ATKINS' PHYSICAL CHEMISTRY-ACCESS
ATKINS' PHYSICAL CHEMISTRY-ACCESS
11th Edition
ISBN: 9780198834700
Author: ATKINS
Publisher: OXF
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Chapter 15, Problem 15A.9P
Interpretation Introduction

Interpretation:

The separation of the {hkl} planes in an orthorhombic crystal with sides a, b and c has to be derived.

Expert Solution & Answer
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Explanation of Solution

As known the trigonometric function as follows,

ATKINS' PHYSICAL CHEMISTRY-ACCESS, Chapter 15, Problem 15A.9P , additional homework tip  1

Figure 1

Consider an orthorhombic crystal with edges a, b and c. Let (hkl) be the Miller indices of the plane ABC.

Let, OP=d be a normal to the plane passing through the origin O and makes angles α',β'andγ' with x, y and z-axes respectively.

ATKINS' PHYSICAL CHEMISTRY-ACCESS, Chapter 15, Problem 15A.9P , additional homework tip  2

Figure 2

Then,

  cosα'=OPOA=d1a/h=d1hacosβ'=OPOB=d1b/h=d1hbcosγ'=OPOC=d1c/h=d1hc

From the law of directions of cosines, it can be written cos2α'+cos2β'+cos2γ'=1

  (d1ha)2+(d1kb)2+(d1lc)2=11d12=h2a2+k2b2+l2c2d1=1h2a2+k2b2+l2c2

Let the next plane parallel to ABC be at a distance OQ from the origin. Then its intercepts are

  OA':OB':OC'=2ah:2bk:2cl

ATKINS' PHYSICAL CHEMISTRY-ACCESS, Chapter 15, Problem 15A.9P , additional homework tip  3

Figure 3

  cosα''=OQOA'=d22a/h=d2h2acosβ''=OQOB'=d22b/h=d2h2bcosγ''=OQOC'=d22c/h=d2h2c

From the law of directions of cosines, it can be written cos2α''+cos2β''+cos2γ''=1

  (d2h2a)2+(d2k2b)2+(d2l2c)2=11d22=h24a2+k24b2+l24c2d2=2h2a2+k2b2+l2c2

Interplanar distance, d:

It is the distance between two planes. Here, the two plane are ABC and A'B'C'.

Then

  d=d2d1=2h2a2+k2b2+l2c21h2a2+k2b2+l2c2=1h2a2+k2b2+l2c2

  OR

  1dhkl2=h2a2+k2b2+l2c2

This is the expression for the separation of the {hkl} planes in an orthorhombic crystal with sides a, b and c.

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Chapter 15 Solutions

ATKINS' PHYSICAL CHEMISTRY-ACCESS

Ch. 15 - Prob. 15A.2AECh. 15 - Prob. 15A.2BECh. 15 - Prob. 15A.3AECh. 15 - Prob. 15A.3BECh. 15 - Prob. 15A.4AECh. 15 - Prob. 15A.4BECh. 15 - Prob. 15A.1PCh. 15 - Prob. 15A.2PCh. 15 - Prob. 15A.3PCh. 15 - Prob. 15A.4PCh. 15 - Prob. 15A.5PCh. 15 - Prob. 15A.6PCh. 15 - Prob. 15A.7PCh. 15 - Prob. 15A.8PCh. 15 - Prob. 15A.9PCh. 15 - Prob. 15B.1DQCh. 15 - Prob. 15B.2DQCh. 15 - Prob. 15B.3DQCh. 15 - Prob. 15B.1AECh. 15 - Prob. 15B.1BECh. 15 - Prob. 15B.2AECh. 15 - Prob. 15B.2BECh. 15 - Prob. 15B.3AECh. 15 - Prob. 15B.3BECh. 15 - Prob. 15B.4AECh. 15 - Prob. 15B.4BECh. 15 - Prob. 15B.5AECh. 15 - Prob. 15B.5BECh. 15 - Prob. 15B.6AECh. 15 - Prob. 15B.6BECh. 15 - Prob. 15B.7AECh. 15 - Prob. 15B.7BECh. 15 - Prob. 15B.11AECh. 15 - Prob. 15B.11BECh. 15 - Prob. 15B.12AECh. 15 - Prob. 15B.12BECh. 15 - Prob. 15B.1PCh. 15 - Prob. 15B.2PCh. 15 - Prob. 15B.3PCh. 15 - Prob. 15B.4PCh. 15 - Prob. 15B.6PCh. 15 - Prob. 15B.7PCh. 15 - Prob. 15C.1DQCh. 15 - Prob. 15C.2DQCh. 15 - Prob. 15C.1AECh. 15 - Prob. 15C.2AECh. 15 - Prob. 15C.2BECh. 15 - Prob. 15C.3AECh. 15 - Prob. 15C.3BECh. 15 - Prob. 15C.4AECh. 15 - Prob. 15C.4BECh. 15 - Prob. 15C.5AECh. 15 - Prob. 15C.5BECh. 15 - Prob. 15C.1PCh. 15 - Prob. 15C.2PCh. 15 - Prob. 15C.3PCh. 15 - Prob. 15C.4PCh. 15 - Prob. 15C.5PCh. 15 - Prob. 15C.7PCh. 15 - Prob. 15C.8PCh. 15 - Prob. 15C.9PCh. 15 - Prob. 15D.1DQCh. 15 - Prob. 15D.1AECh. 15 - Prob. 15D.1BECh. 15 - Prob. 15D.2AECh. 15 - Prob. 15D.2BECh. 15 - Prob. 15D.3AECh. 15 - Prob. 15D.3BECh. 15 - Prob. 15D.1PCh. 15 - Prob. 15D.2PCh. 15 - Prob. 15E.1DQCh. 15 - Prob. 15E.1AECh. 15 - Prob. 15E.1BECh. 15 - Prob. 15E.2AECh. 15 - Prob. 15E.2BECh. 15 - Prob. 15E.3AECh. 15 - Prob. 15E.3BECh. 15 - Prob. 15E.5PCh. 15 - Prob. 15F.1DQCh. 15 - Prob. 15F.1AECh. 15 - Prob. 15F.1BECh. 15 - Prob. 15F.2AECh. 15 - Prob. 15F.2BECh. 15 - Prob. 15F.3AECh. 15 - Prob. 15F.3BECh. 15 - Prob. 15F.4AECh. 15 - Prob. 15F.4BECh. 15 - Prob. 15F.5AECh. 15 - Prob. 15F.5BECh. 15 - Prob. 15G.1DQCh. 15 - Prob. 15G.2DQCh. 15 - Prob. 15G.1AECh. 15 - Prob. 15G.1BECh. 15 - Prob. 15.1IA
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