The separation of the { 133 } and { 399 } planes in the given lattice has to be calculated. Concept Introduction: The separation of planes, d, of the { h k l } planes in an orthogonal lattice is given by, 1 d 2 = h 2 a 2 + k 2 b 2 + l 2 c 2 Where, a, b and c are dimensions of the unit cell.

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Answer 1

Question

Chapter 15, Problem 15A.2ST

Interpretation Introduction

Interpretation:

The separation of the {133} and {399} planes in the given lattice has to be calculated.

Concept Introduction:

The separation of planes, d, of the {hkl} planes in an orthogonal lattice is given by,

1d2=h2a2+k2b2+l2c2

Where,

a, b and c are dimensions of the unit cell.

Expert Solution & Answer

Explanation of Solution

Given data:

The dimensions of the unit cell are a=0.82 nm, b = 0.94 nm and c = 0.75 nm

The planes of orthorhombic lattice is {133} and {399}

Separation of planes:

The separation of planes is determined using the below formula.

1d2=h2a2+k2b2+l2c2

Consider {hkl} indices as {133}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)21d2 =27.6728 nm2d2 =127.6728 nm2 =0.0361 nm2d =0.19 nm

Now, consider {hkl} indices as {399}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =(3×1)2(0.82 nm)2+(3×3)2(0.94 nm)2+(3×3)2(0.75 nm)21d2 =(3)2(0.82 nm)2+(9)2(0.94 nm)2+(9)2(0.75 nm)21d2 =249.0553 nm2d2 =1249.0553 nm2 =0.004015 nm2d =0.63 nm

Therefore,

The separation of the {133} planes in the given lattice is 0.19 nm

The separation of the {399} planes in the given lattice is 0.63 nm

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Answer 2

Question

Chapter 15, Problem 15A.2ST

Interpretation Introduction

Interpretation:

The separation of the {133} and {399} planes in the given lattice has to be calculated.

Concept Introduction:

The separation of planes, d, of the {hkl} planes in an orthogonal lattice is given by,

1d2=h2a2+k2b2+l2c2

Where,

a, b and c are dimensions of the unit cell.

Expert Solution & Answer

Explanation of Solution

Given data:

The dimensions of the unit cell are a=0.82 nm, b = 0.94 nm and c = 0.75 nm

The planes of orthorhombic lattice is {133} and {399}

Separation of planes:

The separation of planes is determined using the below formula.

1d2=h2a2+k2b2+l2c2

Consider {hkl} indices as {133}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)21d2 =27.6728 nm2d2 =127.6728 nm2 =0.0361 nm2d =0.19 nm

Now, consider {hkl} indices as {399}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =(3×1)2(0.82 nm)2+(3×3)2(0.94 nm)2+(3×3)2(0.75 nm)21d2 =(3)2(0.82 nm)2+(9)2(0.94 nm)2+(9)2(0.75 nm)21d2 =249.0553 nm2d2 =1249.0553 nm2 =0.004015 nm2d =0.63 nm

Therefore,

The separation of the {133} planes in the given lattice is 0.19 nm

The separation of the {399} planes in the given lattice is 0.63 nm

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Answer 3

Question

Chapter 15, Problem 15A.2ST

Interpretation Introduction

Interpretation:

The separation of the {133} and {399} planes in the given lattice has to be calculated.

Concept Introduction:

The separation of planes, d, of the {hkl} planes in an orthogonal lattice is given by,

1d2=h2a2+k2b2+l2c2

Where,

a, b and c are dimensions of the unit cell.

Expert Solution & Answer

Explanation of Solution

Given data:

The dimensions of the unit cell are a=0.82 nm, b = 0.94 nm and c = 0.75 nm

The planes of orthorhombic lattice is {133} and {399}

Separation of planes:

The separation of planes is determined using the below formula.

1d2=h2a2+k2b2+l2c2

Consider {hkl} indices as {133}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)21d2 =27.6728 nm2d2 =127.6728 nm2 =0.0361 nm2d =0.19 nm

Now, consider {hkl} indices as {399}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =(3×1)2(0.82 nm)2+(3×3)2(0.94 nm)2+(3×3)2(0.75 nm)21d2 =(3)2(0.82 nm)2+(9)2(0.94 nm)2+(9)2(0.75 nm)21d2 =249.0553 nm2d2 =1249.0553 nm2 =0.004015 nm2d =0.63 nm

Therefore,

The separation of the {133} planes in the given lattice is 0.19 nm

The separation of the {399} planes in the given lattice is 0.63 nm

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Answer 4

Question

Chapter 15, Problem 15A.2ST

Interpretation Introduction

Interpretation:

The separation of the {133} and {399} planes in the given lattice has to be calculated.

Concept Introduction:

The separation of planes, d, of the {hkl} planes in an orthogonal lattice is given by,

1d2=h2a2+k2b2+l2c2

Where,

a, b and c are dimensions of the unit cell.

Expert Solution & Answer

Explanation of Solution

Given data:

The dimensions of the unit cell are a=0.82 nm, b = 0.94 nm and c = 0.75 nm

The planes of orthorhombic lattice is {133} and {399}

Separation of planes:

The separation of planes is determined using the below formula.

