Q15 P - Chemistry: Molecular Nature Of Matter And Change

Question

Write balanced nuclear equations for the following:

(a) Formation of  through positron emission

(b) Formation of polonium-215 through  decay

(c) Formation of  through electron capture

 

Step-by-Step Solution

Answer

The required equations are

$\begin{array}{l}{}_{26}^{52}\text{Fe}{\to }_{+1}^0\text{e}{+}_{25}^{52}\text{Mn}\\ {}_{86}^{219}\text{Rn}{\to }_2^4\text{He}{+}_{84}^{215}\text{Po}\\ {}_{37}^{81}\text{Rb}{+}_{-1}^0\text{e}{\to }_{36}^{81}\text{Kr}\end{array}$

Step 1: Given information

- A: mass number

- Z: atomic number

- X: element on reactant side

Step 3: Formation of 52 Mn (a)

Formation of${}^{52}\text{Mn}$ through positron emission

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

- Atomic number of is 25

${\text{Z}}_{\text{A}}^{\text{A}}{\to }_{+1}^0\text{e}{+}_{25}^{52}\text{Mn}$

First, let us calculate the values of A and Z

$\begin{array}{l}\text{A}=0+52 \\ \text{A}=52\\ \text{Z}=1+25\\ \text{Z}=\text{26}\end{array}$

The element with atomic number 26 is iron, so the element on the reactant side is iron- 52${}_{26}^{52}\text{Fe}{\to }_{+1}^0\text{e}{+}_{25}^{52}\text{Mn}$

Step 4: Formation of polonium-215 (b)

Formation of polonium-215 through alpha decay

- Alpha particle:-${}_2^4\text{He}$ Atomic number of polonium is 84

${}_{\text{Z}}^{\text{A}}\text{X}{\to }_2^4\text{He}{+}_{84}^{215}\text{Po}$

First, let us calculate the values of A and Z

$\begin{array}{l}\text{A}=4+215\\ \text{A}=219\\ \text{Z}=2+84\\ \text{Z}=86\end{array}$

The element with atomic number 86 is radon, so the element on the reactant side is radon-219$${}_{86}^{219}\text{Rn}{\to }_2^4\text{He}{+}_{84}^{215}\text{Po}$

Step 5: Formation of (c)

Formation of ${}^{81}\text{Kr}$through electron capture

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

- Atomic number of krypton is 36

${}_{\text{Z}}^{\text{A}}\text{X}{+}_{-1}^0\text{e}{\to }_{36}^{81}\text{Kr}$

First, let us calculate the values of A and Z

$\begin{array}{r}\text{A}+\text{0}=81\\ \text{A}=81\\ \text{Z}-1=36\\ \text{Z}=37\end{array}$

The element with atomic number 37 is rubidium, so the element on the reactant side is rubidium-81

${}_{37}^{81}\text{Rb}{+}_{-1}^0\text{e}{\to }_{36}^{81}\text{Kr}$