Chapter 4.8, Problem 33AAP

To determine

The number of atoms in a critically sized nucleus for the homogeneous nucleation of pure platinum.

Answer to Problem 33AAP

The number of atoms in a critically sized nucleus for the homogeneous nucleation of pure platinum is $\text{641}\text{atoms}$.

Explanation of Solution

Express the volume of critical size nucleus.

 $V=\frac{4}{3}\pi {\left(r^{\ast }\right)}^3$                                                                                        (I)

Here, critical radius is $r^{\ast }$.

Since platinum has a FCC structure, each unit cell has 4 atoms.

Express the volume of platinum.

 $V_{\text{Pt}}=a^3$                                                                                                  (II)

Here, side of a FCC unit cell is $a$.

Express the volume per atom.

 $V_{\text{atom}}=\frac{V_{\text{Pt}}}{n}$                                                                                                 (III)

Here, number of atoms per FCC unit cell is $n$.

Express the number of atoms in a critically sized nucleus for the homogeneous nucleation of pure platinum.

 $\frac{V}{V_{\text{atom}}}$                                                                                                 (IV)

Conclusion:

For platinum, write the critical radius as:

 $r^{\ast }=1.11\times {10}^{-7}\text{cm}$

Write the side of a FCC platinum unit cell.

 $a=0.3293\times {10}^{-9}\text{m}$

Substitute $1.11\times {10}^{-7}\text{cm}$ for $r^{\ast }$ in Equation (I).

 $\begin{array}{c}V=\frac{4}{3}\pi {\left(1.11\times {10}^{-7}\text{cm}\right)}^3\\ =5.73\times {10}^{-21}{\text{cm}}^3\end{array}$

Substitute $0.3293\times {10}^{-9}\text{m}$ for $a$ in Equation (II).

 $\begin{array}{c}{V}_{\text{Pt}}={\left(0.3293\times {10}^{-9}\text{m}\right)}^3\\ =3.574\times {10}^{-29}{\text{m}}^3\left[\frac{{10}^6{\text{cm}}^{\text{3}}}{{\text{m}}^3}\right]\\ =3.574\times {10}^{-23}{\text{cm}}^3\end{array}$

Substitute $3.574\times {10}^{-23}{\text{cm}}^3$ for $V_{\text{Pt}}$ and $\text{4}\text{atoms}/\text{FCC}\text{unit}\text{cell}$ for $n$ in Equation (III).

 $\begin{array}{c}{V}_{\text{atom}}=\frac{3.574\times {10}^{-23}{\text{cm}}^3}{\text{4}\text{atoms}/\text{FCC}\text{unit}\text{cell}}\\ =8.935\times {10}^{-24}{\text{cm}}^{\text{3}}/\text{atom}\end{array}$

Substitute $5.73\times {10}^{-21}{\text{cm}}^3$ for $V$ and $8.935\times {10}^{-24}{\text{cm}}^{\text{3}}/\text{atom}$ for $V_{\text{atom}}$ in Equation (IV).

 $\begin{array}{c}\frac{V}{V_{\text{atom}}}=\frac{5.73\times {10}^{-21}{\text{cm}}^3}{8.935\times {10}^{-24}{\text{cm}}^{\text{3}}/\text{atom}}\\ =641\text{atoms}\end{array}$

Hence, the number of atoms in a critically sized nucleus for the homogeneous nucleation of pure platinum is $\text{641}\text{atoms}$.