Chapter 3, Problem 3.71P

Interpretation Introduction

Interpretation:

The planar densities for planes $\left\{100\right\},\left\{110\right\},\left\{111\right\}$ in a BCC unit cell needs to be calculated and compared. The cell which is the most close-packed needs to be determined.

Concept introduction:

Planar density is the ratio of the area of the plane to the number of atoms in a plane.

Answer to Problem 3.71P

Planar densities for the plane:

  $\begin{array}{c}\text{Plane }\left\{\text{100}\right\}\text{ =}\frac{\text{3}}{{\text{16R}}^{\text{2}}}\\ \text{Plane }\left\{\text{110}\right\}\text{ =}\frac{\text{3}}{{\text{8R}}^{\text{2}}\sqrt{\text{2}}}\\ \text{Plane }\left\{\text{111}\right\}\text{ =}\frac{\text{0}\text{.11}}{{\text{R}}^{\text{2}}}\end{array}$

Plane (110) is denser (closely packed) than others.

Explanation of Solution

Planar density is expressed as,

P = No. of atom/Area of plane =Z/A

(a) For plane $\left\{100\right\}$ ;

  

  $\begin{array}{l}\text{A number of atoms in plane }\left(\text{100}\right)\text{, Z =1 atom}\\ \text{Length of direction vector, }\\ \text{A =4R/}\sqrt{\text{3}}\text{Area of }\left(\text{100}\right){\text{ plane =A =a}}^{\text{2}}\\ \text{=[4R/}\sqrt{\text{3}}{\text{]}}^{\text{2}}\\ \begin{array}{l}{\text{=16R}}^{\text{2}}\text{/3}\hfill \\ {\left[\text{planar density}\right]}_{\text{1}}\text{=1atom/}\left({\text{16R}}^{\text{2}}\text{/3}\right){\text{ =3/16R}}^{\text{2}}\hfill \end{array}\end{array}$

(b) For plane $\left\{110\right\}$;

  

BCC unit cell with the plane (110) is rectangular.

Area of rectangular $x\times y$

Where $x=4R/\sqrt{3}$

Diagonal length =4R.

Using Pythagoras theorem,

  $x^2+y^2=z^2$

  $y=\sqrt{z^2-x^2}$

  $y=\sqrt{{\left(4R\right)}^2+{\left(\frac{4R}{\sqrt{3}}\right)}^2}$

  $y=4R\sqrt{2}/\surd 3$

  $\begin{array}{l}\text{area of plane }\left(110\right)=A\\ =x\times y \\ =4R\sqrt{3}\times 4R\sqrt{2}/\sqrt{3} \\ =16R^2\sqrt{2}/3\end{array}$

Number of atoms in the plane (110), $z=\text{2 atom}$

1 atom at every four corners and 1 center atom within the cell.

Planar density = $\frac{2\text{ atoms}}{\frac{16R^2\sqrt{2}}{3}}=\frac{6}{16R^2\sqrt{2}}$

[Planar density]₂ = $\frac{3}{8R^2\sqrt{2}}$

(c) For plane $\left\{111\right\}$;

  

Plane (111)inthe BCC unit cell is having a triangular section.

  $\begin{array}{l}\text{Area of rectangle = ½ x}\times \text{y}\\ {\text{Where, x}}^{\text{2}}{\text{=a}}^{\text{2}}{\text{+a}}^{\text{2}}{\text{=2a}}^{\text{2}}\\ \text{x=}\sqrt{\text{2}}\times \text{4R/}\sqrt{\text{3}}\end{array}$

Now, to calculate middle line length y is from the figure,

  $\begin{array}{l}{x}^2={(}^x+y^2\\ y^2=x^2-{\left(x/y\right)}^2\\ y^2=3x^2/4\\ y=\sqrt{3}x/2\end{array}$

Now,

  $\begin{array}{l} x=\frac{\sqrt{2}\times 4R}{\sqrt{3}}=\frac{4R}{\sqrt{2}\sqrt{3} }\\ y=\frac{\sqrt{3}\times 4R\sqrt{2}}{2}=\frac{4R\sqrt{2}}{2}\\ \text{putting values of x and y in area of plane}\\ \text{Area of plane }\left(111\right)=\frac{1}{2\times x\times y } \\ =\frac{\frac{1}{2\times 4R\sqrt{2}}}{\frac{\sqrt{3}\times 4R\sqrt{2}}{2} } \\ =\frac{16\times 2\times R^2}{4\times \sqrt{3} } \\ A=\frac{8R^2}{\sqrt{3}}\\ \text{Number of atoms }\left(\text{111}\right)\text{ = 0}\text{.5 atom}\\ \text{Planar density =}\frac{\text{ 0}\text{.5 atom }}{\left(\frac{\text{8}{R}^{\text{2}}}{\sqrt{\text{3}}}\right)}\\ \text{=}\frac{\text{0}\text{.5}\times \sqrt{\text{3}}}{\text{8}{R}^{\text{2}}}\\ {\left[\text{Planar density}\right]}_{\text{3}}\text{=}\frac{\text{0}\text{.11}}{R^{\text{2}}}\end{array}$

A plane with higher plane density is highly dense.

  $\begin{array}{l}{P}_1>P_2>P_3\\ \frac{3}{16R^2}>\frac{3}{8R^2\sqrt{2}}>\frac{0.11}{R^2}\\ \frac{0.1875}{R^2}>\frac{0.2562}{R^2}>\frac{0.108}{R^2}\end{array}$

Plane density for plane $\left\{110\right\}$ is higher than the others. Hence, it is denser.

Conclusion

By comparing values of planar density, the plane $\left\{110\right\}$ is having a higher volume of dense density. Thus, the plane $\left\{110\right\}$ is denser among others.