Chapter 2, Problem 2.64QP

(a)

Interpretation Introduction

Interpretation: The neutron numbers should be predicted for the given neutron–rich isotope ${}^{\text{40}}\text{Mg}$.

Concept Introduction:

Conversion of atoms to moles: $\text{no}\text{.of moles = }\frac{\text{no}\text{.of atoms}}{\text{Avogadro's number}}.$

Nuclear stability: The nucleus is composed of protons and neutrons. The strongest nuclear force binds the particles tightly. Though the protons repel each other due to no attraction between similar charges, possess short-range attractions made the attraction possible between proton and proton, proton and neutron, neutron and neutron.

The stability of any element is determined by the difference between coulombic repulsion and the short-range attraction. If repulsion outweighs the attraction, the disintegration of nucleus occurs by producing the daughter nuclides. If the attractive forces prevail, the nucleus is stable.

$\begin{array}{l}{\text{The element symbol : }}_{\text{Z}}^{\text{A}}\text{X,}\\ \text{where, A (mass number) = no}\text{.of protons + no}\text{.of neutrons}\text{.}\\ \text{ Z (atomic number) = no}\text{. of protons}\text{. (electrons = protons)}\text{.}\end{array}$

Answer to Problem 2.64QP

${}^{\text{40}}\text{Mg has totally 28 neutrons}\text{.}$

Explanation of Solution

Predict the neutron numbers in the ${}^{\text{40}}\text{Mg}$.

$\begin{array}{l}{\text{The element symbol : }}_{\text{Z}}^{\text{A}}\text{X,}\\ \text{where, A (mass number) = no}\text{.of protons + no}\text{.of neutrons}\text{.}\\ \text{ Z (atomic number) = no}\text{. of protons}\text{. (electrons = protons)}\text{.}\end{array}$

  $\begin{array}{l}{\text{The element symbol : }}_{}^{\text{40}}\text{Mg,}\\ \text{Z (atomic number) of magnesium is 12}\\ \text{No}\text{.of neutrons = mass number - atomic number}\text{.}\\ \text{ no}\text{.of neutrons = 40 - 12 = 28}\\ \text{The total neutrons are : 28}\text{.}\end{array}$

(b)

Interpretation Introduction

Interpretation: The neutron numbers should be predicted for the given neutron–rich isotope ${}^{\text{44}}\text{Si}$.

Concept Introduction:

Conversion of atoms to moles: $\text{no}\text{.of moles = }\frac{\text{no}\text{.of atoms}}{\text{Avogadro's number}}.$

Nuclear stability: The nucleus is composed of protons and neutrons. The strongest nuclear force binds the particles tightly. Though the protons repel each other due to no attraction between similar charges, possess short-range attractions made the attraction possible between proton and proton, proton and neutron, neutron and neutron.

The stability of any element is determined by the difference between coulombic repulsion and the short-range attraction. If repulsion outweighs the attraction, the disintegration of nucleus occurs by producing the daughter nuclides. If the attractive forces prevail, the nucleus is stable.

$\begin{array}{l}{\text{The element symbol : }}_{\text{Z}}^{\text{A}}\text{X,}\\ \text{where, A (mass number) = no}\text{.of protons + no}\text{.of neutrons}\text{.}\\ \text{ Z (atomic number) = no}\text{. of protons}\text{. (electrons = protons)}\text{.}\end{array}$

Answer to Problem 2.64QP

${}^{\text{44}}\text{Si has totally 30 neutrons}\text{.}$

Explanation of Solution

Predict the neutron numbers in the ${}^{\text{44}}\text{Si}$.

  $\begin{array}{l}{\text{The element symbol : }}_{}^{\text{44}}\text{Si,}\\ \text{Z (atomic number) of Silicon is 14}\\ \text{No}\text{.of neutrons = mass number - atomic number}\text{.}\\ \text{ no}\text{.of neutrons = 44 - 14 = 30}\\ \text{The total neutrons are : 30}\text{.}\end{array}$

(c)

Interpretation Introduction

Interpretation: The neutron numbers should be predicted for the given neutron–rich isotope ${}^{\text{48}}\text{Ca}$.

Concept Introduction:

Conversion of atoms to moles: $\text{no}\text{.of moles = }\frac{\text{no}\text{.of atoms}}{\text{Avogadro's number}}.$

Nuclear stability: The nucleus is composed of protons and neutrons. The strongest nuclear force binds the particles tightly. Though the protons repel each other due to no attraction between similar charges, possess short-range attractions made the attraction possible between proton and proton, proton and neutron, neutron and neutron.

The stability of any element is determined by the difference between coulombic repulsion and the short-range attraction. If repulsion outweighs the attraction, the disintegration of nucleus occurs by producing the daughter nuclides. If the attractive forces prevail, the nucleus is stable.

$\begin{array}{l}{\text{The element symbol : }}_{\text{Z}}^{\text{A}}\text{X,}\\ \text{where, A (mass number) = no}\text{.of protons + no}\text{.of neutrons}\text{.}\\ \text{ Z (atomic number) = no}\text{. of protons}\text{. (electrons = protons)}\text{.}\end{array}$

Answer to Problem 2.64QP

${}^{\text{48}}\text{Ca has totally 28 neutrons}\text{.}$

Explanation of Solution

Predict the neutron numbers in the ${}^{\text{48}}\text{Ca}$.

  $\begin{array}{l}{\text{The element symbol : }}_{}^{\text{48}}\text{Ca,}\\ \text{Z (atomic number) of Calcium is 20}\text{.}\\ \text{No}\text{.of neutrons = mass number - atomic number}\text{.}\\ \text{ no}\text{.of neutrons = 48 - 20 = 28}\\ \text{The total neutrons are : 28}\text{.}\end{array}$

(d)

Interpretation Introduction

Interpretation: The neutron numbers should be predicted for the given neutron–rich isotope ${}^{\text{43}}\text{Al}$.

Concept Introduction:

Conversion of atoms to moles: $\text{no}\text{.of moles = }\frac{\text{no}\text{.of atoms}}{\text{Avogadro's number}}.$

Nuclear stability: The nucleus is composed of protons and neutrons. The strongest nuclear force binds the particles tightly. Though the protons repel each other due to no attraction between similar charges, possess short-range attractions made the attraction possible between proton and proton, proton and neutron, neutron and neutron.

The stability of any element is determined by the difference between coulombic repulsion and the short-range attraction. If repulsion outweighs the attraction, the disintegration of nucleus occurs by producing the daughter nuclides. If the attractive forces prevail, the nucleus is stable.

$\begin{array}{l}{\text{The element symbol : }}_{\text{Z}}^{\text{A}}\text{X,}\\ \text{where, A (mass number) = no}\text{.of protons + no}\text{.of neutrons}\text{.}\\ \text{ Z (atomic number) = no}\text{. of protons}\text{. (electrons = protons)}\text{.}\end{array}$

Answer to Problem 2.64QP

${}^{\text{43}}\text{Al has totally 30 neutrons}\text{.}$

Explanation of Solution

Predict the neutron numbers in the ${}^{\text{43}}\text{Al}$.

  $\begin{array}{l}{\text{The element symbol : }}_{}^{\text{43}}\text{Al,}\\ \text{Z (atomic number) of Aluminum is 13}\\ \text{No}\text{.of neutrons = mass number - atomic number}\text{.}\\ \text{ no}\text{.of neutrons = 43 - 13 = 30}\\ \text{The total neutrons are : 30}\text{.}\end{array}$