Chapter 2, Problem 2.82QP

(a)

Interpretation Introduction

Interpretation: The density of the electrons in the atom to be calculated.

Concept Introduction:

Atoms: Atoms consist of tiny particles called protons, neutrons and electrons. Proton and neutrons are present in the nucleus and the electron resides around the nucleus. The protons number will be same as the electrons count in the atom.

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

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.

Answer to Problem 2.82QP

The atom consists of concentrated mass called nucleus at the center and surrounded by the electrons.

Explanation of Solution

Every atom contains a nucleus in which all of its positive charge and most of its mass are concentrated is proven by the experiment.

Experiment: Allowing alpha-particles (positively charged) to bombard with the gold foil, expected all the rays to pass through. In contrast, the rays are deflected with angles is observed.

Rutherford’s Experiment evidences about:

1. 1. The atom is mostly consist of empty space (due to most of the rays went through the foil).

1. 2. Very solid particles; presence of nucleus is revealed by the rays were bounced back.

1. 3. The nucleus consist of positive charge is confirmed by the rays deflected at an angle. The similar charges have no attraction and are deflected away.

Conclusion

The atomic structure was explained.

(b)

Interpretation Introduction

Interpretation: The density of the electrons in the atom to be calculated.

Concept Introduction:

Atoms: Atoms consist of tiny particles called protons, neutrons and electrons. Proton and neutrons are present in the nucleus and the electron resides around the nucleus. The protons number will be same as the electrons count in the atom.

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

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.

Answer to Problem 2.82QP

The density of the electrons is $3.72\times 10^{-4}g/c{m}^3$.

Explanation of Solution

Calculate the density of the nucleus.

Consider, the nucleus is spherical, and the volume of the nucleus is:

$\begin{array}{l}\text{V = }\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\\ \text{ = }\frac{\text{4}}{\text{3}}\text{π}\text{}\text{ (3}{\text{.04×10}}^{\text{-13}}{\text{cm)}}^{\text{3}}\text{=}\text{1}{\text{.177×10}}^{\text{-37}}{\text{cm}}^{\text{3}}\text{.}\\ \\ \text{From which the density of nucleus is calculated by :}\\ \text{d = }\frac{\text{m}}{\text{V}}\text{ = }\frac{\text{3}{\text{.82×10}}^{\text{-23}}\text{g}}{\text{1}{\text{.177×10}}^{\text{-37}}{\text{cm}}^{\text{3}}}\text{ = }3.25\times 10^{14}g/c{m}^3.\end{array}$

In order to calculate the density, the value of volume is must; which is calculated as shown above by considering the nucleus is spherical.

Calculate the density of space occupied by electrons in sodium atom.

To be required: The mass of $11$ electrons and the volume occupied by $11$ electrons.

$\begin{array}{l}\text{The mass of 11 electrons:}\\ \text{11electrons × }\frac{\text{9}{\text{.1094×10}}^{\text{-28}}\text{ g}}{\text{1 electron}}\text{ = 1}{\text{.00203 × 10}}^{\text{-26}}\text{ g}\end{array}$

The volume occupied by the electron is obtained by the differences between the volume of the atom and volume of nucleus.

The volume of atom is:

$\begin{array}{l}\underset{\_}{\text{Conversion of pm into cm:}}\\ \text{186 pm × }\frac{{\text{1×10}}^{\text{-12}}\text{ g}}{\text{1 pm}}\text{ ×}\frac{\text{1cm}}{{\text{1×10}}^{\text{-2}}\text{m}}\text{ = 1}{\text{.86 × 10}}^{\text{-8}}\text{ cm}\text{.}\\ \\ {\text{V}}_{\text{atom}}\text{ = }\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\text{ = }\frac{\text{4}}{\text{3}}\text{ π (1}{\text{.86×10}}^{\text{-8}}{\text{cm)}}^{\text{3}}\text{ = 2}{\text{.695 × 10}}^{\text{-23}}{\text{ cm}}^{\text{3}}\\ \\ {\text{V}}_{\text{electrons}}{\text{ = V}}_{\text{atom}}{\text{- V}}_{\text{nucleus}}\text{= (2}{\text{.695 × 10}}^{\text{-23}}{\text{ cm}}^{\text{3}}\text{) - (1}{\text{.177 × 10}}^{\text{-37}}{\text{ cm}}^{\text{3}}\text{)}\\ \text{ =}\text{2}{\text{.695 × 10}}^{\text{-23}}{\text{ cm}}^{\text{3}}\text{.}\\ \text{The volume occupied by the nucleus is insignificant compared to the space occupied by the atoms}\text{.}\end{array}$

Hence, the required terms are sufficient to calculate the density of the electrons:

$\text{d = }\frac{\text{m}}{\text{V}}\text{ = }\frac{\text{1}{\text{.00203×10}}^{\text{-26}}\text{ g}}{\text{2}{\text{.695×10}}^{\text{-23}}{\text{ cm}}^{\text{3}}}\text{ = }3.72\times 10^{-4}g/c{m}^3$

The density of the electrons is $3.72\times 10^{-4}g/c{m}^3$.

Conclusion

The density of the electrons in the atom is calculated.