Chapter 2, Problem 120AP

(a) Describe Rutherford’s experiment and how the results revealed the nuclear structure of the atom. (b) Consider the ${}^{\text{23}}\text{Na}$ atom. Given that the radius and mass of the nucleus are $\text{3}\text{.04 }\times {\text{ 10}}^{\text{-15}}\text{m and 3}\text{.82 }\times {\text{ 10}}^{\text{-23}}$ g, respectively, calculate the density of the nucleus in g/cm³. The radius of a ${}^{\text{23}}\text{Na}$ atom is 186 pm. Calculate the density of the space occupied by the electrons outside the nucleus in the sodium atom. Do your results support Rutherford's model of an atom? [The volume of a sphere of radius $r\text{ is }\frac{4}{5}\pi r^3$.]

Interpretation Introduction

Interpretation:

The Rutherford’s experiment and its result as nuclear structure of atom is to be described and the density of the nucleus in ${\text{g/cm}}^3$ and the density of the space occupied by an electron in sodium atom is to be calculated. Also, the results support Rutherford’s model or not are to be predicted.

Concept Introduction:

The Rutherford model of an atom is also known as the planetary model of the atom. It is used to describe the structure of the atom.

The density of a substance is defined as the measurement of mass per unit volume and is calculated as $\text{density}=\frac{\text{mass}}{\text{volume}}$

Answer to Problem 120AP

Solution:

(a)

In the Rutherford model, the alpha particles are bombarded on gold foil, where some particles will pass and some will deflect back. The main observation is that the nucleus is present in the center of the atom and the size of the atom is larger than the size of the nucleus.

(b) The density of nucleus is $3.25\times {10}^{14}{\text{ g/cm}}^3$

The density of the electron is $3.72\times {10}^{-4}{\text{ g/cm}}^3$. Yes, the results support the Rutherford model.

Explanation of Solution

a) The Rutherford’s experiment and the results that revealed the nuclear structure of the atom.

The Rutherford model carries out an experiment that when alpha particles are bombarded on thin gold foil, then most of the alpha particles pass through it and some are deflected back through it. The main observation of Rutherford is that a positive charge situated at the center is the nucleus and the size of the nucleus is smaller than the size of the atom. The average of scattering tells about the number of protons, which is based on their electrostatic interactions.

Given information: The radius of the nucleus is $3.04\times {10}^{-15}\text{ m,}$ the mass of the nucleus is $3.82\times {10}^{-23}\text{ g,}$

and the radius of sodium is $186\text{ pm}$.

b) The density of nucleus in ${\text{g/cm}}^{\text{3}}$

and density of the space occupied by the electrons outside the nucleus in the sodium atom. The results come out support Rutherford’s model of an atom.

As the nucleus have a spherical shape, so the volume of a sphere is to be calculated as follows:

$V=\frac{4}{3}\pi r^3$

Here, $r$ is the radius of the sphere and $V$ is the volume of the sphere.

Substitute the values in the above equation as:

$\begin{array}{l}{V}_{\text{nucleus}}=\frac{4}{3}\pi {\left(3.04\times {10}^{-13}\text{ cm}\right)}^3\\ V_{\text{nucleus}}=\frac{4}{3}\times \frac{22}{7}{\left(3.04\times {10}^{-13}\text{ cm}\right)}^3\\ V_{\text{nucleus}}=1.177\times {10}^{-37}{\text{ cm}}^3.\end{array}$

Therefore, the volume of the nucleus is $1.177\times {10}^{-37}{\text{ cm}}^3.$

The density of the nucleus is to be calculated as follows:

$\text{density}=\frac{\text{mass}}{\text{volume}}$.

Substitute the value of the mass of the nucleus $\left(3.82\times {10}^{-23}\text{ g}\right)$ given and the volume of the nucleus in the above equation:

$\begin{array}{l}d=\frac{3.82\times {10}^{-23}\text{ g}}{1.177\times {10}^{-37}{\text{ cm}}^3}\\ d=3.25\times {10}^{14}{\text{ g/cm}}^3.\end{array}$

Therefore, the density of the nucleus is $3.25\times {10}^{14}{\text{ g/cm}}^3$.

