Q8 P - Chemistry: Molecular Nature Of Matter And Change

Question

Write balanced nuclear equations for the following:

(a) Alpha decay of ${}_{\mathbf{92}}^{\mathbf{234}}\mathbf{\text{U}}$

(b) Electron capture by neptunium-232

(c) Positron emission by${}_{\mathbf{7}}^{\mathbf{12}}\mathbf{ }\mathbf{\text{N}}$

Step-by-Step Solution

Answer

The required balanced equations are

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

Step 1: 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.

Step 3: Alpha decay of (a)

- Alpha particle: ${}_2^4\text{He}$

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

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

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

Step 4: Electron capture by neptunium- (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

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

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

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

Step 5: Positron emission by (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

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

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

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