Chapter 9, Problem 9.28P

Interpretation Introduction

(a)

Interpretation:

Identify the missing particles by completing the given nuclear reaction.

$\text{O}\text{ + }\text{O}\stackrel{}{\to }?\text{+ }\text{H}\text{e}$

Concept Introduction:

In a nuclear reaction we have to balance the following two.

1. Number of protons
2. Mass number

The sum of protons on reactant side should be equal to sum of protons on product side.

Similarly the total mass number on both the side should be equal.

Answer to Problem 9.28P

The complete reaction will be.

$\text{O}\text{ + }\text{O}\stackrel{}{\to }\text{S}\text{i+ }\text{H}\text{e}$.

Explanation of Solution

The sum of protons on reactant side $\text{8 + 8 = 16}$

The total mass number on reactant side = $\text{16 + 16 = 32}$

Let number of protons in missing species on product side is “x”.

Therefore.

$\begin{array}{l}\text{x + 2 = 16}\\ \text{x = 14}\end{array}$

This is atomic number of silicon hence symbol will be “Si”.

Let mass number of missing species on product side is “y”.

Therefore.

$\begin{array}{l}\text{y + 4 = 32}\\ \text{y = 28}\end{array}$

Hence, the complete equation will be.

$\text{O}\text{ + }\text{O}\stackrel{}{\to }\text{S}\text{i+ }\text{H}\text{e}$.

Interpretation Introduction

(b)

Interpretation:

Identify the missing particles by completing the given nuclear reaction.

$\text{U}\text{ + }\text{n}\stackrel{}{\to }\text{S}\text{r + ? + 3}\text{n}$

Concept Introduction:

In a nuclear reaction we have to balance the following two.

1. Number of protons
2. Mass number

The sum of protons on reactant side should be equal to sum of protons on product side.

Similarly the total mass number on both the side should be equal.

Answer to Problem 9.28P

The complete reaction will be.

$\text{U}\text{ + }\text{n}\stackrel{}{\to }\text{S}\text{r + }\text{X}\text{e + 3}\text{n}$.

Explanation of Solution

The sum of protons on reactant side $\text{92 + 0 = 92}$

The total mass number on reactant side = $\text{235 + 1 = 236}$

Let number of protons in missing species on product side is “x”.

Therefore.

$\begin{array}{l}\text{38 + x + 0 =92}\\ \text{x = 54}\end{array}$

This is atomic number of Xenon hence symbol will be “Xe”.

Let mass number of missing species on product side is “y”.

Therefore.

$\begin{array}{l}\text{y + 90 + 3 = 236}\\ \text{y = 143}\end{array}$

Hence the complete equation will be.

$\text{U}\text{ + }\text{n}\stackrel{}{\to }\text{S}\text{r + }\text{X}\text{e + 3}\text{n}$.

Interpretation Introduction

(c)

Interpretation:

Identify the missing particles by completing the given nuclear reaction.

$\text{C}\text{+ }\text{H}\text{e }\stackrel{}{\to }\text{O}\text{ + ? }$

Concept Introduction:

In a nuclear reaction we have to balance the following two.

1. Number of protons
2. Mass number

The sum of protons on reactant side should be equal to sum of protons on product side.

Similarly the total mass number on both the side should be equal.

Answer to Problem 9.28P

The complete reaction will be.

$\text{C}\text{+ }\text{H}\text{e }\stackrel{}{\to }\text{O}\text{ + }\text{n}$.

Explanation of Solution

The sum of protons on reactant side $\text{6 + 2= 8}$

The total mass number on reactant side = $\text{13 + 4 = 17}$

Let number of protons in missing species on product side is “x”.

Therefore.

$\begin{array}{l}\text{6 + x = 6}\\ \text{x = 0}\end{array}$

Hence it is not an element.

Let mass number of missing species on product side is “y”.

Therefore.

$\begin{array}{l}\text{y + 16 = 17}\\ \text{y = 1}\end{array}$

Hence it should be a neutron.

Hence the complete equation will be.

$\text{C}\text{+ }\text{H}\text{e }\stackrel{}{\to }\text{O}\text{ + }\text{n}$.

Interpretation Introduction

(d)

Interpretation:

Identify the missing particles by completing the given nuclear reaction.

$\text{B}\text{i }\stackrel{}{\to }\text{ ? + }\text{e}$

Concept Introduction:

In a nuclear reaction we have to balance the following two.

1. Number of protons
2. Mass number

The sum of protons on reactant side should be equal to sum of protons on product side.

Similarly the total mass number on both the side should be equal.

In beta emission one neutron is converted to a proton and one electron.

Hence there is an increase in atomic number while the mass number remains the same.

Answer to Problem 9.28P

The complete reaction will be.

$\text{B}\text{i }\stackrel{}{\to }\text{P}\text{o + }\text{e}$

Explanation of Solution

As it is beta elimination here a neutron is converted to proton and electron.

So there is increase in atomic number.

The parent nucleus is Bismuth, on increase its proton number the element formed will have atomic number 84, this is polonium (symbol Po).

Hence the complete reaction will be.

$\text{B}\text{i }\stackrel{}{\to }\text{P}\text{o + }\text{e}$.

Interpretation Introduction

(e)

Interpretation:

Identify the missing particles by completing the given nuclear reaction.

$\text{C}\text{+ }\text{H}\stackrel{}{\to }?\text{+ γ}$

Concept Introduction:

In a nuclear reaction we have to balance the following two.

1. Number of protons
2. Mass number

The sum of protons on reactant side should be equal to sum of protons on product side.

Similarly the total mass number on both the side should be equal.

In Gamma radiation emission there is no change in the total number of protons on reactant and product side.

Similarly there is no change in mass number on reactant and product side.

Answer to Problem 9.28P

The complete reaction will be.

$\text{C}\text{+ }\text{H}\stackrel{}{\to }\text{N}\text{+ γ}$

Explanation of Solution

The gamma radiation has not mass number and no proton.

The sum of protons on reactant side $\text{6 + 1= 7}$

The total mass number on reactant side = $\text{12 + 1 = 13}$

Let number of protons in missing species on product side is “x”.

Therefore.

$\begin{array}{l}\text{0 + x = 7}\\ \text{x = 7}\end{array}$

Hence the new element will have atomic number 7, it is nitrogen.

Let mass number of missing species on product side is “y”.

Therefore.

$\begin{array}{l}\text{y + 0 = 13}\\ \text{y = 13}\end{array}$

Hence the complete equation will be.

$\text{C}\text{+ }\text{H}\stackrel{}{\to }\text{N}\text{+ γ}$.