Chapter 4.8, Problem 35AAP

Calculate the number of atoms in a critically sized nucleus for the homogeneous nucleation of pure iron.

To determine

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

Answer to Problem 35AAP

The number of atoms in a critically sized nucleus for the homogeneous nucleation of pure iron is $\text{329}\text{atoms}$.

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 pure iron has a BCC structure, each unit cell has 2 atoms.

Express the volume of pure iron.

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

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

Express the volume per atom.

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

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

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

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

Conclusion:

For pure iron, write the critical radius as:

 $r^{\ast }=9.72\times {10}^{-8}\text{cm}$

Write the side of a BCC unit cell.

 $a=0.286\times {10}^{-9}\text{m}$

Substitute $9.72\times {10}^{-8}\text{cm}$ for $r^{\ast }$ in Equation (I).

 $\begin{array}{c}V=\frac{4}{3}\pi {\left(9.72\times {10}^{-8}\text{cm}\right)}^3\\ =3.85\times {10}^{-21}{\text{cm}}^3\end{array}$

Substitute $0.286\times {10}^{-9}\text{m}$ for $a$ in Equation (II).

 $\begin{array}{c}{V}_{\text{Pt}}={\left(0.286\times {10}^{-9}\text{m}\right)}^3\\ =2.34\times {10}^{-29}{\text{m}}^3\left[\frac{{10}^6{\text{cm}}^{\text{3}}}{{\text{m}}^3}\right]\\ =2.34\times {10}^{-23}{\text{cm}}^3\end{array}$

Substitute $2.34\times {10}^{-23}{\text{cm}}^3$ for $V_{\text{Pt}}$ and $\text{2}\text{atoms}/\text{BCC}\text{unit}\text{cell}$ for $n$ in Equation (III).

 $\begin{array}{c}{V}_{\text{atom}}=\frac{2.34\times {10}^{-23}{\text{cm}}^3}{2\text{atoms}/\text{BCC}\text{unit}\text{cell}}\\ =1.17\times {10}^{-23}{\text{cm}}^{\text{3}}/\text{atom}\end{array}$

Substitute $3.85\times {10}^{-21}{\text{cm}}^3$ for $V$ and $1.17\times {10}^{-23}{\text{cm}}^{\text{3}}/\text{atom}$ for $V_{\text{atom}}$ in Equation (IV).

 $\begin{array}{c}\frac{V}{V_{\text{atom}}}=\frac{3.85\times {10}^{-21}{\text{cm}}^3}{1.17\times {10}^{-23}{\text{cm}}^{\text{3}}/\text{atom}}\\ =329\text{atoms}\end{array}$

Hence, the number of atoms in a critically sized nucleus for the homogeneous nucleation of pure iron is $\text{329}\text{atoms}$.