this BCC (110) plane. The planar section represented in the above figure
is a rectangle, as noted in the figure below. 48 From this figure, the area
of the rectangle is the product of x and y. The length x is just the unit
cell edge length, which for BCC

bonds will be satisfied for the (111) plane, giving rise to a lower surface
energy. 4.27 (a) The surface energy will be greater than the grain
boundary energy since some atoms on one side of the boundary will
bond to atoms on the other side-i.e., there wi

VCNA n = 18.91 g/cm3 ( )(2.0)(2.5)(3.0) x 10-24 cm3 ( /unit cell)6.023 x
1023 ( atoms/mol) 4 atoms/unit cell = 42.7 g/mol 3.52 Although each
individual grain in a polycrystalline material may be anisotropic, if the
grains have random orientations, then th

A ave we have Aave = 100 CV AV + (100 CV) AFe = 100 CV 50.94 g /mol
+ (100 CV) 55.85 g /mol whereas for ave 83 ave = 100 CV V + (100
CV) Fe = 100 CV 6.10 g /cm3 + (100 CV) 7.87 g /cm3 Within the BCC
unit cell there are 2 equivalent atoms, and thus, the v

(z) 0.650 0.6420 y 0.6667 0.700 0.6778 y 0.650 0.700 0.650 = 0.6667
0.6420 0.6778 0.6420 from which y = 0.6844 = 393.8 s t And, solving
for t gives t = 3.31 x 105 s = 92 h 5.15 This problem calls for an estimate
of the time necessary to achieve a carbon

2 log M 100 log 2 + 1 From Figure 9.22a, NM is measured to
be approximately 4, which leads to n = log 4 + 2 log 90 100
log 2 + 1 = 2.7 80 4.32 (a) This part of problem asks that we compute
the number of grains per square inch for an ASTM grain size of 6

x 100 = 41.9 at% Also, 67 CZn ' = 0.719 mol 0.519 mol + 0.719 mol x 100
= 58.1 at% 4.11 In this problem we are asked to determine the
concentrations, in atom percent, of the Ag-Au-Cu alloy. It is first
necessary to convert the amounts of Ag, Au, and Cu in

radius we must first determine the lattice parameter, a, using Equation
(3.3W), 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 53 And R = a 2 2 = 0.3804 nm 2 2 = 0

which atomic bonding is predominantly ionic in nature is less likely to
form a noncrystalline solid upon solidification than a covalent material
because covalent bonds are directional whereas ionic bonds are
nondirectional; it is more difficult for the at

g/cm3, respectively; and, therefore CSi ' = 0.25 0.25 2.33 g / cm3 +
99.75 7.87 g / cm3 x 103 70 = 19.6 kg/m3 4.16 We are asked in this
problem to determine the approximate density of a high-leaded brass
that has a composition of 64.5 wt% Cu, 33.5 wt% Zn,

employment of Equation (4.18) is necessary, using the following values:
N1 = NAu = 5.5 x 1021 atoms/cm3 1 = Au = 19.32 g/cm3 2 = Ag =
10.49 g/cm3 A1 = AAu = 196.97 g/mol A2 = Ag = 107.87 g/mol 74 Thus
CAu = 100 1 + NAAg NAuAAu Ag Au = 100 1 + 6.023 x 1023

applied stress. Normal and shear stresses are defined by Equations (6.1)
and (6.3), respectively. However, we now chose to express these
stresses in terms (i.e., general terms) of normal and shear forces (P and
V) as = P A = V A For static equilibrium in

temperature to which the diffusion couple must be heated so as to
produce a concentration of 3.0 wt% Ni at the 2-mm position, we must
first utilize Equation (5.6b) with time t being a constant. That is x2 D =
constant Or x1000 2 D1000 = xT 2 DT Now, solvi

76.15 kJ R T The x's cancel out, which means that the
process is independent of sheet thickness. Now solving the above
expression for the absolute temperature T gives T = 3467 K which value
is extremely high (surely above the vaporization point of iron).

V will not be exactly equal to (V1 + V2). Each of V1 and V2 may be
expressed in terms of its mass density, which when substituted into the
above equation ave = m1 + m2 m1 1 + m2 2 Furthermore, from
Equation (4.3) m1 = C1 m1 ( + m2) 100 m2 = C2 m1 ( + m2)

= 1.2 x 10-14 m2/s. Solving for T from Equation (5.9a) yields T = Qd R ln
D o ( ) ln D = 272,000 J/mol (8.31 J/mol-K) ln 1.1 x 10-4 ( )- ln 1.2 x 1014 ( ) = 1427 K = 1154C 5.20 In this problem we are given Qd for
the diffusion of Cu in Ag (i.e., 193,000 J

of the problem calls for us to estimate the ASTM grain size number for
this same material. The average grain size number, n, is related to the
number of grains per square inch, N, at a magnification of 100x
according to Equation 4.16. Inasmuch as the magn

