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

Beta- decay of silicon-32

- Electron:${}_{−1}^0\text{e}$

- Atomic number of silicon is 14

${}_{14}^{32}\text{Si}{→}_{−1}^0\text{e}{+}_{\text{Z}}^{\text{A}}\text{X}$

First, let us calculate the values of A and Z

$\begin{array}{c}32=0+\text{A}\\ \text{A}=32\\ 14=−1+\text{Z}\\ \text{Z}=15\end{array}$

The element with atomic number 15 is phosphorus, so the product is phosphorus-32${}_{14}^{32}\text{Si}{→}_{−1}^0\text{e}{+}_{15}^{32}\text{P}$

Alpha decay of polonium--218 Alpha particle:${}_2^4\text{He}$

- Atomic number of polonium is 84

${}_{84}^{218}\text{Po}{→}_2^4\text{He}+{\text{Z}}_{\text{A}}^{\text{X}}$

First, let us calculate the values of A and Z

$\begin{array}{c}218=4+\text{A}\\ \text{A}=218−4\\ =214\\ 84=2+\text{Z}\end{array}$

The element with atomic number 82 is lead, so the product is lead-214

${}_{84}^{218}\text{Po}{→}_2^4\text{He}{+}_{82}^{214}\text{Pb}$

Electron capture by${}_{49}^{110}\ln$

- Electron:${}_{−1}^0\text{e}$

${}_{49}^{110}\text{In}{+}_{−1}^0\text{e}{→}_{\text{Z}}^{\text{A}}\text{X}$

First, let us calculate the values of A and Z

$\begin{array}{c}110+0=\text{A}\\ \text{A}=110\\ 49−1=\text{Z}\\ \text{Z}=48\end{array}$

The element with atomic number 48 is cadmium, so the product is cadmium-110

${}_{49}^{110}\text{In}{+}_{−1}^0\text{e}{→}_{48}^{110}\text{Cd}$