Chapter 18, Problem 18.48QA

Interpretation Introduction

To find:

The unit cell that is consistent with the given data for molybdenum

Answer to Problem 18.48QA

Solution:

(c) Face-centered cubic 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 molybdenum. 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 $10.28g/cm^3$ 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 $d=10.28g/c{m}^3$

ii) Radius $r=139 pm$

4) Calculations:

i) For a simple cubic unit cell:

Number of atoms per cell = $1$

Mass of $1$ atom of molybdenum is calculated as

$m=1 atom Mo \times \frac{1 mol Mo}{6.0221 \times {10}^{23}atoms Mo} \times \frac{95.96 g Mo }{1 mol Mo }=1.593 \times {10}^{-22}g Mo$

Calculating the edge length:

$r=0.500 \times l$

$139 pm=0.500 \times l$

$l=\frac{139 pm}{0.500}= 278 pm$

Volume of the unit cell:

$V=l^3={\left(278 pm\right)}^3 \times \frac{{\left({10}^{-10}cm \right)}^3}{{\left( 1 pm\right)}^3}=2.148 \times {10}^{-23}c{m}^3$

Calculating the density of the unit cell:

$d=\frac{m}{V_{unit cell}}= \frac{1.593 \times {10}^{-22}g}{2.148 \times {10}^{-23}c{m}^3}=7.416\frac{g}{c{m}^3}$

ii) For a bcc unit cell:

Number of atoms per cell = $2$

Mass of $2$ atoms of molybdenum is calculated as

$m=2 atom Mo \times \frac{1 mol Mo}{6.0221 \times {10}^{23}atoms Mo} \times \frac{95.96 g Mo }{1 mol Mo }=3.187 \times {10}^{-22}g Mo$

Calculating the edge length:

$r=0.4330 \times l$

$139 pm=0.4330 \times l$

$l=\frac{139 pm}{0.4330}= 321.016 pm$

Volume of the unit cell:

$V=l^3={\left(321.016 pm\right)}^3 \times \frac{{\left({10}^{-10}cm \right)}^3}{{\left( 1 pm\right)}^3}=3.308 \times {10}^{-23}c{m}^3$

Calculating the density of the unit cell:

$d=\frac{m}{V_{unit cell}}= \frac{3.187 \times {10}^{-22}g}{3.308 \times {10}^{-23}c{m}^3}=9.634 \frac{g}{c{m}^3}$

iii) For an fcc unit cell:

Number of atoms per cell = $4$

Mass of 4 atoms of molybdenum is calculated as

$m=4 atom Mo \times \frac{1 mol Mo}{6.0221 \times {10}^{23}atoms Mo} \times \frac{95.96 g Mo }{1 mol Mo }=6.374 \times {10}^{-22}g Mo$

Calculating the edge length:

$r=0.3536 \times l$

$139 pm=0.3536 \times l$

$l=\frac{139 pm}{0.3536}= 393.099 pm$

Volume of the unit cell:

$V=l^3={\left(393.099 pm\right)}^3 \times \frac{{\left({10}^{-10}cm \right)}^3}{{\left( 1 pm\right)}^3}=6.074 \times {10}^{-23}c{m}^3$

Calculating the density of the unit cell:

$d=\frac{m}{V_{unit cell}}= \frac{6.374 \times {10}^{-22}g}{6.074\times {10}^{-23}c{m}^3}=10.49 \frac{g}{c{m}^3}$

Thus, the calculated density

$10.49 \frac{g}{c{m}^3}$

in case of fcc unit cell is very close the given density

$10.28\frac{g}{c{m}^3}.$

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 molybdenum in each type of unit cell and comparing it with the given density.