Chapter 2, Problem 153GQ

Consider an atom of ⁶⁴Zn.

(a) Calculate the density of the nucleus in grams per cubic centimeter, knowing that the nuclear radius is 4.8 × 10^–6 nm and the mass of the ⁶⁴Zn atom is 1.06 × 10^–22 g. (Recall that the volume of a sphere is [4/3] πτ³.)

(b) Calculate the density of the space occupied by the electrons in the zinc atom, given that the atomic radius is 0.125 nm and the electron mass is 9.11 × 10^–28 g.

(c) Having calculated these densities, what state­ment can you make about the relative densities of the parts of the atom?

Interpretation Introduction

Interpretation: For given atom of ${}^{64}Zn$, the density of nucleus in grams should be calculated in per cubic centimeter using the given nuclear radium and the mass of the atom.

Concept introduction:

Equation for density from volume and mass is,

    $Density=\frac{Mass}{Volume}$

Equation for finding Volume of sphere is,

    $Volume=\left(4/3\right)\pi r^3$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.

Answer to Problem 153GQ

The density of zinc nucleus is $2.3\times {10}^{14}g/c{m}^3$

Explanation of Solution

The radius of the zinc nucleus is given that $4.8\times {10}^{-6}nm=4.8\times {10}^{-13}cm$.

Equation for finding Volume of sphere is,

    $Volume=\left(4/3\right)\pi r^3$

Therefore, the volume of zinc nucleus is,

    $\begin{array}{l}Volume=\left(4/3\right)3.14\times {(4.8\times {10}^{-13}cm)}^3\\ =463.01\times {10}^{-39}c{m}^3\end{array}$

Mass of the zinc atom is $1.06\times {10}^{-22}g$

Equation for density from volume and mass is,

    $Density=\frac{Mass}{Volume}$

Therefore, the density of zinc nucleus is,

    $Density=\frac{1.06\times {10}^{-22}g}{463.01\times {10}^{-39}c{m}^3}=2.3\times {10}^{14}g/c{m}^3$

Interpretation Introduction

Interpretation: For given atom of ${}^{64}Zn$, the density of space occupied by electrons in atom with atomic radius and the mass of electron should be calculated. 

Concept introduction:

Equation for density from volume and mass is,

    $Density=\frac{Mass}{Volume}$

Equation for finding Volume of sphere is,

    $Volume=\left(4/3\right)\pi r^3$

Answer to Problem 153GQ

The density of space occupied by electrons in Zn atom is $3.34\times {10}^{-3}g/c{m}^3$

Explanation of Solution

The radius of the zinc atom is given that $0.125nm=1.25\times {10}^{-8}cm$.

Equation for finding Volume of sphere is,

    $Volume=\left(4/3\right)\pi r^3$

Therefore, the volume of zinc nucleus is,

    $\begin{array}{l}Volume=\left(4/3\right)3.14\times {(1.25\times {10}^{-8}cm)}^3\\ =8.17\times {10}^{-24}c{m}^3\end{array}$

The volume of zinc nucleus is $463.01\times {10}^{-39}c{m}^3$, which negligibly small quantity, so the volume occupied by electrons in Zn atom is approximately equals to the volume of atom,

    $Volume=8.17\times {10}^{-24}c{m}^3$

Mass of the zinc atom is the sum of mass of nucleus and mass of total number of electrons, that is $30\times 9.11\times {10}^{-28}g=2.73\times {10}^{-26}g$

Equation for density from volume and mass is,

    $Density=\frac{Mass}{Volume}$

Therefore, the density of space occupied by electrons in Zn atom is,

    $Density=\frac{2.73\times {10}^{-26}g}{8.17\times {10}^{-24}c{m}^3}=3.34\times {10}^{-3}g/c{m}^3$

Interpretation Introduction

Interpretation: Considering the density of nucleus and the density of space occupied by electrons in given atoms the statement that describes about relative densities of parts of given atom should be identified.

Concept introduction:

Equation for density from volume and mass is,

    $Density=\frac{Mass}{Volume}$

Equation for finding Volume of sphere is,

    $\text{Volume}\text{=}\left(\text{4/3}\right){\text{πr}}^{\text{3}}$

Answer to Problem 153GQ

The density of nucleus is far greater than the density of space occupied by electrons (outside nucleus in atom).

Explanation of Solution

The density of zinc nucleus is found that $2.3\times {10}^{14}g/c{m}^3$

The density of space occupied by electrons in Zn atom is found that $3.34\times {10}^{-3}g/c{m}^3$.

The ratio of density of zinc nucleus to density of space occupied by electrons is,

    $\frac{2.3\times {10}^{14}g/c{m}^3}{3.34\times {10}^{-3}g/c{m}^3}=688.6$

Hence, the density of nucleus is far greater than the density of space occupied by electrons (outside nucleus in atom).