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Resolução Callister Capítulo 3 Engenharia dos Materiais

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=
1
2 R2 3
=
1
2 3 (0.145 nm)2
= 13.73 nm−2 = 1.373 × 1019 m−2 
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 3-71 
 Polycrystalline Materials 
 
 3.56 Although each individual grain in a polycrystalline material may be anisotropic, if the grains have 
random orientations, then the solid aggregate of the many anisotropic grains will behave isotropically. 
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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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 3-72 
 X-ray Diffraction: Determination of Crystal Structures 
 
 3.57 From the Table 3.1, aluminum has an FCC crystal structure and an atomic radius of 0.1431 nm. 
Using Equation 3.1, the lattice parameter a may be computed as 
 
 a=2 R 2 = (2) (0.1431 nm) 2 = 0.4048 nm 
 
Now, the interplanar spacing d110 maybe determined using Equation 3.14 as 
 
 
d110 =
a
(1)2 + (1)2 + (0)2
= 0.4048 nm
2
= 0.2862 nm 
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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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 3-73 
 3.58 We first calculate the lattice parameter using Equation 3.3 and the value of R (0.1249 nm) cited in 
Table 3.1, as follows: 
 
 
a = 4 R
3
= (4) (0.1249 nm)
3
= 0.2884 nm 
 
Next, the interplanar spacing for the (310) set of planes may be determined using Equation 3.14 according to 
 
 
d310 = 
a
(3)2 + (1)2 + (0)2
= 0.2884 nm
10
= 0.0912 nm 
 
And finally, employment of Equation 3.13 yields the diffraction angle as 
 
 
sin θ = nλ
2d310
= (1)(0.0711 nm)
(2)(0.0912 nm)
= 0.390 
 
Which leads to 
 
 θ = sin
-1(0.390) = 22.94° 
 
And, finally 
 
 2θ = (2)(22.94°) = 45.88° 
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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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 3-74 
 3.59 From the table, α-iron has a BCC crystal structure and an atomic radius of 0.1241 nm. Using 
Equation 3.3 the lattice parameter, a, may be computed as follows: 
 
 
a = 4 R
3
= (4) (0.1241 nm)
3
= 0.2866 nm 
 
Now, the d111 interplanar spacing may be determined using Equation 3.14 as 
 
 
d111 = 
a
(1)2 + (1)2 + (1)2
= 0.2866 nm
3
= 0.1655 nm 
 
And, similarly for d211 
 
 
d211 =
a
(2)2 + (1)2 + (1)2
= 0.2866 nm
6
= 0.1170 nm 
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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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 3-75 
 3.60 (a) From the data given in the problem, and realizing that 36.12° = 2θ, the interplanar spacing for the 
(311) set of planes for rhodium may be computed using Equation 3.13 as 
 
 
d311 =
nλ
2 sin θ
= (1)(0.0711 nm)
(2) sin 36.12°
2
⎛ 
⎝ 
⎜ 
⎞ 
⎠ 
⎟ 
= 0.1147 nm 
 
 (b) In order to compute the atomic radius we must first determine the lattice parameter, a, using Equation 
3.14, and then R from Equation 3.1 since Rh has an FCC crystal structure. Therefore, 
 
 a = d311 (3)
2 + (1)2 + (1)2 = (0.1147 nm)( 11) = 0.3804 nm 
 
And, from Equation 3.1 
 
 
R = a
2 2
= 0.3804 nm
2 2
= 0.1345 nm 
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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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 3-76 
 3.61 (a) From the data given in the problem, and realizing that 75.99° = 2θ, the interplanar spacing for the 
(211) set of planes for Nb may be computed using Equation 3.13 as follows: 
 
 
d211 =
nλ
2 sin θ
= (1)(0.1659 nm)
(2) sin 75.99°
2
⎛ 
⎝ 
⎜ 
⎞ 
⎠ 
⎟ 
= 0.1348 nm 
 
 (b) In order to compute the atomic radius we must first determine the lattice parameter, a, using Equation 
3.14, and then R from Equation 3.3 since Nb has a BCC crystal structure. Therefore, 
 
 a = d211 (2)
2 + (1)2 + (1)2 = (0.1347 nm)( 6) = 0.3300 nm 
 
And, from Equation 3.3 
 
 
R =
a 3
4
=
(0.3300 nm) 3
4
= 0.1429 nm 
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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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 3-77 
 3.62 The first step to solve this problem is to compute the interplanar spacing using Equation 3.13. Thus, 
 
 
dhkl =
nλ
2 sin θ
= (1)(0.1542 nm)
(2) sin 44.53°
2
⎛ 
⎝ 
⎜ 
⎞ 
⎠ 
⎟ 
= 0.2035 nm 
 
Now, employment of both Equations 3.14 and 3.1 (since Ni’s crystal structure is FCC), and the value of R for nickel 
from Table 3.1 (0.1246 nm) leads to 
 
 
 h2 + k 2 + l2 = a
dhkl
=
2R 2
dhkl
 
 
 
= 
(2)(0.1246 nm) 2
(0.2035 nm)
 = 1.732 
 
This means that 
 
 h
2 + k 2 + l2 = (1.732)2 = 3.0 
 
By trial and error, the only three integers that are all odd or even, the sum of the squares of which equals 3.0 are 1, 
1, and 1. Therefore, the set of planes responsible for this diffraction peak is the (111) set. 
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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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 3-78 
 3.63 For each peak, in order to compute the interplanar spacing and the lattice parameter we must employ 
Equations 3.14 and 3.13, respectively. The first peak of Figure 3.21, which results from diffraction by the (111) set 
of planes, occurs at 2θ = 31.3°; the corresponding interplanar spacing for this set of planes, using Equation 3.13, is 
equal to 
 
 
d111 =
nλ
2 sin θ
= (1)(0.1542 nm)
(2) sin 31.3°
2
⎛ 
⎝ 
⎜ 
⎞ 
⎠ 
⎟ 
 = 0.2858 nm 
 
And, from Equation 3.14, the lattice parameter a is determined as 
 
 a = dhkl (h)
2 + (k)2 + (l)2 = d111 (1)
2 + (1)2 + (1)2 
 
 = (0.2858 nm) 3 = 0.4950 nm 
 
Similar computations are made for the other peaks