Chapter 2, Problem 2.7P

To determine

(a)

Thenumbers of atoms for FCC nickel (Ni).

Answer to Problem 2.7P

The numbers of atoms for FCC nickel (Ni) is $91700\text{atoms}$.

Explanation of Solution

Given:

The side of cube is $10\text{nm}$

Lattice parameter for nickel is $0.352\text{nm}$

Formula used:

Write the expression to find the number of atoms per unit volume.

$n_{\text{a}}=\frac{\text{Number}\text{of}\text{atoms}}{a^3}$ …… (I)

Here, $n_{\text{a}}$ is the number of atoms per unit volume and the $\text{Number}\text{of}\text{atoms}$ isthe numbers of atoms in a unit cell.

Write the expression to find the number of atom for the volume.

$N=n_{\text{a}}{a}_1{}^3$ …… (II)

Here $N$ is the number of atoms for the cube, $a_1$ is the side of cube.

Calculation:

Nickel has Face Centered Cubic (FCC) structure with $4$ number of atoms.

Substitute $4\text{atoms}$ for $\text{Number}\text{of}\text{atoms}$ and $0.352\text{nm}$ for $a$ in equation (I) to find $n_{\text{a}}$.

$\begin{array}{c}{n}_{\text{a}}=\frac{4\text{atoms}}{{\left(0.352\text{nm}\right)}^3}\\ =\frac{4\text{atoms}}{{\left(0.352\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}^3}\\ =\frac{4\text{atoms}}{0.044\times {10}^{-27}{\text{m}}^3}\\ =91.70\times {10}^{27}\text{atoms}/{\text{m}}^{\text{3}}\end{array}$

Substitute $91.70\times {10}^{27}\text{atoms}/{\text{m}}^{\text{3}}$ for $n_a$, and $10\text{nm}$ for $a_1$ in equation (II) to find $N$.

$\begin{array}{c}N=91.70\times {10}^{27}\text{atoms}/{\text{m}}^{\text{3}}{\left(10\text{nm}\right)}^3\\ =91.70\times {10}^{27}\text{atoms}/{\text{m}}^{\text{3}}{\left(10\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}^3\\ =\left(91.70\times {10}^{27}\text{atoms}/{\text{m}}^{\text{3}}\right)\times \left({10}^3\times {10}^{-27}\text{m}\right)\\ =91700\text{atoms}\end{array}$

Conclusion:

Therefore, the numbers of atoms for FCC nickel (Ni) is $91700\text{atoms}$.

To determine

(b)

The number of atoms by using the density of nickel.

Answer to Problem 2.7P

The number of atoms by using the density of nickel is $91300\text{atoms}$.

Explanation of Solution

Formula used:

Write the formula to find theatoms per unit volume.

$N_{\text{V}}=\frac{{\rho }_{\gamma \text{Ni}}\times N_{\text{A}}}{M_{\text{Ni}}}$ …… (III)

Here, $N_{\text{V}}$ is the atom per unit volume, ${\rho }_{\gamma \text{Ni}}$ is thedensity of Nickel, $M_{\text{Ni}}$ is the molecular weight of nickel, and $N_{\text{A}}$ is the Avogadro’s number.

Write the expression to find the number of atom for the volume.

$N=N_{\text{V}}{a}_1{}^3$ ……. (IV)

Here, $N$ is the number of atoms for the given volume.

Calculation:

The molecular weight of nickel is $58.71\times {10}^{-3}\text{kg}/\text{mole}$.

The density of nickel is $8.902\text{g}/{\text{cm}}^{\text{3}}$.

Avogadro’s number is $6.02\times {10}^{23}\text{atoms}/\text{mole}$.

Substitute $8.902\text{g}/{\text{cm}}^{\text{3}}$ for ${\rho }_{\gamma \text{Ni}}$, $58.71\times {10}^{-3}\text{kg}/\text{mole}$ for $M_{\text{Ni}}$ and $6.02\times {10}^{23}\text{atoms}/\text{mole}$ for $N_{\text{A}}$ in equation (II) to find $N_{\text{V}}$.

$\begin{array}{c}{N}_{\text{V}}=\frac{8.902\text{g}/{\text{cm}}^{\text{3}}\times 6.02\times {10}^{23}\text{atoms}/\text{mole}}{58.71\times {10}^{-3}\text{kg}/\text{mole}}\\ =\frac{8.902\text{g}/{\text{cm}}^{\text{3}}\times \left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\times \left(\frac{1{\text{cm}}^3}{{10}^{-6}{\text{m}}^{\text{3}}}\right)\times 6.02\times {10}^{23}\text{atoms}/\text{mole}}{58.71\times {10}^{-3}\text{kg}/\text{mole}}\\ =0.913\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}\end{array}$

Substitute $0.913\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}$ for $N_{\text{V}}$ and $10\text{nm}$ for $a_1$ in equation (IV) to find $N$.

$\begin{array}{c}N=0.913\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}{\left(10\text{nm}\right)}^3\\ =0.913\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}{\left(10\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}^3\\ =\left(91.3\times {10}^{27}\text{atoms}/{\text{m}}^{\text{3}}\right)\times \left({10}^3\times {10}^{-27}{\text{m}}^3\right)\\ =91300\text{atoms}\end{array}$

Conclusion:

Therefore, the number of atoms by using the density of nickel is $91300\text{atoms}$.