91影视

- 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.

Betaemission by magnesium-27

- Electron:${}_{鈭�1}^0\text{e}$

- Atomic number of magnesium is 12

${}_{12}^{27}\text{Mg}{鈫�}_{鈭�1}^0\text{e}{+}_{\text{Z}}^{\text{A}}\text{X}$

First, let us calculate the values of A and Z

$\begin{array}{c}27=0+\text{A}\\ \text{A}=27\\ 12=鈭�1+\text{Z}\\ \text{Z}=13\end{array}$

The element with atomic number 13 is aluminium, so the product is aluminium-27

${}_{12}^{27}\text{Mg}{鈫�}_{鈭�1}^0\text{e}{+}_{13}^{27}\text{Al}$

Betaemission by${}_{12}^{23}\text{Mg}$

- Positron:${}_{+1}^0\text{e}$

${}_{12}^{23}\text{Mg}{鈫�}_{+1}^0\text{e}{+}_{\text{Z}}^{\text{A}}\text{X}$

First, let us calculate the values of A and Z

$\begin{array}{c}23=0+\text{A}\\ \text{A}=23\\ 12=1+\text{Z}\\ \text{Z}=12鈭�1\\ =11\end{array}$

The element with atomic number 11 is sodium, so the product is sodium-23

${}_{12}^{23}\text{Mg}{鈫�}_{+1}^0\text{e}{+}_{11}^{23}\text{Na}$

Electron capture by${}_{46}^{103}\text{Pd}$

- Electron:${}_{鈭�1}^0\text{e}$

${}_{46}^{103}\text{Pd}{+}_{鈭�1}^0\text{e}{鈫�}_{\text{Z}}^{\text{A}}\text{X}$

First, let us calculate the values of A and Z

$\begin{array}{c}103+0=\text{A}\\ \text{A}=103\\ 46鈭�1=\text{Z}\\ \text{Z}=45\end{array}$

The element with atomic number is rhodium, so the product is rhodium-103

${}_{46}^{103}\text{Pd}{+}_{鈭�1}^0\text{e}{鈫�}_{45}^{103}\text{Rh}$