2.5 times that of B2 , and concentrations are proportional to the square
root of the partial pressure, the diffusion flux of A, JA, is the square root
of 2.5 times the diffusion flux of nitrogen JB-i.e. JA = 2.5 JB 107 Thus,
equating the Fick's law expres

1017 m2 While for the (110) plane PD110(Mo) = 3 8R2 2 = 3 8 (0.136
nm)2 2 = 14.34 nm2 = 1.434 x 1017 m2 3.49 (a) A (0001) plane for
an HCP unit cell is show below. 49 Each of the 6 perimeter atoms in this
plane is shared with three other unit cells, where

incomplete solubility. All these metals have either BCC or HCP crystal
structures, and/or the difference between their atomic radii and that
for Cu are greater than 15%, and/or have a valence different than 2+.
(c) C, H, and O form interstitial solid solu

(44.8)(107.87 g /mol) + (46.2)(196.97 g /mol) + (9.0)(63.55 g /mol) x 100
69 = 62.7 wt% CCu = CCu ' ACu CAg ' AAg + CAu ' AAu + CCu ' ACu x 100
= (9.0)(63.55 g /mol) (44.8)(107.87 g /mol) + (46.2)(196.97 g /mol) +
(9.0)(63.55 g /mol) x 100 = 4.0 wt% 4.14

)/s exp 138,300 J/mol (8.31 J/ mol- K)(873 K) = 1.05 x 10-5 m
2 /s (b) Using these values of D o and Qd, D at 1123 K (850C) is just D =
1.05 x 10-5 m2 ( )/s exp 138,300 J/mol (8.31 J /mol - K)(1123 K)
= 3.8 x 10-12 m 2 /s 5.23 This problem asks us to de

solve this problem, we must first compute the value of Do from the data
given at 1200C (1473 K); this requires the combining of both Equations
(5.3) and (5.8). Solving for Do from these expressions gives Do = J C /
x exp Qd RT = 7.8 x 108 kg/m2 - s 500 kg

are constructed below from the three crystallographic planes provided
in the problem. 50 (a) This unit cell belongs to the tetragonal system
since a = b = 0.40 nm, c = 0.55 nm, and = = = 90. (b) This crystal
structure would be called body-centered tetrago

(3.2W). Thus, dhkl n 2 sin = (1)(0.1542 nm) (2) sin 44.53 2
= 0.2035 nm Now, employment of both Equations (3.3W) and (3.1), and
the value of R for nickel from Table 3.1 (0.1246 nm) leads to h2 + k2 + l 2
= a dhkl = 2R 2 d hkl = (2)(0.1246 nm) 2 (0.2035 n

the atom fraction of component 1 ( times the total number of atoms
per cubic centimeter in the alloy (N). Thus, using the equivalent of
Equation (4.2), we may write c1 ' ) N1 = c1 ' N = c1 ' NA ave A ave
Realizing that c1 ' = C1 ' 100 and C2 ' = 100 C1 '

= 212,200 J/mol Rather than trying to make a graphical
extrapolation to determine Do, a more accurate value is obtained
analytically using Equation (5.9b) taking a specific value of both D and T
(from 1/T) from the plot given in the problem; for example,

nm2 = C2 ' n m1 + n ( m2) 100 Thus Aave = C1 ' A1 nm1 + n ( m2) 100 +
C2 ' A2 n m1 + n ( m2) 100 n m1 + nm2 = C1 'A1 + C2 ' A2 100 which is
the desired result. 4.7 In order to compute composition, in atom
percent, of a 92.5 wt% Ag-7.5 wt% Cu alloy, we emp

concentration, in weight percent, of an element in an alloy may be
computed using a modification of Equation (4.3). For this alloy, the
concentration of iron (CFe) is just 66 CFe = mFe mFe + mC + mCr x 100 =
105 kg 105 kg + 0.2 kg + 1.0 kg x 100 = 98.87 w

= NACMo CMo AMo Mo + AMo W 100 C ( ) Mo = 6.023 x 1023 ( )
atoms/mol (16.4) (16.4)(95.94 g /mol) (10.22 g / cm3) + 95.94 g /mol
19.3 g /cm3 ( ) 100 16.4 = 1.73 x 1022 atoms/cm3 4.19 This problem
asks us to determine the number of niobium atoms per cubic
c

68 4.12 We are asked to compute the composition of an alloy in atom
percent. Employment of Equation (4.6) leads to CPb ' = CPbASn CPbASn
+ CSnAPb x 100 = 5.5(118.69 g/mol) 5.5(118.69 g/mol) + 94.5(207.2
g/mol) x 100 = 3.2 at% CSn ' = CSnAPb CSnAPb + CPbAS

directions. Therefore, the Burgers vector is b = a 2 [100] (b) For Cu
which has an FCC crystal structure, R = 0.1278 nm (Table 3.1) and a = 2R
2 = 0.3615 nm [Equation (3.1)]; therefore b = a 2 h2 + k2 + l2 = 0.3615
nm 2 (1 ) 2 + (1 ) 2 + (0) 2 = 0.2556 nm