Chapter 12, Problem 12.29QP

Interpretation Introduction

Interpretation: The density of the FCC form of sodium $\left(\text{Na}\right)$  metal is to be calculated.

Concept introduction: 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: The density of the FCC form of $\text{Na}$ metal.

Answer to Problem 12.29QP

Solution

The density of the FCC form of sodium $\left(\text{Na}\right)$  metal is $\underset{\_}{1.06g/c{m}^3}$.

Explanation of Solution

Explanation

The given density of crystalline form of sodium atom $\left(d\right)$ in BCC form is $0.971{\text{g/cm}}^{\text{3}}$.

The given radius of sodium $\left(r\right)$ is $186\text{pm}$.

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

Then the given radius will be written as $186\times {10}^{-10}{\text{cm}}^3$.

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

Sodium metal crystallizes with a BCC structure at normal atmospheric pressure.

The value of edge length of a body-centered cubic unit cell will be calculated by the formula given below,

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

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

Therefore by rearranging the above formula to calculate edge length will be,

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

The given radius of sodium atoms $\left(r\right)$ is $186\times {10}^{-10}{\text{cm}}^3$.

Substitute this value of radius in the edge length formula.

$\begin{array}{c}l=\frac{186\times {10}^{-10}\text{cm}}{0.4330}\\ =429.6\times {10}^{-10}\text{cm}\end{array}$

The volume of the sphere is calculated by the formula.

$V=l^3$

Where,

• $V$ is the volume of the sphere.
• $l$ is the edge length of the unit cell.

Substitute the value of $l$ in the above formula.

$\begin{array}{c}V=l^3\\ ={\left(321\times {10}^{-10}\text{cm}\right)}^3\\ =7.93\times {10}^{-23}{\text{cm}}^{\text{3}}\end{array}$

The formula used for density is,

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

Where,

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

Rearrange the above formula of density to calculate the mass of the unit cell.

$\begin{array}{l}\rho =\frac{m}{V}\\ m=\rho \times V\end{array}$

Substitute the value of volume $\left(V\right)$ and the given density of $\text{Na}$ in above expression.

$\begin{array}{c}m=\rho \times V\\ =0.971{\text{g/cm}}^{\text{3}}\times 7.93\times {10}^{-23}{\text{cm}}^{\text{3}}\\ =7.70\times {10}^{-23}\text{g}\end{array}$

In BCC the number of atoms present in one unit is $2$.

Therefore the exact mass of one atom of sodium atom will be calculated by dividing the mass obtained with the number of atoms present in BCC structure of $\text{Na}$ atom as,

$\begin{array}{c}\text{Mass}\text{of}\text{one}\text{atom}\text{of}\text{Na}=\frac{\text{obtained}\text{mass}}{\text{number}\text{of}\text{atoms}\text{present}\text{in}\text{BCC}\text{form}}\\ =\frac{7.70\times {10}^{-23}\text{g}}{2}\\ =3.85\times {10}^{-23}\text{g}\end{array}$

Therefore, mass $\left(m\right)$ of $\text{Na}$ atom is $3.85\times {10}^{-23}\text{g}$.

At $613,000\text{atm}$ the structure of sodium metal becomes FCC.

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 value of edge length of a face-centered cubic unit cell will be calculated by the formula given below,

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

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

Therefore, by rearranging the above formula to calculate edge length will be,

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

The given radius of sodium atoms $\left(r\right)$ is $186\times {10}^{-10}{\text{cm}}^3$.

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

$\begin{array}{c}l=\frac{186\times {10}^{-10}\text{cm}}{0.3536}\\ =526.01\times {10}^{-10}\text{cm}\end{array}$

Substitute the value of $l$ in the above formula of volume as,

$\begin{array}{c}V=l^3\\ ={\left(526.01\times {10}^{-10}\text{cm}\right)}^3\\ =1.45\times {10}^{-22}{\text{cm}}^{\text{3}}\end{array}$

The number of atoms present in one FCC unit cell is $4$.

Substitute the value of volume $\left(V\right)$, number of atoms present in FCC unit cell and the calculated mass $\left(m\right)$ of $\text{Na}$ in above formula.

$\begin{array}{c}\rho =\frac{m}{V}\\ \rho =\frac{4\times 3.85\times {10}^{-23}\text{g}}{1.45\times {10}^{-22}{\text{cm}}^3}\\ =\underset{\_}{1.06g/c{m}^3}\end{array}$

Therefore, the density of the FCC form of sodium $\left(\text{Na}\right)$  metal is $\underset{\_}{1.06g/c{m}^3}$.

Conclusion

The density of the FCC form of sodium $\left(\text{Na}\right)$  metal is $\underset{\_}{1.06g/c{m}^3}$.