Essentials of Materials Science and Engineering, SI Edition
Essentials of Materials Science and Engineering, SI Edition
4th Edition
ISBN: 9781337672078
Author: ASKELAND, Donald R., WRIGHT, Wendelin J.
Publisher: Cengage Learning
Question
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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 Ao. 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 =3/4a0

Here, a0 is edge length.

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

Expert Solution & Answer
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Answer to Problem 3.77P

  For plane ( 100)=Planar density=8.12× 10 14 /cm2                          Packing fraction=0.589For plane ( 110)=Planar density=1.149× 10 15 /cm2                          Packing fraction=0.833For plane ( 111)=Planar density=4.69× 10 14 /cm2                          Packing fraction=0.34Plaine ( 110) is more close-packed.

Explanation of Solution

  Lattice parameter of BCC lithium = 3.5089AoRadius of atom=r =3/4a0                              =3×305089/4                              =1.519375                            =1.5193×10-8cm

For plane (100):

Essentials of Materials Science and Engineering, SI Edition, Chapter 3, Problem 3.77P , additional homework tip  1

From figure,

  Number of atoms in the plane ( 100)=1atom Area of plane A=a02= ( 3.5089× 10 -8 )2Planer density=atom per plane/area of the plane                      =1/ ( 3.5089× 10 -8 )2                      =8012× 10 14 /cm2Packing fraction=area of atom/area of plane                           =πr2/ ( 3.5089× 10 -8 )2                          =0.589

For plane (110):

Essentials of Materials Science and Engineering, SI Edition, Chapter 3, Problem 3.77P , additional homework tip  2

From figure,

  Number of atoms in the plane (110)=2 atoms Area of plane A=a0×2a0                        =2a02                       =2×(3.5089× 10 -8)2                         =9.862× 10 -13 cm 2 Planer density=atom per plane/area of plane                       =2/9.862× 10 -13                       =1.149× 10 15 /cm 2 Packing factor=area of atom/area of plane                       =2πr 2 /9.862× 10 -13                       =0.833

For plane(111):

Essentials of Materials Science and Engineering, SI Edition, Chapter 3, Problem 3.77P , additional homework tip  3

From figure,

  A number of atoms in the plane (111)=1/2 atoms Area of plane A=2/2×3/2a02                       =3/2(3.5089× 10 -8)2                          =1.066× 10 -15 cm 2 Planer density=atom per plane/area of plane                       =1/2/1.066× 10 -15                       =4.69× 10 14 /cm 2 Packing factor=area of atom/area of plane                       =2pr 2 /1.066× 10 -15                       =0.34

Higher the packing fraction, more dense it will be.

Conclusion

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

  0.34<0.589<0.833Plane ( 111)<Plane( 100)<Plane( 110)

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

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Chapter 3 Solutions

Essentials of Materials Science and Engineering, SI Edition

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