Chapter 18, Problem 18.47QA

Interpretation Introduction

To find:

The type of unit cell of the copper.

Answer to Problem 18.47QA

Solution:

The crystalline form of copper has a face-centered unit cell.

Explanation of Solution

1) Concept:

For each type of unit cell, we can determine the number of atoms in the unit cell and calculate the mass of the unit cell using the molar mass of copper. The radius of the atom can be used to calculate the edge length. The volume of the unit cell can be determined from the edge length. Using the mass of the unit cell and its volume, we can determine the density for each type of unit cell. The calculated density that matches with the given density $8.95\frac{g}{c{m}^3}$ will determine the type of unit cell.

2) Formula:

i) $V=l^3$

ii) $d=\frac{m}{V_{unit cell}}$

iii) $r=0.500 l$

iv) $r=0.4330 l$

v) $r=0.3536 l$

3) Given:

i) Density $d=8.95\frac{g}{c{m}^3}$

ii) Radius $r=127.8 pm$

4) Calculations:

a. For a simple cubic unit cell:

Number of atoms per unit cell = 1

Mass of 1 atom of copper is calculated as

$m=1 atom Cu \times \frac{1 mol Cu}{6.0221 \times {10}^{23}atoms Mo} \times \frac{63.55 g Cu }{1 mol Cu }=1.055 \times {10}^{-22}g Cu$

Calculating the edge length:

$r=0.500 \times l$

$127.8 pm=0.500 \times l$

$l=\frac{127.8 pm}{0.500}= 255.6 pm$

Volume of unit cell:

$V=l^3={\left(255.6 pm\right)}^3 \times \frac{{\left({10}^{-10}cm \right)}^3}{{\left( 1 pm\right)}^3}=1.669 \times {10}^{-23}c{m}^3$

Calculating the density of unit cell:

$d=\frac{m}{V_{unit cell}}= \frac{1.055 \times {10}^{-22}g}{1.669\times {10}^{-23}c{m}^3}=6.321\frac{g}{c{m}^3}$

b. For a bcc unit cell:

Number of atoms per cell = 2

Mass of 2 atoms of copper is calculated as

$m=2 atom Cu \times \frac{1 mol Cu}{6.0221 \times {10}^{23}atoms Cu} \times \frac{63.55 g Cu }{1 mol Cu }=2.110 \times {10}^{-22}g Cu$

Calculating the edge length:

$r=0.4330 \times l$

$127.8 pm=0.4330 \times l$

$l=\frac{127.8 pm}{0.4330}= 295.150 pm$

Volume of unit cell:

$V=l^3={\left(295.150 pm\right)}^3 \times \frac{{\left({10}^{-10}cm \right)}^3}{{\left( 1 pm\right)}^3}=2.571 \times {10}^{-23}c{m}^3$

Calculating the density of unit cell:

$d=\frac{m}{V_{unit cell}}= \frac{2.110 \times {10}^{-22}g}{2.571 \times {10}^{-23}c{m}^3}=8.206 \frac{g}{c{m}^3}$

c. For an fcc unit cell:

Number of atoms per cell = 4

Mass of 4 atoms of copper is calculated as

$m=4 atom Cu \times \frac{1 mol Cu}{6.0221 \times {10}^{23}atoms Cu} \times \frac{63.55 g Cu }{1 mol Cu }=4.221 \times {10}^{-22}g Cu$

Calculating the edge length:

$r=0.3536 \times l$

$127.8 pm=0.3536 \times l$

$l=\frac{127.8 pm}{0.3536}= 361.425 pm$

Volume of unit cell:

$V=l^3={\left(361.425pm\right)}^3 \times \frac{{\left({10}^{-10}cm \right)}^3}{{\left( 1 pm\right)}^3}=4.721\times {10}^{-23}c{m}^3$

Calculating the density of unit cell:

$d=\frac{m}{V_{unit cell}}= \frac{4.221\times {10}^{-22}g}{4.721\times {10}^{-23}c{m}^3}=8.94 \frac{g}{c{m}^3}$

Thus, the calculated density

$8.94 \frac{g}{c{m}^3}$

in case of fcc unit cell matches with the given density

$8.95\frac{g}{c{m}^3}$

Thus, the face-centered cubic unit cell is consistent with the given data.

Conclusion:

The type of unit cell is determined by calculating the density of copper in each type of unit cell and comparing it with the given density.