Chapter 12, Problem 12.26QP

(a)

Interpretation Introduction

Interpretation: The density and the radius of the simple cubic phase of $\text{Ca}$ atom is to be calculated.

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 one lattice point is present at each of the eight 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.

To determine: The density of Ca metal in simple cubic phase.

Answer to Problem 12.26QP

Solution

The density of Ca metal in simple cubic phase is $\underset{\_}{0.809g/c{m}^3}$.

Explanation of Solution

Explanation

The number of atom for simple cubic unit cell, $\text{Z}$ is $1$.

The molar mass of Calcium metal, $\text{M}$ is, $40.08\text{g/mol}$.

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

The standard value of radius of Calcium metal in FCC phase is $231\text{pm}$.

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

Therefore, the radius of $\text{Ca}$ metal will be written as $231\times {10}^{-10}\text{cm}$.

The value of edge length of a simple cubic unit cell is Calculated by the expression,

$a=2r$

Where,

• $a$ is the edge of simple unit cell
• $r$ is the radius

The radius of Calcium atoms $\left(r\right)$ is $231\times {10}^{-10}\text{cm}$.

Substitute this value of radius in the edge length expression.

$\begin{array}{c}a=2\times 231\times {10}^{-10}\text{cm}\\ =462\times {10}^{-10}\text{cm}\end{array}$

The density of simple unit cell is calculated by the formula,

$\rho =\frac{\text{Z×M}}{a^3{\text{×N}}_{\text{A}}}$

Where,

• $\rho$ is the density of the unit cell.
• $\text{Z}$ is the number of atoms present per unit cell.
• $\text{M}$ is the molar mass of the element.
• $a$ is the edge length of the unit cell.

Substitute the values of $\text{Z}$, $\text{M}$, ${\text{N}}_{\text{A}}$ and $a$ in the above formula and Calculate the density of Calcium metal in simple cubic phase.

$\rho =\frac{\text{Z×M}}{a^3{\text{×N}}_{\text{A}}}$

$\begin{array}{c}\rho =\frac{1\text{×48}\text{.08}\text{g/mol}}{9.86\times \text{10}{\text{cm}}^3\text{×6}\text{.022}\times {\text{10}}^{23}{\text{mol}}^{-1}}\\ =\frac{\text{48}\text{.08}\text{g}}{59.38{\text{cm}}^3}\\ =\underset{\_}{0.809g/c{m}^3}\end{array}$

Therefore, The density of Ca metal in simple cubic phase is $\underset{\_}{0.809g/c{m}^3}$.

Conclusion

The density of Ca metal in simple cubic phase is $\underset{\_}{0.809g/c{m}^3}$.

(b)

Interpretation Introduction

To determine: The radius of Ca metal in simple cubic phase.

Answer to Problem 12.26QP

Solution

The radius of Ca metal in simple cubic phase is $\underset{\_}{1.86\times 10^{-8}cm}$.

Explanation of Solution

Explanation

The values of $\text{Z}$, $\text{M}$, and ${\text{N}}_{\text{A}}$ will remain the same for Calcium metal in simple cubic phase as above.

The standard value of density of $\text{Ca}$ metal in FCC is $1.55{\text{g/cm}}^{\text{3}}$

The value of edge length of a simple cubic unit cell is Calculated by the expression,

$a=2r$

Where,

• $a$ is the edge of simple unit cell.
• $r$ is the radius.

The density of simple unit cell is Calculated by the formula,

$\rho =\frac{\text{Z×M}}{a^3{\text{×N}}_{\text{A}}}$

Where,

• $\rho$ is the density of the unit cell.
• $\text{Z}$ is the number of atoms present per unit cell.
• $\text{M}$ is the molar mass of the element.
• $a$ is the edge length of the unit cell.

Substitute the values of $\text{Z}$, $\text{M}$, ${\text{N}}_{\text{A}}$ and $a$ in the above formula and Calculate the radius of Calcium metal in simple cubic phase.

$\begin{array}{c}\rho =\frac{\text{Z×M}}{a^3{\text{×N}}_{\text{A}}}\\ 1.55{\text{g/cm}}^{\text{3}}=\frac{1\text{×48}\text{.08}\text{g/mol}}{{\left(2r\right)}^3\text{×6}\text{.022}\times {\text{10}}^{23}{\text{mol}}^{-1}}\\ r^3=\frac{48.08\text{g}}{8\times 1.55{\text{g/cm}}^{\text{3}}\times \text{6}\text{.022}\times {\text{10}}^{23}}\\ r=\underset{\_}{1.86\times 10^{-8}cm}\end{array}$

Therefore, the radius of Ca metal in simple cubic phase is $\underset{\_}{1.86\times 10^{-8}cm}$.

Conclusion

The radius of Ca metal in simple cubic phase is $\underset{\_}{1.86\times 10^{-8}cm}$.