Chapter 2, Problem 2.78QP

(a)

Interpretation Introduction

Interpretation:

Rutherford’s experiment has to be described and how it led to the structure of the atom should be explained.  Also, how was he able to estimate the number of protons in a nucleus from the scattering of alpha particles should be explained.

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.

The symbol of an element is ${}_{\text{Z}}^{\text{A}}\text{X}$.

Where,

  $\begin{array}{l}\text{A (mass number) = no}\text{.of protons + no}\text{.of neutrons}\text{.}\\ \text{Z (atomic number) = no}\text{. of protons}\text{. (electrons = protons)}\text{.}\end{array}$

Explanation of Solution

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

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.

Figure 1

Rutherford’s Experiment evidences about:

• The atom mostly consist of empty space (due to empty space most of the rays went through the foil).
• Very solid particles; presence of nucleus is revealed by the rays that bounced back.
• The nucleus consisting of positive charge is confirmed by the rays deflected at an angle.  The similar charges have no attraction and are deflected away.

(b)

Interpretation Introduction

Interpretation:

The density of the nucleus $\left({\text{g/cm}}^{\text{3}}\right)$ and density of the space occupied by the electrons in $\text{Na}$ atom has to be calculated.  Also whether the results support Rutherford’s model of an atom or not should be explained.

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.

The symbol of an element is ${}_{\text{Z}}^{\text{A}}\text{X}$.

Where,

  $\begin{array}{l}\text{A (mass number) = no}\text{.of protons + no}\text{.of neutrons}\text{.}\\ \text{Z (atomic number) = no}\text{. of protons}\text{. (electrons = protons)}\text{.}\end{array}$

Explanation of Solution

Given:

Radius of the nucleus is $\text{3}{\text{.04×10}}^{\text{-15}}\text{m}$

Mass of the nucleus is $\text{3}{\text{.82×10}}^{\text{-23}}\text{g}$

Calculation of density of the nucleus.

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

  $\begin{array}{l}\text{V }\text{= }\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\\ \text{= }\frac{\text{4}}{\text{3}}\text{π}\text{}\text{ (3}{\text{.04×10}}^{\text{-13}}{\text{cm)}}^{\text{3}}\\ \text{=}\text{1}{\text{.177×10}}^{\text{-37}}{\text{cm}}^{\text{3}}\end{array}$

From which density of the nucleus is,

  $\begin{array}{l}\text{d }\text{= }\frac{\text{m}}{\text{V}}\\ \text{= }\frac{\text{3}{\text{.82×10}}^{\text{-23}}\text{g}}{\text{1}{\text{.177×10}}^{\text{-37}}{\text{cm}}^{\text{3}}}\\ \text{= }3.25\times 10^{14}g/c{m}^3\end{array}$

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

Calculation of density of space occupied by electrons in sodium atom.

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

Mass of 11 electrons are:

$\text{11electrons × }\frac{\text{9}{\text{.1094×10}}^{\text{-28}}\text{ g}}{\text{1 electron}}\text{ = 1}{\text{.00203 × 10}}^{\text{-26}}\text{ g}$

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:

Conversion of pm to cm:

  $\begin{array}{l}\text{186 pm × }\frac{{\text{1×10}}^{\text{-12}}\text{ g}}{\text{1 pm}}\text{ ×}\frac{\text{1cm}}{{\text{1×10}}^{\text{-2}}\text{m}}\text{ = 1}{\text{.86 × 10}}^{\text{-8}}\text{ cm}\text{.}\\ \\ {\text{V}}_{\text{atom}}\text{ = }\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\\ \text{= }\frac{\text{4}}{\text{3}}\text{ π (1}{\text{.86×10}}^{\text{-8}}{\text{cm)}}^{\text{3}}\\ \text{= 2}{\text{.695 × 10}}^{\text{-23}}{\text{ cm}}^{\text{3}}\\ \\ {\text{V}}_{\text{electrons}}{\text{ = V}}_{\text{atom}}{\text{- V}}_{\text{nucleus}}\\ \text{= (2}{\text{.695 × 10}}^{\text{-23}}{\text{ cm}}^{\text{3}}\text{) - (1}{\text{.177 × 10}}^{\text{-37}}{\text{ cm}}^{\text{3}}\text{)}\\ \text{ =}\text{2}{\text{.695 × 10}}^{\text{-23}}{\text{ cm}}^{\text{3}}\text{.}\end{array}$

The volume occupied by the nucleus is significant compared to the space occupied by the atoms.

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

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

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