Materials Science and Engineering: An Introduction, 10e WileyPLUS + Abridged Loose-leaf
10th Edition
ISBN: 9781119472070
Author: William D. Callister Jr., David G. Rethwisch
Publisher: Wiley (WileyPLUS Products)
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Question
Chapter 3, Problem 67QAP
(a)
To determine
The interplanar spacing for
To determine
The atomic radius of the rubidium atom.
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Students have asked these similar questions
Qu 4 Silver has FCC crystal structure at room temperature, and a lattice constant, a, of 0.407 nm.
Draw a reduced sphere silver unit cell in the grids provided below, clearly label the lattice dimensions.
Within the unit cell you drew, shade the (1 0 0) plane.
How many atoms are contained within the (1 0 0) plane?
Calculate the area of (1 0 0) plane in [nm?]. Express your answer in [nm?] to three significant figures.
Calculate the planar density of the (1 0 0) plane in [atoms/nm?]. Express the answer in atoms/nm to three significant figures. show all work step by step
Crystallographic planes
Crystallographic planes are denoted by Miller indices.
5b
Algorithm for Miller indices
1. Read off intercepts of plane with axes in
terms of a, b, c
2. Take reciprocals of intercepts
3. Reduce to smallest integer values
4. Enclose in parentheses, no commas.
353
1/3 1/5 1/3
535
(535)
In the cubic system, a plane and a
direction with the same indices are
orthogonal. E.g. [100] direction is
perpendicular to (100) plane.
Correspondingly, [123] direction is
perpendicular to (123) plane.
[2,3,3]
Plane intercepts axes at 3a, 2b, 2c
2
11 1
Reciprocal numbers are:
3'2'2
b.
Indices of the plane (Miller): (2,3,3)
2
a
Indices of the direction: [2,3,3]
X
(200)
(100)
(110)
(111)
(100)
Indices of crystallographic plane can be found from cross product of indices of
any two non-parallel directions in this plane.
Crystallographic positions
Crystallographic position
is denoted by three
numbers, which are
coefficients of the
position vector, e.g. ½½½
for the red atom.
Here the 'new' atom is at a/2 + b/2 + c/2
Silicon crystal has so-called "diamond type lattice".
Each Si atom has 4 nearest neighbors.
Diamond lattice starts with a FCC lattice and then
adds four additional INTERNAL atoms at locations
r = a/4+b/4+c/4 away from each of the atoms. In
other words, diamond lattice is formed by two FCC
lattices sifted by the vector r.
Chapter 3 Solutions
Materials Science and Engineering: An Introduction, 10e WileyPLUS + Abridged Loose-leaf
Ch. 3 - Prob. 1QAPCh. 3 - Prob. 2QAPCh. 3 - Prob. 3QAPCh. 3 - Prob. 4QAPCh. 3 - Prob. 5QAPCh. 3 - Prob. 6QAPCh. 3 - Prob. 7QAPCh. 3 - Prob. 8QAPCh. 3 - Prob. 9QAPCh. 3 - Prob. 10QAP
Ch. 3 - Prob. 11QAPCh. 3 - Prob. 12QAPCh. 3 - Prob. 14QAPCh. 3 - Prob. 15QAPCh. 3 - Prob. 16QAPCh. 3 - Prob. 17QAPCh. 3 - Prob. 19QAPCh. 3 - Prob. 20QAPCh. 3 - Prob. 21QAPCh. 3 - Prob. 23QAPCh. 3 - Prob. 24QAPCh. 3 - Prob. 25QAPCh. 3 - Prob. 26QAPCh. 3 - Prob. 27QAPCh. 3 - Prob. 28QAPCh. 3 - Prob. 29QAPCh. 3 - Prob. 34QAPCh. 3 - Prob. 35QAPCh. 3 - Prob. 36QAPCh. 3 - Prob. 37QAPCh. 3 - Prob. 39QAPCh. 3 - Prob. 40QAPCh. 3 - Prob. 41QAPCh. 3 - Prob. 43QAPCh. 3 - Prob. 44QAPCh. 3 - Prob. 45QAPCh. 3 - Prob. 46QAPCh. 3 - Prob. 47QAPCh. 3 - Prob. 48QAPCh. 3 - Prob. 49QAPCh. 3 - Prob. 50QAPCh. 3 - Prob. 51QAPCh. 3 - Prob. 52QAPCh. 3 - Prob. 53QAPCh. 3 - Prob. 59QAPCh. 3 - Prob. 60QAPCh. 3 - Prob. 61QAPCh. 3 - Prob. 62QAPCh. 3 - Prob. 63QAPCh. 3 - Prob. 64QAPCh. 3 - Prob. 65QAPCh. 3 - Prob. 66QAPCh. 3 - Prob. 67QAPCh. 3 - Prob. 68QAPCh. 3 - Prob. 69QAPCh. 3 - Prob. 72QAPCh. 3 - Prob. 1SSPCh. 3 - Prob. 1FEQPCh. 3 - Prob. 2FEQPCh. 3 - Prob. 3FEQP
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