Q14 P - Chemistry: Molecular Nature Of Matter And Change

Question

Write balanced nuclear equations for the following:

(a) Formation of ${}^{186}\text{Ir}$ through electron capture

(b) Formation of francium- 221 through  decay

(c) Formation of iodine- 129 through ${\mathbf{\text{β}}}^{\mathbf{\text{-}}}$decay.

 

Step-by-Step Solution

Answer

The required balanced equations are

$\begin{array}{l}{}_{78}^{186}\text{Pt}{+}_{-1}^0\text{e}{\to }_{77}^{186}\text{Ir}\\ {}_{89}^{225}\text{Ac}{\to }_2^4\text{He}{+}_{87}^{221}\text{Fr}\\ {}_{52}^{129}\text{Te}{\to }_{-1}^0\text{e}{+}_{53}^{129}\text{I}\end{array}$

Step 1: Given information

- A: mass number

- Z: atomic number

- X: element on reactant side

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:Formation of (a)

Formation of${}^{186}\text{Ir}$ through electron capture

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

- Atomic number of Ir is77

${}_{\text{Z}}^{\text{A}}\text{X}{+}_{-1}^0\text{e}{\to }_{77}^{186}\text{Ir}$

First, let us calculate the values of A and Z

$\begin{array}{c}\text{A}+\text{0}=186\\ \text{A}=186\\ \text{Z}-1=77\\ \text{Z}=78\end{array}$

The element with atomic number 78 is platinum, so the element on the reactant side is platinum-186

${}_{78}^{186}\text{Pt}{+}_{-1}^0\text{e}{\to }_{77}^{186}\text{Ir}$

Step 4: Formation of francium- (b)

Formation of francium- 221 through alpha decay

- Alpha particle:-${}_2^4\text{He}$ Atomic number of francium is 87

$\underset{\text{Z}}{\text{A}}\text{X}{\to }_2^4\text{He}{+}_{87}^{221}\text{Fr}$

First, let us calculate the values of  A and Z

$\begin{array}{c}\text{A}=4+221 \\ \text{A}=225\\ \text{Z}=2+87\\ \text{Z}=89\end{array}$

The element with atomic number 89 is actinium, so the element on the reactant side is actinium-225

${}_{89}^{225}\text{Ac}{\to }_2^4\text{He}{+}_{87}^{221}\text{Fr}$

 

Step 5: Formation of iodine- (c)

Formation of iodine- 129 through Beta- decay

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

- Atomic number of iodine is53

${}_{\text{Z}}^{\text{A}}\text{X}{\to }_{-1}^0\text{e}{+}_{53}^{129}\text{I}$

First, let us calculate the values of A and Z

$\begin{array}{c}\text{A}=0+129 \\ \text{A}=129\\ \text{Z}=-1+53\\ \text{Z}=52\end{array}$

The element with atomic number 52 is tellurium, so the element on the reactant side is tellurium-129

${}_{52}^{129}\text{Te}{\to }_{-1}^0\text{e}{+}_{53}^{129}\text{I}$