Chapter 12, Problem 12.28QP

(a)

Interpretation Introduction

Interpretation: The given data of molybdenum atom is consistent with the data of whether simple cubic, body-centered cubic or face-centered cubic are to be identified.

Concept introduction: In simple cubic crystal structure there is one lattice point on each corner of the cubic unit cell that is there are $8$ corners and $1$ lattice point is present at each of the $8$ corners.

Body centered cubic cell form consists of one lattice point at the center with eight corner lattice points.

Face centered cubic cell form consists of lattice point on all the six faces of the cubic unit cell with eight corner lattice points.

The density is defined as mass per volume of the substance.

To determine: If the given data of molybdenum atom is consistent with the data of a simple cubic unit cell or not.

Answer to Problem 12.28QP

Solution

The given data of molybdenum atom is not consistent with the data of a simple cubic unit cell.

Explanation of Solution

Explanation

The given density of crystalline form of molybdenum $\left(d\right)$ is $10.28{\text{g/cm}}^{\text{3}}$

The given radius of molybdenum $\left(r\right)$ is $139\text{pm}$

As $1\text{pm}={10}^{-10}\text{cm}$

Then the given radius will be written as $139\times {10}^{-10}\text{cm}$

The data of molybdenum unit cell is consistent with the data of simple cubic unit cell or not will be identified by comparing the values of given density with the calculated density value for simple cubic unit cell.

Density is calculated by the formula,

$\begin{array}{c}\text{Density}\left(d\right)=\frac{\text{Mass}\left(m\right)}{\text{Volume}\left(V\right)}\\ d=\frac{m}{V}\end{array}$ (1)

Where,

• $d$ is the density of the substance.
• $m$ is the mass of the substance.
• $V$ is the volume of the substance.

The density of simple cubic unit cell is calculated by calculating the value of mass and volume.

The value of atomic mass of molybdenum $\left(\text{A}\right)$ is $95.94\text{g}$.

The standard value of Avogadro’s number, ${\text{N}}_{\text{A}}$ is, $6.022\times {10}^{23}\text{atoms}$.

The mass of simple cubic unit cell is calculated by the formula,

$m=\frac{n\text{A}}{{\text{N}}_{\text{A}}}$ (2)

Where,

• $n$ is the number of atoms per unit cell.
• $\text{A}$ is the atomic mass of an atom.
• ${\text{N}}_{\text{A}}$ is Avogadro’s number.

Substitute the formula of mass, obtained in equation $\left(2\right)$ in equation $\left(1\right)$

$d=\left[\frac{n\text{A}}{{\text{N}}_{\text{A}}}\right]\frac{1}{V}$ (3)

In simple cubic unit cell there is one lattice point on each corner of the cubic unit cell that is there are $8$ lattice points.

The number of atoms $\left(n\right)$ is calculated by the formula,

$n=\frac{1}{8}\times \left(\text{Number}\text{of}\text{lattice}\text{points}\right)$

Substitute the value of the number of lattice points in the above expression.

$\begin{array}{c}n=\frac{1}{8}\times 8\\ =1\end{array}$

Thus, the value of $n$ is $1$. Therefore, one atom is present in one unit cell of simple cubic.

The volume is calculated by the formula,

Volume $\left(V\right)=l^3$ (4)

Where,

• $l$ is the edge length of the unit cell

The value of edge length of a simple cubic unit cell is calculated by the formula,

$l=2r$

Where,

• $r$ is the radius.

The given radius of molybdenum atoms $\left(r\right)$ is $139\times {10}^{-10}\text{cm}$.

Substitute this value of radius in the edge length formula.

$\begin{array}{l}l=2r\\ l=2\times 139\times {10}^{-10}\text{cm}\\ =278\times {10}^{-10}\text{cm}\end{array}$

Substitute the value of $l$ in equation (4).

