Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Concept explainers

Question
Chapter 21, Problem 21.64E
Interpretation Introduction

(a)

Interpretation:

The angles of diffraction for a cubic crystal for the given incoming X radiation are 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. The parameters of a crystal can be obtained experimentally by X-ray diffraction technique.

Expert Solution
Check Mark

Answer to Problem 21.64E

The table that represents the miller indices and corresponding value of diffraction angle is shown below as,

Miller indices (hkl) Diffraction angle (θ=sin1(nλh2+k2+l22a))
(100) 6.853688
(110) 9.715979
(111) 11.92863
(200) 13.80798
(210) 15.4763
(211) 16.9962
(220) 19.72637
(300),(221) 20.97761
(310) 22.17077
(311) 23.31509
(222) 24.41774
(320) 25.48445
(321) 26.51988
(400) 28.51165

Explanation of Solution

The wavelength of the given X-ray is 1.5418A.

The given lattice parameter is 6.46A.

The value of n for first-order diffractions is 1.

The Bragg equation for diffraction of X rays is given by an expression as shown below.

nλ=2(ah2+k2+l2)sinθ …(1)

Where,

h,k,lare the Miller indices.

a is the side of the unit cell.

λ is the wavelength of the incident ray.

θ is the Bragg’s angle.

n is the order of diffraction.

Rearrange the equation (1) for the value of θ.

θ=sin1(nλh2+k2+l22a) …(2)

The value of λ, a, n and different Miller indices are substituted in the equation (2). The table that represents the miller indices and corresponding value of diffraction angle is shown below as,

Miller indices (hkl) (h2+k2+l2) Diffraction angle (θ=sin1(nλh2+k2+l22a))
(100) 1 6.853688
(110) 2 9.715979
(111) 3 11.92863
(200) 4 13.80798
(210) 5 15.4763
(211) 6 16.9962
(220) 8 19.72637
(300),(221) 9 20.97761
(310) 10 22.17077
(311) 11 23.31509
(222) 12 24.41774
(320) 13 25.48445
(321) 14 26.51988
(400) 16 28.51165
Conclusion

The table that represents the miller indices and corresponding value of diffraction angle is shown below as,

Miller indices (hkl) Diffraction angle (θ=sin1(nλh2+k2+l22a))
(100) 6.853688
(110) 9.715979
(111) 11.92863
(200) 13.80798
(210) 15.4763
(211) 16.9962
(220) 19.72637
(300),(221) 20.97761
(310) 22.17077
(311) 23.31509
(222) 24.41774
(320) 25.48445
(321) 26.51988
(400) 28.51165
Interpretation Introduction

(b)

Interpretation:

The diffractions that would be absent if the given crystal were body-centered cubic or face centered cubic are to be determined.

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. The parameters of a crystal can be obtained experimentally by X-ray diffraction technique.

Expert Solution
Check Mark

Answer to Problem 21.64E

The diffractions that would be absent if the given crystal was body-centered cubic are represented in the table shown below.

Miller indices (hkl) Diffraction angle (θ=sin1(nλh2+k2+l22a))
(100) 6.853688
(111) 11.92863
(210) 15.4763
(300),(221) 20.97761
(311) 23.31509
(320) 25.48445

The diffractions that would be absent if the given crystal was face-centered cubic are represented in the table shown below.

Miller indices (hkl) Diffraction angle (θ=sin1(nλh2+k2+l22a))
(100) 6.853688
(110) 9.715979
(210) 15.4763
(211) 16.9962
(300),(221) 20.97761
(310) 22.17077
(320) 25.48445
(321) 26.51988

Explanation of Solution

From Table 21.3,

The Miller indices of diffractions those are present in body centered cubic crystal are (110), (200), (211), (220), (310), (222), (321) and (400).

Therefore, the diffractions that would be absent if the given crystal was body-centered cubic are represented in the table shown below as,

Miller indices (hkl) (h2+k2+l2) Diffraction angle (θ=sin1(nλh2+k2+l22a))
(100) 1 6.853688
(111) 3 11.92863
(210) 5 15.4763
(300),(221) 9 20.97761
(311) 11 23.31509
(320) 13 25.48445

From Table 21.3,

The Miller indices of diffractions those are present in face centered cubic crystal are (111), (200), (220), (311), (222), and (400).

Therefore, the diffractions that would be absent if the given crystal was face-centered cubic are represented in the table shown below as,

Miller indices (hkl) (h2+k2+l2) Diffraction angle (θ=sin1(nλh2+k2+l22a))
(100) 1 6.853688
(110) 2 9.715979
(210) 5 15.4763
(211) 6 16.9962
(300),(221) 9 20.97761
(310) 10 22.17077
(320) 13 25.48445
(321) 14 26.51988
Conclusion

The diffractions that would be absent if the given crystal was body-centered cubic are represented in the table shown below.

Miller indices (hkl) Diffraction angle (θ=sin1(nλh2+k2+l22a))
(100) 6.853688
(111) 11.92863
(210) 15.4763
(300),(221) 20.97761
(311) 23.31509
(320) 25.48445

The diffractions that would be absent if the given crystal was face-centered cubic are represented in the table shown below.

Miller indices (hkl) Diffraction angle (θ=sin1(nλh2+k2+l22a))
(100) 6.853688
(110) 9.715979
(210) 15.4763
(211) 16.9962
(300),(221) 20.97761
(310) 22.17077
(320) 25.48445
(321) 26.51988

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Students have asked these similar questions
The density of tantalum at 20oC is 16.69 g/cm3 and its unit cell is cubic. Given that the first five observed Bragg diffraction angles are θ =19.31o, 27.88o, 34.95o, 41.41o and 47.69o, find the type of unit cell and its length. Take the wavelength of the x-ray radiation to be λ =154.433 pm.
3 Rubidium (Rb) adopts a body-centred cubic, BCC, structure at a pressure of 1 atm. Upon application of high pressure (of the order of 30,000 atm), Rb undergoes a structural phase transition to a face-centred-cubic, FCC, structure. (i) Describe what could be the main factor that induces the phase transformation. (ii) Describe how the diffraction patterns of the two forms would differ.
Calculate the volume of the unit cell of chromium in cubic meters, given that chromium has a body-centered cubic crystal structure and an atomic radius of 0.200 nm

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