Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Chapter 22, Problem 22.35E
Interpretation Introduction

(a)

Interpretation:

The closest Na+Na+ distance for a surface made by the (100) plane is to be calculated.

Concept introduction:

A unit cell of the crystal is the three-dimensional arrangement of the atoms present in the crystal. The unit cell is the smallest and simplest unit of the crystal which on repetition forms an entire crystal. Unit cell can be a cubic unit cell or hexagonal unit cell. The classification of a unit cell depends on the lattice site occupied by the atoms.

Expert Solution
Check Mark

Answer to Problem 22.35E

The closest Na+Na+ distance for a surface made by the (100) plane is 3.988A.

Explanation of Solution

The structure of a face-centered cubic lattice is shown below.

Physical Chemistry, Chapter 22, Problem 22.35E , additional homework tip  1

Figure 1

The plane of face-centered cubic lattice that has (100) Miller indices is the outer plane. The outer plane of the face-centered cubic lattice is shown below.

Physical Chemistry, Chapter 22, Problem 22.35E , additional homework tip  2

Figure 2

The closest Na+Na+ distance is denoted by the symbol b in the Figure (2).

The lattice parameter (a) of face-centered cubic lattice is 5.640A.

The Pythagoras theorem is shown below.

H2=B2+P2…(1)

Where,

H is the hypotenuse of the triangle.

B is the base of the triangle.

P is the perpendicular of the triangle.

The hypotenuse of the triangle shown in Figure (2) is 2b.

The base and perpendicular of the triangle shown in Figure (2) are a.

Substitute the value of hypotenuse, base, and perpendicular in the equation (1).

(2b)2=a2+a24b2=2a22b2=a2b=a2

Substitute the value of a in the above expression.

b=5.640A2=3.988A

Therefore, the closest Na+Na+ distance for a surface made by the (100) plane is 3.988A.

Conclusion

The closest Na+Na+ distance for a surface made by the (100) plane is 3.988A.

Interpretation Introduction

(b)

Interpretation:

The closest Na+Na+ distance for a surface made by the (110) plane is to be calculated.

Concept introduction:

A unit cell of the crystal is the three-dimensional arrangement of the atoms present in the crystal. The unit cell is the smallest and simplest unit of the crystal which on repetition forms an entire crystal. Unit cell can be a cubic unit cell or hexagonal unit cell. The classification of a unit cell depends on the lattice site occupied by the atoms.

Expert Solution
Check Mark

Answer to Problem 22.35E

The closest Na+Na+ distance for a surface made by the (110) plane is 3.988A.

Explanation of Solution

The structure of a face-centered cubic lattice is shown below.

Physical Chemistry, Chapter 22, Problem 22.35E , additional homework tip  3

Figure 1

The plane of face-centered cubic lattice that has (110) Miller indices is the diagonal plane. The diagonal plane of the face-centered cubic lattice is shown below.

Physical Chemistry, Chapter 22, Problem 22.35E , additional homework tip  4

Figure 3

The lattice parameter (a) of face-centered cubic lattice is 5.640A.

The relation between the length of the edge of a cube (a) and diagonal (d) of the cube is shown below.

d=2a

The diagonal of the triangle shown in Figure (3) is 2b.

Substitute the value of d, and a in the above equation.

2b=(2)(5.640A)b=5.640A2=3.988A

Therefore, the closest Na+Na+ distance for a surface made by the (110) plane is 3.988A.

Conclusion

The closest Na+Na+ distance for a surface made by the (110) plane is 3.988A.

Interpretation Introduction

(c)

Interpretation:

The closest Na+Na+ distance for a surface made by the (111) plane is to be calculated.

Concept introduction:

A unit cell of the crystal is the three-dimensional arrangement of the atoms present in the crystal. The unit cell is the smallest and simplest unit of the crystal which on repetition forms an entire crystal. Unit cell can be a cubic unit cell or hexagonal unit cell. The classification of a unit cell depends on the lattice site occupied by the atoms.

Expert Solution
Check Mark

Answer to Problem 22.35E

The closest Na+Na+ distance for a surface made by the (111) plane is 3.988A.

Explanation of Solution

The structure of a face-centered cubic lattice is shown below.

Physical Chemistry, Chapter 22, Problem 22.35E , additional homework tip  5

Figure 1

The plane of face-centered cubic lattice that has (111) Miller indices is the trigonal plane. The outer plane of the face-centered cubic lattice is shown below.

Physical Chemistry, Chapter 22, Problem 22.35E , additional homework tip  6

Figure 4

The side of the trigonal plane is same as the diagonal of the of the cube.

The lattice parameter (a) of face-centered cubic lattice is 5.640A.

The relation between the length of the edge of a cube (a) and diagonal (d) of the cube is shown below.

d=2a

The diagonal of the triangle shown in Figure (4) is 2b.

Substitute the value of d, and a in the above equation.

2b=(2)(5.640A)b=5.640A2=3.988A

Therefore, the closest Na+Na+ distance for a surface made by the (111) plane is 3.988A.

Conclusion

The closest Na+Na+ distance for a surface made by the (111) plane is 3.988A.

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Students have asked these similar questions
In the early days of X-ray crystallography there was an urgent need to know the wavelengths of X-rays. One technique was to measure the diffraction angle from a mechanically ruled grating. Another method was to estimate the separation of lattice planes from the measured density of a crystal. The mass density of NaCl is 2.17 g cm−3 and the (100) reflection using radiation of a certain wavelength occurred at 6.0°. Calculate the wavelength of the X-rays.
Express the relationship between atomic radius (r) and the edge length (a) in the bcc unit cell.
A solid is thought to have an orthorhombic structure. Calculate the positions D and 2θ for the plane 111 expected in the diffraction pattern as a result of CuKa radiation(λ=1.54 Å), since the edges of the unit cell are a= 3.50 Å, B=4.0 Å, C=5.5 Å.
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