1d2=h2a2+k2b2+l2c2

Consider {hkl} indices as {133}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)21d2 =27.6728 nm2d2 =127.6728 nm2 =0.0361 nm2d =0.19 nm

Now, consider {hkl} indices as {399}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =(3×1)2(0.82 nm)2+(3×3)2(0.94 nm)2+(3×3)2(0.75 nm)21d2 =(3)2(0.82 nm)2+(9)2(0.94 nm)2+(9)2(0.75 nm)21d2 =249.0553 nm2d2 =1249.0553 nm2 =0.004015 nm2d =0.63 nm

Therefore,

The separation of the {133} planes in the given lattice is 0.19 nm

The separation of the {399} planes in the given lattice is 0.63 nm

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Answer 5

Chapter 15, Problem 15A.2ST

Interpretation Introduction

Interpretation:

The separation of the {133} and {399} planes in the given lattice has to be calculated.

Concept Introduction:

The separation of planes, d, of the {hkl} planes in an orthogonal lattice is given by,

1d2=h2a2+k2b2+l2c2

Where,

a, b and c are dimensions of the unit cell.

Expert Solution & Answer

Explanation of Solution

Given data:

The dimensions of the unit cell are a=0.82 nm, b = 0.94 nm and c = 0.75 nm

The planes of orthorhombic lattice is {133} and {399}

Separation of planes:

The separation of planes is determined using the below formula.

1d2=h2a2+k2b2+l2c2

Consider {hkl} indices as {133}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)21d2 =27.6728 nm2d2 =127.6728 nm2 =0.0361 nm2d =0.19 nm

Now, consider {hkl} indices as {399}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =(3×1)2(0.82 nm)2+(3×3)2(0.94 nm)2+(3×3)2(0.75 nm)21d2 =(3)2(0.82 nm)2+(9)2(0.94 nm)2+(9)2(0.75 nm)21d2 =249.0553 nm2d2 =1249.0553 nm2 =0.004015 nm2d =0.63 nm

Therefore,

The separation of the {133} planes in the given lattice is 0.19 nm

The separation of the {399} planes in the given lattice is 0.63 nm

Answer 6

Chapter 15, Problem 15A.2ST

Interpretation Introduction

Interpretation:

The separation of the {133} and {399} planes in the given lattice has to be calculated.

Concept Introduction:

The separation of planes, d, of the {hkl} planes in an orthogonal lattice is given by,

1d2=h2a2+k2b2+l2c2

Where,

a, b and c are dimensions of the unit cell.

Expert Solution & Answer

Explanation of Solution

Given data:

The dimensions of the unit cell are a=0.82 nm, b = 0.94 nm and c = 0.75 nm

The planes of orthorhombic lattice is {133} and {399}

Separation of planes:

The separation of planes is determined using the below formula.

1d2=h2a2+k2b2+l2c2

Consider {hkl} indices as {133}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)21d2 =27.6728 nm2d2 =127.6728 nm2 =0.0361 nm2d =0.19 nm

Now, consider {hkl} indices as {399}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =(3×1)2(0.82 nm)2+(3×3)2(0.94 nm)2+(3×3)2(0.75 nm)21d2 =(3)2(0.82 nm)2+(9)2(0.94 nm)2+(9)2(0.75 nm)21d2 =249.0553 nm2d2 =1249.0553 nm2 =0.004015 nm2d =0.63 nm

Therefore,

The separation of the {133} planes in the given lattice is 0.19 nm

The separation of the {399} planes in the given lattice is 0.63 nm

Answer 7

Chapter 15, Problem 15A.2ST

Interpretation Introduction

Interpretation:

The separation of the {133} and {399} planes in the given lattice has to be calculated.

Concept Introduction:

The separation of planes, d, of the {hkl} planes in an orthogonal lattice is given by,

1d2=h2a2+k2b2+l2c2

Where,

a, b and c are dimensions of the unit cell.

Answer 8

Expert Solution & Answer

Explanation of Solution

Given data:

The dimensions of the unit cell are a=0.82 nm, b = 0.94 nm and c = 0.75 nm

The planes of orthorhombic lattice is {133} and {399}

Separation of planes:

The separation of planes is determined using the below formula.

1d2=h2a2+k2b2+l2c2

Consider {hkl} indices as {133}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =12(0.82 nm)2+32(0.94 nm)2+32(0.75 nm)21d2 =27.6728 nm2d2 =127.6728 nm2 =0.0361 nm2d =0.19 nm

Now, consider {hkl} indices as {399}

Substitute the given data in the below equation.

1d2 =h2a2+k2b2+l2c21d2 =(3×1)2(0.82 nm)2+(3×3)2(0.94 nm)2+(3×3)2(0.75 nm)21d2 =(3)2(0.82 nm)2+(9)2(0.94 nm)2+(9)2(0.75 nm)21d2 =249.0553 nm2d2 =1249.0553 nm2 =0.004015 nm2d =0.63 nm