Chapter 3, Problem 3.9P

To determine

(a)

The comparison between the magnitude of the total surface energy to the total bond energy.

Answer to Problem 3.9P

The magnitude of total energy of sphere is $-0.16$ times the total bond energy.

Explanation of Solution

Given:

The surface energy of $\text{Ni}$ is $1.90\text{J}/{\text{m}}^{\text{2}}$.

The cohesive energy of $\text{Ni}$ is $-4.435\text{eV}/\text{atom}$.

The lattice parameter of FCC Ni is $0.352\text{nm}$.

The total radius of the sphere is $1\text{atomic}\text{diameter}$.

Formula used:

The total surface energy of the sphere is given by

$E=\gamma \left(4\pi r^2\right)$   ....... (I)

Here, $E$ is the total surface energy of the sphere, $\gamma$ is the surface energy and $r$ is the atomic radius.

The atomic diameter is given by

$d_{\text{a}}=\frac{a}{2}\left[110\right]$   ....... (II)

Here, $d_{\text{a}}$ is the atomic diameter of a nickel and $a$ is the lattice parameter.

The total bond energy is given by,

$E_{\text{b}}=N{E}_{\text{C}}$   ....... (III)

Here, $E_{\text{b}}$ is the total bond energy, $N$ is the number of atoms and $E_{\text{C}}$ is the cohesive energy of Ni atom.

The ratio of total surface energy and bond energy is given by

$\frac{E}{E_{\text{b}}}$   ....... (IV)

Calculation:

The atomic diameter of the Ni atom is calculated as

Substitute $0.352\text{nm}$ for $a$ in equation (II).

$\begin{array}{c}{d}_{\text{a}}=\frac{0.352\text{nm}}{2}{\left[1^2+1^2+0^2\right]}^{\frac{1}{2}}\\ =\frac{0.352\text{nm}}{2}\left(\sqrt{2}\right)\\ =\frac{0.352\text{nm}}{\sqrt{2}}\\ =0.249\text{nm}\end{array}$

The atomic diameter of $\text{Ni}$ is equal to the radius of the sphere.

$r=0.249\text{nm}$

The total surface energy of the sphere is calculated as,

Substitute $0.249\text{nm}$ for $r$. $1.90\text{J}/{\text{m}}^{\text{2}}$ for $\gamma$ in equation (I).

$\begin{array}{c}E=\left(1.90\text{J}/{\text{m}}^{\text{2}}\right)\left[4\pi {\left(0.249\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}^2\right]\\ =\left(1.90\text{J}/{\text{m}}^{\text{2}}\right)\left(7.79\times {10}^{-19}{\text{m}}^{\text{2}}\right)\\ =1.48\times {10}^{-18}\text{J}\end{array}$

The total bond energy is calculated as

Substitute $-4.435\text{eV}/\text{atom}$ for $E_{\text{C}}$ and $13$ for $N$ in equation (III).

$\begin{array}{c}{E}_{\text{b}}=\left(13\text{atoms}\right)\left(-4.435\text{eV}/\text{atom}\times \frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)\\ =13\left(-7.105\times {10}^{-19}\text{J}\right)\\ =-9.236\times {10}^{-18}\text{J}\end{array}$

Substitute $-9.236\times {10}^{-18}\text{J}$ for $E_{\text{b}}$ and $1.48\times {10}^{-18}\text{J}$ for $E$ in equation (IV).

$\begin{array}{c}\frac{E}{E_{\text{b}}}=\frac{1.48\times {10}^{-18}\text{J}}{-9.236\times {10}^{-18}\text{J}}\\ \frac{E}{E_{\text{b}}}=-0.16\\ E=-0.16E_{\text{b}}\end{array}$

Conclusion:

Therefore, the magnitude of the total energy of the sphere is $-0.16$ times the total bond energy.

To determine

(b)

Whether the $\text{13-atom}$ cluster of atoms is stable.

Answer to Problem 3.9P

Yes, the $\text{13-atom}$ cluster of atoms stable as the total energy of the sphere is negative.

Explanation of Solution

Formula used:

The total energy of the Ni atom is given by,

$E_T=E+E_{\text{b}}$   ....... (V)

Calculation:

The total energy is calculated as,

Substitute $-9.236\times {10}^{-18}\text{J}$ for $E_{\text{b}}$ and $1.48\times {10}^{-18}\text{J}$ for $E$ in equation (V).

$\begin{array}{c}{E}_T=1.48\times {10}^{-18}\text{J}+\left(-9.236\times {10}^{-18}\text{J}\right)\\ =-7.756\times {10}^{-18}\text{J}\end{array}$

The total energy of the sphere is negative which signifies that the $13$ Ni atom cluster is stable.

Conclusion:

Therefore, the $13$ Ni atom cluster is stable as the total energy of the sphere is negative.