$\begin{array}{c}\text{Volume}\left(V\right)=l^3\\ V={\left(278\times {10}^{-10}\text{cm}\right)}^3\\ V=2.15\times {10}^{-23}{\text{cm}}^{\text{3}}\end{array}$

Substitute the values of $n$, $\text{A}$, ${\text{N}}_{\text{A}}$ and $V$ in equation (3).

$\begin{array}{l}d=\left[\frac{n\text{A}}{{\text{N}}_{\text{A}}}\right]\frac{1}{V}\\ d=\left[\frac{1\text{atom}\times 95.94\text{g}}{6.022\times {10}^{23}\text{atoms}}\right]\frac{1}{2.15\times {10}^{-23}{\text{cm}}^{\text{3}}}\\ d=\left[\frac{1\text{atom}\times 95.94\text{g}}{6.022\times {10}^{23}\text{atoms}\times 2.15\times {10}^{-23}{\text{cm}}^{\text{3}}}\right]\\ d=7.41{\text{g/cm}}^{\text{3}}\end{array}$

Thus, the calculated value of density of simple cubic unit cell is $7.41{\text{g/cm}}^{\text{3}}$.

This does not match the density of molybdenum unit cells that is $10.28{\text{g/cm}}^{\text{3}}$.

Hence, it is proven that the given data of molybdenum atom is not consistent with the data of a simple cubic unit cell.

(b)

Interpretation Introduction

To determine: If the given data of molybdenum atom is consistent with the data of a body-centered cubic unit cell or not.

Answer to Problem 12.28QP

Solution

The given data of molybdenum atom is not consistent with the data of a body-centered cubic unit cell.

Explanation of Solution

Explanation

The BCC structure consists of atoms at each corner of the cubic unit cell and one atom at the center of the unit cell. There are $8$ corners, that is eight lattice points with $1$ atoms per corner and $1$ atom in the body center.

The number of atoms present is calculated by the formula,

$\text{Number}\text{of}\text{atoms}=\frac{1}{8}\times \left(\text{Number}\text{of}\text{lattice}\text{points}\right)+1$

The number of atoms present is,

$8\times \frac{1}{8}+1=2$

Substitute the value of the number of lattice points in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{atoms}=\frac{1}{8}\times \left(\text{Number}\text{of}\text{lattice}\text{points}\right)+1\\ =\frac{1}{8}\times \left(\text{8}\right)+1\\ =2\end{array}$

Thus, there are $2$ atoms are present in one unit cell.

Hence, $n=2$.

The value of edge length of a body-centered cubic unit cell is calculated by the formula,

$r=\frac{l\sqrt{3}}{4}$

The standard value of $\frac{\sqrt{3}}{4}=0.4330$

Rearrange the above formula to calculate edge length.

$l=\frac{r}{0.4330}$

The given radius of molybdenum atoms $\left(r\right)$ is $139\times {10}^{-10}\text{cm}$.

Substitute this value of radius in the above expression.

$\begin{array}{l}l=\frac{139\times {10}^{-10}\text{cm}}{0.4330}\\ =321\times {10}^{-10}\text{cm}\end{array}$

Substitute the value of $l$ in equation (4).

$\begin{array}{c}\text{Volume}\left(V\right)=l^3\\ V={\left(321\times {10}^{-10}\text{cm}\right)}^3\\ V=3.31\times {10}^{-23}{\text{cm}}^{\text{3}}\end{array}$

Substitute the values of $n$, $\text{A}$, ${\text{N}}_{\text{A}}$ and $V$ in equation (3).

$\begin{array}{l}d=\left[\frac{n\text{A}}{{\text{N}}_{\text{A}}}\right]\frac{1}{V}\\ d=\left[\frac{2\text{atoms}\times 95.94\text{g}}{6.022\times {10}^{23}\text{atoms}}\right]\frac{1}{3.31\times {10}^{-23}{\text{cm}}^{\text{3}}}\\ d=\left[\frac{2\text{atoms}\times 95.94\text{g}}{6.022\times {10}^{23}\text{atoms}\times 3.31\times {10}^{-23}{\text{cm}}^{\text{3}}}\right]\\ d=9.63{\text{g/cm}}^{\text{3}}\end{array}$

Thus, the calculated value of density of body-centered cubic unit cell is $9.63{\text{g/cm}}^{\text{3}}$.

