Q13 P - Chemistry: Molecular Nature Of Matter And Change

Question

Write balanced nuclear equations for the following:

(a) Formation of ${}_{\mathbf{95}}^{\mathbf{241}}\mathbf{\text{Am}}$ through${\mathbf{\text{β}}}^{\mathbf{\text{-}}}$ decay

(b) Formation of  ${}_{\mathbf{89}}^{\mathbf{228}}\mathbf{\text{Ac}}$through ${\mathbf{\text{β}}}^{\mathbf{\text{-}}}$ decay

(c) Formation of ${}_{\mathbf{83}}^{\mathbf{203}}\mathbf{\text{Bi}}$ through $\mathbf{\text{α}}$ decay

 

Step-by-Step Solution

Answer

The required equations are

$\begin{array}{l}{}_{23}^{48} \text{V}\to \underset{+1}{0}\text{e}{+}_{22}^{48}\text{Ti}\\ {}_{48}^{107}\text{Cd}{+}_{-1}^0\text{e}{\to }_{47}^{107}\text{Ag}\\ {}_{86}^{222}\text{Rn}{\to }_2^4\text{He}{+}_{84}^{218}\text{Po}\end{array}$

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 ${}_{22}^{48}\text{Ti}$through positron emission

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

${}_{\text{Z}}^{\text{A}}\text{X}{\to }_{+1}^0\text{e}{+}_{22}^{48}\text{Ti}$

First, let us calculate the values of A  and Z

$\begin{array}{c}\text{A}=0+48\\ \text{A}=48\\ \text{Z}=1+22\\ \text{Z}=23\end{array}$

The element with atomic number 23 is vanadium, so the element on the reactant side is vanadium-48$${}_{23}^{48} \text{V}{\to }_{+1}^0\text{e}{+}_{22}^{48}\text{Ti}$

Step 4: Formation of silver-107 (b)

Formation of silver-107 through electron capture

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

- Atomic number of silveris 47

${}_{\text{Z}}^{\text{A}}\text{X}{+}_{-1}^0\text{e}{\to }_{47}^{107}\text{Ag}$

First, let us calculate the values of A and Z

$\begin{array}{c}\text{A}+\text{0}=107\\ \text{A}=107\\ \text{Z}-1=47\\ \text{Z}=48\end{array}$

The element with atomic number 48 is cadmium, so the element on the reactant side is cadmium-107

${}_{48}^{107}\text{Cd}{+}_{-1}^0\text{e}{\to }_{47}^{107}\text{Ag}$

Step 5: Formation of polonium-206 (c)

 Formation of polonium-206 through alpha decay

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

- Atomic number of polonium is84

${}_{\text{Z}}^{\text{A}}\text{X}{\to }_2^4\text{He}{+}_{84}^{218}\text{Po}$

First, let us calculate the values of A and Z

$\begin{array}{c}\text{A}=4+218 \\ \text{A}=222\\ \text{Z}=2+84\\ \text{Z}=86\end{array}$

The element with atomic number is radon, so the element on the reactant side is radon-222${}_{86}^{222}\text{Rn}{\to }_2^4\text{He}{+}_{84}^{218}\text{Po}$