# material science chapter 3

```3-3
3.3 This problem calls for a demonstration of the relationship a =
4R
for BCC. Consider the BCC unit
3
cell shown below
Using the triangle NOP
(NP) 2 = a 2 + a 2 = 2a 2
And then for triangle NPQ,
(NQ) 2 = (QP) 2 + ( NP) 2
But NQ = 4R, R being the atomic radius. Also, QP = a. Therefore,
(4R) 2 = a 2 + 2a 2
or
a =
4R
3
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3-9
3.8 We are asked to determine the radius of a palladium atom, given that Pd has an FCC crystal structure.
For FCC, n = 4 atoms/unit cell, and VC = 16R 3 2 (Equation 3.4). Now,
ρ =
=
nAPd
VC N A
nAPd
(16R 3 2 ) N A
And solving for R from the above expression yields
⎛ nA
⎞1/3
Pd
⎟⎟
R = ⎜⎜
⎝ 16 ρN A 2 ⎠
⎤1/3
⎡
(4 atoms/unit cell) (106.4 g/mol)
⎥
⎢
=
⎢⎣ (16)(12.0 g/cm3)(6.023 x 1023 atoms/mol)( 2 ) ⎥⎦
= 1.38 x 10-8 cm = 0.138 nm
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
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3-10
3.9 This problem asks for us to calculate the radius of a tantalum atom. For BCC, n = 2 atoms/unit cell,
and
⎛ 4 R ⎞3
64 R 3
VC = ⎜
⎟ =
⎝ 3⎠
3 3
Since, from Equation 3.5
ρ =
=
nATa
VC N A
nATa
⎛ 64 R 3 ⎞
⎜⎜
⎟⎟ N A
⎝ 3 3 ⎠
and solving for R the previous equation
⎛ 3 3nA ⎞1/3
Ta ⎟
R = ⎜⎜
⎟
⎝ 64ρ N A ⎠
⎡ (3 3) (2 atoms/unit cell) (180.9 g/mol) ⎤1/3
⎥
= ⎢
⎢⎣ (64) (16.6 g/cm3)(6.023 x 1023 atoms/mol) ⎥⎦
= 1.43 x 10-8 cm = 0.143 nm
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
3-18
3.15 In order to determine the APF for U, we need to compute both the unit cell volume (VC) which is just
the product of the three unit cell parameters, as well as the total sphere volume (VS) which is just the product of the
volume of a single sphere and the number of spheres in the unit cell (n). The value of n may be calculated from
Equation 3.5 as
n =
=
ρVC N A
AU
(19.05 g/cm3 )(2.86)(5.87)(4.95)(× 10-24 cm3)(6.023 × 1023 atoms / mol)
238.03 g/mol
= 4.01 atoms/unit cell
Therefore
⎛4
⎞
(4) ⎜ π R 3 ⎟
⎝3
⎠
APF =
=
(a)(b)(c)
VC
VS
=
⎡4
⎤
(4)⎢ (π)(1.385Źx 10-8 cm) 3⎥
⎣3
⎦
(2.86) (5.87)(4.95)(xŹ10-24 cm3)
= 0.536
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
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