Chapter 2, Problem 125AP

Interpretation Introduction

Interpretation:

The statement, “Radius $r$ is proportional to the cube root of a mass number,” is to be proven, and the volume of the Li nucleus and the fraction of the atom’s volume are to be calculated.

Concept introduction:

The mass of an atom is contained in its nucleus, which is very small and dense.

In the nucleus, the positivelycharged particles are known as protons and the particles that are electrically neutral are known as neutrons.

The Rutherford model of an atom is also called theplanetary model of anatom. It is used to describe the structure of anatom.

The volume of the nucleus is represented by the expression:

$V=kA$

A nucleus is spherical in shape. The volume of a sphere is represented by the expression:

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

The volume of nucleus is calculated as

$V_{\text{nucleus}}=\frac{4}{3}\pi {\left(r_{\text{o}}{A}^{\frac{1}{3}}\right)}^3$

The nucleus occupies a fraction of the atomic radius is as

$\text{Fraction}=\frac{\text{ Volume of Li nucleus}}{\text{Atomic volume of Li nucleus}}$

Answer to Problem 125AP

Solution:

(a)The statement,“Radius $r$ is proportional to the cube root of a mass number,”has beenproved.

(b)The volume of the Li nucleus is $5.1\times {10}^{-44}{\text{ m}}^3$.

(c) The fraction of the atom’s volume is $3.4\times {10}^{-15}$. Yes, the results support the Rutherford model.

Explanation of Solution

a)The radius $r$ is proportional to the cube root of the mass number $A$.

Nucleons are defined as the sum of the protons and neutrons present in the nucleus of an atom.

The volume of the nucleusis represented by the expression:

$V=kA$

A nucleus is spherical in shape. The volume of a sphere is represented by the expression:

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

Therefore, by equating the expressionsas

$V=kA$ ….. (1)

Here, $A$ is the number of nucleons and $V$ is the volume of the nucleus.

$V=\frac{4}{3}\pi r^3$ …... (2)

Here, $r$ is the radius of the sphere.

By equating equation (1) and (2), we get

$\begin{array}{l}kA=\frac{4}{3}\pi r^3\\ \left(\frac{3}{4\pi }\right)kA=r^3\\ {\left[\left(\frac{3}{4\pi }\right)kA\right]}^{\frac{1}{3}}=r\\ {\left[\left(\frac{3}{4\pi }\right)k\right]}^{\frac{1}{3}}\left(A^{\frac{1}{3}}\right)=r\end{array}$

On solvingfurther, we get

$\text{c}{A}^{\frac{1}{3}}=r$

Here, $\text{c}$ is the constant.

Therefore, the radius $r$ is proportional to the cube root of a mass number.

b) The volume of the Li nucleus.

The radius of the nucleus is $r=r_{\text{o}}{A}^{\frac{1}{3}}$, where $r_{\text{o}}=1.2\times {10}^{-15}\text{ m}$.

Anucleus hasa spherical shape. The volume of a sphere is to be calculated as

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

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

The volume of nucleus is calculated as

$V_{\text{nucleus}}=\frac{4}{3}\pi {\left(r_{\text{o}}{A}^{\frac{1}{3}}\right)}^3$

Here, $A$ is the atomic number. For lithium $-7$, the atomic number is $7$.

Substitute the values of the radius and the atomic number in the above expression

$\begin{array}{l}{V}_{\text{nucleus}}=\left[\frac{4}{3}\pi \left(r_{\text{o}}^3\right)\right]\left(A\right)\\ V_{\text{nucleus}}=\left[\frac{4}{3}\pi {\left(1.2\times {10}^{-15}\text{ m}\right)}^3\right]\left(7\right)\\ V_{\text{nucleus}}=\left[\frac{4}{3}\frac{22}{7}{\left(1.2\times {10}^{-15}\text{ m}\right)}^3\right]\left(7\right)\\ V_{\text{nucleus}}=5.1\times {10}^{-44}{\text{ m}}^3\end{array}$

Therefore, the volume of the Linucleus is $5.1\times {10}^{-44}{\text{ m}}^3$.

c) The fraction of atom’s volume is occupied by its nucleus. The results that support the Rutherford model of an atomor not.

The radius of the ${}_3^7\text{Li}$ atom is $152\text{ pm}$.

The conversion ofpm into m is as

$152\text{ pm}=152\times {10}^{-12}\text{ m}$

The atomic volume is calculated as

A nucleus has a spherical shape. The volume of a sphere is

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

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

Substitute the values in the above equation

$\begin{array}{l}V=\frac{4}{3}\pi {\left(152\times {10}^{-12}\text{ m}\right)}^3\\ V=\frac{4}{3}\frac{22}{7}{\left(152\times {10}^{-12}\text{ m}\right)}^3\\ V=1.5\times {10}^{-29}{\text{ m}}^3\end{array}$

Thus, the atomic volume of lithium is $1.5\times {10}^{-29}{\text{ m}}^3$.

The nucleus occupies afraction of the atomic radius, which is as follows:

$\text{Fraction}=\frac{\text{ Volume of Li nucleus}}{\text{Atomic volume of Li nucleus}}$

Substitute the values in the above equation

$\begin{array}{l}\text{Fraction}=\left(\frac{5.1\times {10}^{-44}\cancel{{\text{m}}^3}}{1.5\times {10}^{-29}\cancel{{\text{m}}^3}}\right)\\ \text{Fraction}=3.4\times {10}^{-15}\end{array}$

Thus, the fraction of the atom’s volume is $3.4\times {10}^{-15}$.

As the nucleus contains a small region in the atom, therefore the results support the Rutherford model of an atom.