Question

Question: Write balanced nuclear equations for the following:

(a) Alpha decay of

(b) Electron capture by neptunium-

(c) Positron emission by

Short answer

The required balanced equations are

$$\begin{gathered} \text{(a}\left){ }_{92}^{234}\text{U}{\to }_2^4\text{He}{+}_{90}^{230}\text{Th} \\ \right(\text{b}){ }_{93}^{232}\text{Np}{+}_{-1}^0\text{e}{\to }_{92}^{232}\text{U} \\ (\text{c}){ }_7^{12}\text{N}{\to }_{+1}^0\text{e}{+}_6^{12}\text{C} \end{gathered}$$

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.

Alpha decay of U92234(a)

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

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

First, let us calculate the values of and

$\begin{array}{rcl}234& =& 4+\text{A}\\ \text{A}& =& 234-4\\ & =& 230\\ 92& =& 2+\text{Z}\end{array}$

Solve further

$\begin{array}{rcl}\text{Z}& =& 92-2\\ & =& 90\end{array}$

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

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

Electron capture by neptunium-232(b)

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

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

First, let us calculate the values of A and Z.

$\begin{array}{rcl}232+0& =& \text{A}\\ \text{A}& =& 232\\ 93-1& =& \text{Z}\\ \text{Z}& =& 92\end{array}$

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

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

Step 5: Positron emission by  712N(c)

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

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

First, let us calculate the values of A and Z.

$\begin{array}{rcl}12& =& 0+\text{A}\\ \text{A}& =& 127\\ & =& 1+\text{Z}\\ \text{Z}& =& 7-1\\ & =& 6\end{array}$

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

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