This does not match the density of molybdenum unit cells that is $10.28{\text{g/cm}}^{\text{3}}$

Hence, it is proven that the given data of molybdenum atom is not consistent with the data of a body-centered cubic unit cell

(c)

Interpretation Introduction

To determine: If the given data of molybdenum atom is consistent with the data of a face-centered cubic unit cell or not.

Answer to Problem 12.28QP

Solution

The given data of molybdenum atom is consistent with the data of a face-centered cubic unit cell.

Explanation of Solution

Explanation

The FCC structure consists of atoms at each of the $8$ corner in addition to the atoms present at all faces of the cube. The atoms that are present at the face of the FCC structure are shared by two adjacent unit cells which provide $\frac{1}{2}$ atom to one of the unit cell.

The number of atoms in FCC structure is calculated by the formula,

$\text{Number}\text{of}\text{atoms}=\frac{1}{8}\times \left(\text{Number}\text{of}\text{corner}\text{atoms}\right)+\frac{1}{2}\times \left(\text{Number}\text{of}\text{atoms}\text{of}\text{the}\text{faces}\right)$

Substitute the required values in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{atoms}=\frac{1}{8}\times \left(\text{Number}\text{of}\text{corner}\text{atoms}\right)+\frac{1}{2}\times \left(\text{Number}\text{of}\text{atoms}\text{of}\text{the}\text{faces}\right)\\ =\left(\frac{1}{8}\times 8\right)+\left(\frac{1}{2}\times 6\right)\\ =4\end{array}$

Thus, there are $4$  atoms are present in one unit cell.

Hence, $n=4$

The value of edge length of a face-centered cubic unit cell is calculated by the formula,

$r=\frac{l\sqrt{2}}{4}$

The standard value of $\frac{\sqrt{2}}{4}=0.3536$

Rearrange the above formula to calculate edge length.

$l=\frac{r}{0.3536}$

The given radius of molybdenum atoms $\left(r\right)$ is $139\times {10}^{-10}\text{cm}$.

Substitute this value of radius in the edge length formula as,

$\begin{array}{l}l=\frac{139\times {10}^{-10}\text{cm}}{0.3536}\\ =393.1\times {10}^{-10}\text{cm}\end{array}$

Substitute the value of $l$ in equation (4).

$\begin{array}{c}\text{Volume}\left(V\right)=l^3\\ V={\left(393.1\times {10}^{-10}\text{cm}\right)}^3\\ V=6.10\times {10}^{-23}{\text{cm}}^{\text{3}}\end{array}$

Substitute the values of $n$, $\text{A}$, ${\text{N}}_{\text{A}}$ and $V$ in equation (3).

$\begin{array}{l}d=\left[\frac{n\text{A}}{{\text{N}}_{\text{A}}}\right]\frac{1}{V}\\ d=\left[\frac{4\text{atoms}\times 95.94\text{g}}{6.022\times {10}^{23}\text{atoms}}\right]\frac{1}{6.10\times {10}^{-23}{\text{cm}}^{\text{3}}}\\ d=\left[\frac{4\text{atoms}\times 95.94\text{g}}{6.022\times {10}^{23}\text{atoms}\times 6.10\times {10}^{-23}{\text{cm}}^{\text{3}}}\right]\\ d=10.45{\text{g/cm}}^{\text{3}}\end{array}$

Thus, the calculated value of density of face-centered cubic unit cell is $10.45{\text{g/cm}}^{\text{3}}$.

This matches approximately with the density of molybdenum unit cells that is $10.28{\text{g/cm}}^{\text{3}}$.

Hence, it is proven that the given data of molybdenum atom is consistent with the data of a face-centered cubic unit cell.

Conclusion

The given data of molybdenum atom is consistent with the data of a face-centered cubic unit cell.