The separation of the { 133 } planes in an orthorhombic lattice has to be determined. Concept Introduction: The separation of planes in an orthorhombic lattice is given by 1 d h k l 2 = h 2 a 2 + k 2 b 2 + l 2 c 2 Where, d h k l = Separation between planes a, b and c are unit cell parameters. h, k and l are Miller indices.

Question

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Answer 1

Question

Chapter 15, Problem 15A.1ST

(a)

Interpretation Introduction

Interpretation:

The separation of the {133} planes in an orthorhombic lattice has to be determined.

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

(a)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 1, k= 3 and l= 3

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d1332=12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)2 = 27.6728 nm−2d133= 0.19 nm.

Therefore, the separation between the given planes is found out to be 0.19 nm.

(b)

Interpretation Introduction

Interpretation:

The separation of the {399} planes in an orthorhombic lattice has to be determined

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

(b)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 3, k= 9 and l= 9

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d3992=32(0.82 nm)2+92(0.94 nm)2+92(0.75 nm)2 = 249.0552 nm−2d399= 0.063nm.

Therefore, the separation between the given planes is found out to be 0.063 nm.

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Answer 2

Question

Chapter 15, Problem 15A.1ST

(a)

Interpretation Introduction

Interpretation:

The separation of the {133} planes in an orthorhombic lattice has to be determined.

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

(a)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 1, k= 3 and l= 3

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d1332=12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)2 = 27.6728 nm−2d133= 0.19 nm.

Therefore, the separation between the given planes is found out to be 0.19 nm.

(b)

Interpretation Introduction

Interpretation:

The separation of the {399} planes in an orthorhombic lattice has to be determined

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

(b)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 3, k= 9 and l= 9

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d3992=32(0.82 nm)2+92(0.94 nm)2+92(0.75 nm)2 = 249.0552 nm−2d399= 0.063nm.

Therefore, the separation between the given planes is found out to be 0.063 nm.

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Answer 3

Question

Chapter 15, Problem 15A.1ST

(a)

Interpretation Introduction

Interpretation:

The separation of the {133} planes in an orthorhombic lattice has to be determined.

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

(a)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 1, k= 3 and l= 3

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d1332=12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)2 = 27.6728 nm−2d133= 0.19 nm.

Therefore, the separation between the given planes is found out to be 0.19 nm.

(b)

Interpretation Introduction

Interpretation:

The separation of the {399} planes in an orthorhombic lattice has to be determined

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

(b)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 3, k= 9 and l= 9

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d3992=32(0.82 nm)2+92(0.94 nm)2+92(0.75 nm)2 = 249.0552 nm−2d399= 0.063nm.

Therefore, the separation between the given planes is found out to be 0.063 nm.

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Answer 4

Question

Chapter 15, Problem 15A.1ST

(a)

Interpretation Introduction

Interpretation:

The separation of the {133} planes in an orthorhombic lattice has to be determined.

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

(a)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 1, k= 3 and l= 3

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d1332=12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)2 = 27.6728 nm−2d133= 0.19 nm.

Therefore, the separation between the given planes is found out to be 0.19 nm.

(b)

Interpretation Introduction

Interpretation:

The separation of the {399} planes in an orthorhombic lattice has to be determined

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

(b)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 3, k= 9 and l= 9

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d3992=32(0.82 nm)2+92(0.94 nm)2+92(0.75 nm)2 = 249.0552 nm−2d399= 0.063nm.

Therefore, the separation between the given planes is found out to be 0.063 nm.

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Answer 5

Chapter 15, Problem 15A.1ST

(a)

Interpretation Introduction

Interpretation:

The separation of the {133} planes in an orthorhombic lattice has to be determined.

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

(a)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 1, k= 3 and l= 3

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d1332=12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)2 = 27.6728 nm−2d133= 0.19 nm.

Therefore, the separation between the given planes is found out to be 0.19 nm.

(b)

Interpretation Introduction

Interpretation:

The separation of the {399} planes in an orthorhombic lattice has to be determined

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

(b)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 3, k= 9 and l= 9

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d3992=32(0.82 nm)2+92(0.94 nm)2+92(0.75 nm)2 = 249.0552 nm−2d399= 0.063nm.

Therefore, the separation between the given planes is found out to be 0.063 nm.

Answer 6

Chapter 15, Problem 15A.1ST

(a)

Interpretation Introduction

Interpretation:

The separation of the {133} planes in an orthorhombic lattice has to be determined.

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

(a)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 1, k= 3 and l= 3

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d1332=12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)2 = 27.6728 nm−2d133= 0.19 nm.

Therefore, the separation between the given planes is found out to be 0.19 nm.

Answer 7

Chapter 15, Problem 15A.1ST

(a)

Interpretation Introduction

Interpretation:

The separation of the {133} planes in an orthorhombic lattice has to be determined.

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

Answer 8

(a)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 1, k= 3 and l= 3

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d1332=12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)2 = 27.6728 nm−2d133= 0.19 nm.

Therefore, the separation between the given planes is found out to be 0.19 nm.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The separation of the {399} planes in an orthorhombic lattice has to be determined

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

(b)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 3, k= 9 and l= 9

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d3992=32(0.82 nm)2+92(0.94 nm)2+92(0.75 nm)2 = 249.0552 nm−2d399= 0.063nm.

Therefore, the separation between the given planes is found out to be 0.063 nm.

Answer 14

(b)

Interpretation Introduction

Interpretation:

The separation of the {399} planes in an orthorhombic lattice has to be determined

Concept Introduction:

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Where,

dhkl= Separation between planes

a, b and c are unit cell parameters.

h, k and l are Miller indices.

Answer 15

(b)

Expert Solution

Answer to Problem 15A.1ST

The separation between the given planes is found out to be 0.19 nm.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given data:

a=0.82 nmb= 0.94 nmc= 0.75 nmh= 3, k= 9 and l= 9

The separation of planes in an orthorhombic lattice is given by

1dhkl2=h2a2+k2b2+l2c2

Now,

1d3992=32(0.82 nm)2+92(0.94 nm)2+92(0.75 nm)2 = 249.0552 nm−2d399= 0.063nm.

Therefore, the separation between the given planes is found out to be 0.063 nm.

Answer 20

Ch. 15 - Prob. 15A.1ST Ch. 15 - Prob. 15B.1ST Ch. 15 - Prob. 15B.3ST Ch. 15 - Prob. 15C.1ST Ch. 15 - Prob. 15C.2ST Ch. 15 - Prob. 15A.1DQ Ch. 15 - Prob. 15A.2DQ Ch. 15 - Prob. 15A.3DQ Ch. 15 - Prob. 15A.1AE Ch. 15 - Prob. 15A.1BE

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Answer to Problem 15A.1ST

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