Q11 P - Chemistry: Molecular Nature Of Matter And Change

Question

Write balanced nuclear equations for the following:

(a) ${\mathit{\beta }}^{\mathbf{-}}$decay of silicon-$\mathbf{32}$

(b) Alpha decay of polonium 218

(c) Electron capture by${}_{\mathbf{49}}^{\mathbf{110}}\mathbf{\text{In}}$

 

Step-by-Step Solution

Answer

$\begin{array}{l}\text{a}){}_{14}^{32}\text{Si}{\to }_{-1}^0\text{e}{+}_{15}^{32}\text{P}\\ \text{b}){}_{84}^{218}\text{Po}{\to }_2^4\text{He}{+}_{82}^{214} \text{Pb}\\ \text{c})49\text{In}{+}_{-1}^0\text{e}{\to }_{48}^{110}\text{Cd}\end{array}$

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: Beta- decay of silicon-32 (a)

Beta- decay of silicon-32

- Electron: ${}_{-1}^0\text{e}$

- Atomic number of silicon is 14

${}_{14}^{32}\text{Si}{\to }_{-1}^0\text{e}{+}_{\text{Z}}^{\text{A}}\text{X}$

First, let us calculate the values of  A and  Z

$\begin{array}{c}32=0+\text{A}\\ \text{A}=32\\ 14=-1+\text{Z}\\ \text{Z}=15\end{array}$

The element with atomic number  15 is phosphorus, so the product is phosphorus-32${}_{14}^{32}\text{Si}{\to }_{-1}^0\text{e}{+}_{15}^{32}\text{P}$

 

Step 4: Alpha decay of polonium-218 (b)

Alpha decay of polonium--218 Alpha particle: ${}_2^4\text{He}$

- Atomic number of polonium is 84

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

First, let us calculate the values of  A and  Z

$\begin{array}{c}218=4+\text{A}\\ \text{A}=218-4\\ =214\\ 84=2+\text{Z}\end{array}$

The element with atomic number 82 is lead, so the product is lead-214

${}_{84}^{218}\text{Po}{\to }_2^4\text{He}{+}_{82}^{214} \text{Pb}$

 

Step 5: Electron capture by 49 110 l n (c)

 Electron capture by${}_{49}^{110}\ln$

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

${}_{49}^{110}\text{In}{+}_{-1}^0\text{e}{\to }_{\text{Z}}^{\text{A}}\text{X}$

First, let us calculate the values of A and Z

$\begin{array}{c}110+0=\text{A}\\ \text{A}=110\\ 49-1=\text{Z}\\ \text{Z}=48\end{array}$

The element with atomic number 48 is cadmium, so the product is cadmium-110

${}_{49}^{110}\text{In}{+}_{-1}^0\text{e}{\to }_{48}^{110}\text{Cd}$