Chapter 3, Problem 3.77P

Interpretation Introduction

Interpretation:

The planar density and packing fraction for the planes (100), (110), and (111) in BCC lithium needs to be calculated and the plane which is close packed needs to be identified.

Concept introduction:

Lithium has a body-centered cubic structure with a lattice parameter of 3.5089 $\stackrel{\text{o}}{\text{A}}$. The atomic radius of FCC structure is used to calculate the planar density and packing fraction at given planes:

The atomic radius can be calculated as follows:

  $r=\sqrt{3}/4a_0$

Here, $a_0$ is edge length.

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

Answer to Problem 3.77P

  $\begin{array}{l}\text{For plane }\left(\text{100}\right)\text{=Planar density=8}\text{.12}\times {\text{10}}^{\text{14}}{\text{/cm}}^{\text{2}}\hfill \\ \text{ Packing fraction=0}\text{.589}\hfill \\ \text{For plane }\left(\text{110}\right)\text{=Planar density=1}\text{.149}\times {\text{10}}^{\text{15}}{\text{/cm}}^{\text{2}}\hfill \\ \text{ Packing fraction=0}\text{.833}\hfill \\ \text{For plane }\left(\text{111}\right)\text{=Planar density=4}\text{.69}\times {\text{10}}^{\text{14}}{\text{/cm}}^{\text{2}}\hfill \\ \text{ Packing fraction=0}\text{.34}\hfill \\ \text{Plaine }\left(\text{110}\right)\text{ is more close-packed}\text{.}\hfill \end{array}$

Explanation of Solution

  $\begin{array}{l}\text{Lattice parameter of BCC lithium }\\ \text{= 3}\text{.5089}\stackrel{\text{o}}{\text{A}}\\ \text{Radius of atom}\\ \text{=r =}\sqrt{\text{3}}{\text{/4a}}_{\text{0}}\\ \text{=}\sqrt{\text{3}}\times \text{305089/4 }\\ \text{=1}\text{.519375 }\\ \text{=1}\text{.5193}\times {\text{10}}^{\text{-8}}\text{cm}\end{array}$

For plane (100):

From figure,

  $\begin{array}{l}\text{Number of atoms in the plane }\left(\text{100}\right)\text{=1atom}\hfill \\ {\text{Area of plane A=a}}_{\text{0}}{}^{\text{2}}\hfill \\ \text{=}{\left(\text{3}\text{.5089}\times {\text{10}}^{\text{-8}}\right)}^{\text{2}}\hfill \\ \text{Planer density=atom per plane/area of the plane}\hfill \\ \text{ =1/}{\left(\text{3}\text{.5089}\times {\text{10}}^{\text{-8}}\right)}^{\text{2}}\hfill \\ \text{ =8012}\times {\text{10}}^{\text{14}}{\text{/cm}}^{\text{2}}\hfill \\ \text{Packing fraction=area of atom/area of plane}\hfill \\ {\text{ =πr}}^{\text{2}}\text{/}{\left(\text{3}\text{.5089}\times {\text{10}}^{\text{-8}}\right)}^{\text{2}}\hfill \\ \text{ =0}\text{.589}\hfill \end{array}$

For plane (110):

From figure,

  $\begin{array}{l}\text{Number of atoms in the plane }\left(\text{110}\right)\\ \text{=2 atoms Area of plane A}\\ {\text{=a}}_{\text{0}}\times \sqrt{\text{2}}{\text{a}}_{\text{0}}\\ \text{=}\sqrt{\text{2}}{\text{a}}_{\text{0}}{}^{\text{2}}\\ \text{=}\sqrt{\text{2}}\times {\left(\text{3}\text{.5089}\times {\text{10}}^{\text{-8}}\right)}^{\text{2}}\\ \begin{array}{l}\text{ =9}\text{.862}\times {\text{10}}^{\text{-13}}{\text{cm}}^{\text{2}}\hfill \\ \text{Planer density=atom per plane/area of plane}\hfill \\ \text{ =2/9}\text{.862}\times {\text{10}}^{\text{-13}}\hfill \\ \text{ =1}\text{.149}\times {\text{10}}^{\text{15}}{\text{/cm}}^{\text{2}}\hfill \\ \text{Packing factor=area of atom/area of plane}\hfill \\ {\text{ =2πr}}^{\text{2}}\text{/9}\text{.862}\times {\text{10}}^{\text{-13}}\hfill \\ \text{ =0}\text{.833}\hfill \end{array}\end{array}$

For plane(111):

From figure,

  $\begin{array}{l}\text{A number of atoms in the plane }\left(\text{111}\right)\\ \text{=1/2 atoms }\\ \text{Area of plane A=}\sqrt[]{\text{2}}\text{/2}\times \sqrt{\text{3}}{\text{/2a}}_{\text{0}}{}^{\text{2}}\\ \text{ =}\sqrt{\text{3}}\text{/2}{\left(\text{3}\text{.5089}\times {\text{10}}^{\text{-8}}\right)}^{\text{2}}\\ \begin{array}{l}\text{ =1}\text{.066}\times {\text{10}}^{\text{-15}}{\text{cm}}^{\text{2}}\hfill \\ \text{Planer density=atom per plane/area of plane}\hfill \\ \text{ =1/2/1}\text{.066}\times {\text{10}}^{\text{-15}}\hfill \\ \text{ =4}\text{.69}\times {\text{10}}^{\text{14}}{\text{/cm}}^{\text{2}}\hfill \\ \text{Packing factor=area of atom/area of plane}\hfill \\ {\text{ =2pr}}^{\text{2}}\text{/1}\text{.066}\times {\text{10}}^{\text{-15}}\hfill \\ \text{ =0}\text{.34}\hfill \end{array}\end{array}$

Higher the packing fraction, more dense it will be.

Conclusion

From the above calculation, packing fraction of all the planes are:

  $\begin{array}{l}\text{0}\text{.34<0}\text{.589<0}\text{.833}\hfill \\ \text{Plane }\left(\text{111}\right)\text{<Plane}\left(\text{100}\right)\text{<Plane}\left(\text{110}\right)\hfill \end{array}$

Hence, plane (110) is highly denser among all the planes because it has a higher value of packing fraction.