Question

Write balanced nuclear equations for the following:

(a)${\mathbf{\text{β}}}^{\mathbf{\text{-}}}$decay of sodium-$\mathbf{26}$

(b)${\mathbf{\text{β}}}^{\mathbf{\text{-}}}$decay of francium- 223

(c) Alpha decay of${}_{\mathbf{83}}^{\mathbf{212}}\mathbf{\text{Bi}}$

Short answer

The required balanced nuclear equations are

$\begin{array}{l}\left(\text{a}\right){}_{11}^{26}\text{Na}{\to }_{-1}^0\text{e}{+}_{12}^{26}\text{Mg}\\ \left(\text{b}\right){}_{87}^{223}\text{Fr}{\to }_{-1}^0\text{e}{+}_{88}^{223}\text{Ra}\\ \left(\text{c}\right){}_{83}^{212}\text{Bi}{\to }_2^4\text{He}{+}_{81}^{208}\text{Tl}\end{array}$

Step-by-step solution

Given information

- A: mass number

- Z: atomic number

- X: element produced

Step 2: Balanced nuclear equation

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 sodium- (a)

Beta decay of sodium-26

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

- Atomic number of sodium is 11

${}_{11}^{26}\text{Na}{\to }_{-1}^0\text{e}{+}_Z^{\text{A}}\text{X}$

First, let us calculate the values of A and Z

$\begin{array}{c}26=0+\text{A}\\ \text{A}=26\\ 11=-1+\text{Z}\\ \text{Z}=12\end{array}$

The element with atomic number 12 is magnesium, so the product is magnesium-26

${}_{11}^{26}\text{Na}{\to }_{-1}^0\text{e}{+}_{12}^{26}\text{Mg}$

Beta decay of francium-223 (b) 

Beta decay of francium-223

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

- Atomic number of francium is 223

${}_{87}^{223}\text{Fr}{\to }_{-1}^0\text{e}+{\text{Z}}_{\text{Z}}^{\text{A}}\text{X}$

First, let us calculate the values of A and Z

$\begin{array}{c}223=0+\text{A}\\ \mathbf{A}=223\\ 87=-1+Z\\ Z=88\end{array}$

The element with atomic number 88 is radium, so the product is radium-223

${}_{87}^{223}\text{Fr}{\to }_{-1}^0\text{e}{+}_{88}^{223}\text{Ra}$

Alpha decay of 83212Bi (c) 

Alpha decay of ${}_{83}^{212}\text{Bi}$

- Alpha particle: ${}_2^4\text{He}$${}_{83}^{212}\text{Bi}{\to }_2^4\text{He}+\frac{\text{A}}{\text{Z}}\text{X}$

First, let us calculate the values of A and Z

$\begin{array}{c}212=4+\text{A}\\ \mathbf{A}=212-4\\ =208\\ 83=2+\mathbf{Z}\end{array}$

Solve further

$\begin{array}{c}\text{Z}=83-2\\ =81\end{array}$

The element with atomic number is thallium, so the product is thallium- 208

${}_{83}^{212}\text{Bi}{\to }_2^4\text{He}{+}_{81}^{208}\text{Tl}$