Chapter 25, Problem 11PS

(a)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Concept Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of $He$ nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

Answer to Problem 11PS

$\text{F}\text{e}\text{+}\text{H}\text{e}\to \text{2}\text{n}\text{+}{}_{\text{28}}^{\text{56}}\text{Ni}$.

Explanation of Solution

The radioactive isotope of iron-54 when irradiated with alpha particle forms ${}_{\text{28}}^{\text{56}}\text{Ni}$ by the emission of 2 neutrons. In ${}_{\text{28}}^{\text{56}}\text{Ni}$ has atomic number 28 and mass number 56.

(b)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of $He$ nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

Answer to Problem 11PS

$\text{A}\text{l}\text{+}\text{H}\text{e}\to \text{P}\text{+}\text{n}$

Explanation of Solution

The radioactive isotope of Aluminium-27 when irradiated with alpha particle forms $\text{P}$ by the emission of one neutrons. In $\text{P}$ has atomic number 15 and mass number 30.

(c)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of $He$ nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

Answer to Problem 11PS

$\text{S}\text{+}\text{n}\to \text{H}\text{+}{}_{\text{15}}^{\text{32}}\text{P}$

Explanation of Solution

The radioactive isotope of Sulphur-32 is irradiated with neutron forms $\text{P}$ by the emission of a proton. In $\text{P}$ has atomic number 15 and mass number 32.

(d)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of $He$ nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

Answer to Problem 11PS

$\text{M}\text{o}\text{+}\text{H}\to \text{n}\text{+}\text{T}\text{c}$

Explanation of Solution

The radioactive isotope of $\text{M}\text{o}$ is irradiated with deuterium forms $\text{T}\text{c}$ by the emission of a neutron. In $\text{T}\text{c}$ has atomic number 97 and mass number 43.

(e)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of $He$ nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

Answer to Problem 11PS

$\text{M}\text{o}\text{+}\text{n}\to \text{T}\text{c}\text{+}\text{β}$

Explanation of Solution

The radioactive isotope of $\text{M}\text{o}$ is irradiated with neutron forms $\text{T}\text{c}$ by the emission of a electron. In $\text{T}\text{c}$ has atomic number 99 and mass number 43.

(f)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of $He$ nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

Answer to Problem 11PS

$\text{F}\to \text{O}\text{+}\text{e}$

Explanation of Solution

The radioactive isotope of $\text{F}$ is irradiated forms $\text{O}$ by the emission of a positron. In $\text{O}$ has atomic number 8 and mass number 18.