate the values of A and Z

$\begin{array}{c}12=0+\text{A}\\ \text{A}=127\\ =1+\text{Z}\\ \text{Z}=7−1\\ =6\end{array}$

The element with atomic number 6 is carbon, so the product is carbon-12

${}_7^{12}\text{N}{→}_{+1}^0\text{e}{+}_6^{12}\text{C}$