The volume occupied by the electrons is to be calculated. The volume of the electrons is calculated by measuring the amount of difference in the volume of the nucleus and the volume of the atom.

The volume of an atom is calculated as follows:

Conversion of the radius of pm into cm is as follows:

$\begin{array}{l}\text{radius}=186\text{ pm}\left(\frac{\text{1}\times {\text{10}}^{-12}}{1\text{ pm}}\right)\left(\frac{1\text{ cm}}{1\times {10}^{-2}\text{ m}}\right)\\ \text{radius}=1.86\times {10}^{-8}\text{ cm}\text{.}\end{array}$

The volume of a sphere is $V=\frac{4}{3}\pi r^3.$

Here, $r$ is the radius of the sphere $\left(1.86\times {10}^{-8}\text{ cm}\right)$

and $V$ is the volume of the sphere.

Substitute the values in the above equation:

$\begin{array}{l}V=\frac{4}{3}\pi {\left(1.86\times {10}^{-8}\text{ cm}\right)}^3\\ V_{\text{atom}}=\frac{4}{3}\times \frac{22}{7}{\left(1.86\times {10}^{-8}\text{ cm}\right)}^3\\ V_{\text{atom}}=2.695\times {10}^{-23}{\text{ cm}}^3.\end{array}$

Therefore, the volume of the atom calculated is $2.695\times {10}^{-23}{\text{ cm}}^3$.

The volume of the electron is calculated by calculating the difference in the volume of the atom and the volume of the nucleus.

The volume of the electron is calculated as follows:

$V_{\text{electron}}=V_{\text{atom}}-V_{\text{nucleus}}$

Substitute the values in the above equation:

$\begin{array}{l}{V}_{\text{electron}}={\left(2.695\times {10}^{-23}\text{ cm}\right)}^3-\left(1.177\times {10}^{-37}{\text{ cm}}^3\right)\\ V_{\text{electron}}=2.695\times {10}^{-23}{\text{ cm}}^3.\end{array}$

Thus, the volume of the electron is $2.695\times {10}^{-23}{\text{ cm}}^3.$

The mass consists of $11$ electrons, which is to be calculated as follows:

$\text{mass}=\text{11 electron}\times \text{mass of 1 electron}\text{.}$

Substitute the value in the above equation:

$\begin{array}{l}\text{mass}=11\text{ electron}\times \frac{9.1094\times {10}^{-28}\text{ g}}{1\text{ electron}}\\ \text{mass}=1.00203\times {10}^{-26}\text{ g}\text{.}\end{array}$

Therefore, the mass of $11$ electrons is $1.00203\times {10}^{-26}\text{ g}\text{.}$

Now, the mass and volume of the electrons are calculated. The density of space occupied by an electron is to be calculated, which is as follows:

$\text{density}=\frac{\text{mass}}{\text{volume}}$.

Substitute the value of the mass of electron $\left(1.00203\times {10}^{-26}\text{ g}\right)$ and the volume of the electron $\left(2.695\times {10}^{-23}{\text{ cm}}^3\right)$ in the above equation:

$\begin{array}{l}d=\frac{1.00203\times {10}^{-26}\text{ g}}{2.695\times {10}^{-23}{\text{ cm}}^3}\\ d=3.72\times {10}^{-4}{\text{ g/cm}}^3.\end{array}$

Therefore, the density of the electron is $3.72\times {10}^{-4}{\text{ g/cm}}^3.$

Yes, the calculated result supports the Rutherford model of an atom. From the calculation, the space occupied by the electron is very less as compared to the space occupied by the nucleus of an atom. Therefore, it is proved that most of the space is empty, and the calculated density of the space occupied by the electrons also supports the Rutherford model of an atom.