Q-1 The atomic packing
factor of magnesium, like that of all HCP metals, is 0.74. What is the volume
of magnesium’s unit cell? [Given Rmg,=0.161 nm.
Soln: HCP
metal has 6 atoms per unit cell.
Vol =
No of atom X Vol of atom /Atomic
factor
= 6 X 4 π /3 (0.161)3/0.74
= 0.14 nm3 = 1.4 X 10-28 m3
Q-2 The unit cell
volume of HCP titanium at 20oC is 0.106 nm3. The c/a
ratio is 1.59.
(a) What are the
values of c and a ?
(b) What is the
radius of the Ti in a direction that lies in the base of the unit cell?
Soln: (a) Base area A= 6 X (1/2) (a) (a sin60o)
= 2.6 a2.
Volume V = (1.5a) (2.6 a2)
= 4.13 a3 = 0.106nm3
=> a3 = 0.02566 nm3
= 0.295 nm
c = 1.59 (0.295) = 0.469 nm
(b) a= 2 Rbase ; Rbase = 0.1475 nm
Q-3: Iron changes
from the BCC to the FCC metal structure at 912oC. At this
temperature, the atomic radii of the iron atoms in the two structures are 0.126
nm and 0.129 nm, respectively.
(a)
What is the percent of volume
change as the structure changes?
(b)
What is the percent of liner
changes, I/O?
Soln: Using a common basis for the two structure,
the simplest is four iron atoms, therefore, two unit cells of BCC iron and one
unit cell of FCC.
Also, aBCC
√3 = 4 RBCC and aFCC √2 = 4 RFCC
(a)
2 VBCC = 2a3BCC
= 2 [ 4 (0.126)/ √3]3 = 0.0493 nm3
VFCC= a3FCC = [ 4 (0.126)/ √2]3
= 0.0486 nm3
ΔV/V = (0.0486-0.0493)/0.0493 =
-0.014 = 1.4% Shrink
(b)
(1+ ΔL/L)3 = 1+ ΔV/V
ΔL/L = -0.0047 = 0.47% shrink
Q-4: Consider the
following four points within a unit cell:
¼,
¼, ¾ 1/8, 1/8, 1/8 7/8, ¾. 1/5 0.53,
0.25, 0.17
(a)
Identify a second identical
site within the reference unit cell (bcc) for each of the four points.
(b)
Identify three identical sites
within an FCC unit cell for each of the four points
Soln: (a) The
translation in BCC cells are ±0.5a, ±0.5a, ±0.5a, Therefore
For
¼, ¼, ¾ (0.25+0.5), (0.25+0.5),
(0.75-0.5) = ¾, ¾, ¼
For 1/8, 1/8, 1/8 (0.125+0.5),
(0.125+0.5), (0.125+0.5) = 5/8, 5/8, 5/8
For 7/8, ¾. 1/5 (0.875-0.5), (0.75-0.5), (0.2 +0.5) = 3/8,
¼, 0.7
For 0.53, 0.25, 0.17 (0.53-0.5), (0.25+0.5), (0.17 +0.5) = 0.03,
0.75, 0.67
(b) The translation in FCC cells are permutations of ±0.5a, ±0.5a,
0.
For ¼, ¼, ¾ ¾,
¾, ¾ ¾, ¼, ¼ ¼, ¾, ¼
For 1/8, 1/8, 1/8 5/8, 5/8, 1/8 5/8, 1/8, 5/8 1/8,
5/8, 5/8
For 7/8, ¾. 1/5 3/8,
¼, 1/5 3/8, ¾, 0.7 7/8, ¼, 0.7
For 0.53, 0.25, 0.17 0.03,0.75,0.17 0.03,0.25,0.67 0.53,
0.75, 0.67
Q-5: X-rays of an
unknown wavelength are diffracted 43.4o by copper (FCC), whose
lattice constant a is 0.3615 nm. Separate determinations indicate that this
diffraction line for copper is the first order (n=1) line for d111.
(a)
What is the wavelength of the
X-ray
(b)
The same X-rays are used to
analyze tungsten (BCC). What is the angle, 2θ, for the second order
diffraction lines of the d010 spacing? [Given Rw= 0.1367
nm
Soln: (a) Since 2θ= 43.4o, Applying Bragg’s law of diffraction nλ = 2 d sinθ
λ= 0.1543 nm
(c)
Lattice parameter aw=
4 x (0.1367)/ √3 = 0.3157 nm, d2010=(02+12+02)=1
Applying Bragg’s Law, Sinθ= 0.488 ; 2θ= 58.4o
Q1 Sterling silver, an alloy containing
approximately 92.5% silver and 7.5% Cu, is heated slowly from room temperature
to 1000oC. (a) What phases will be present as heating progresses?
(b) locate the liquidus and solidus. (c) How many phases are present where the
two meets?
Soln: (a) Room temperature to 740oC : α + β
740oC
to 810oC :
only α
810oC
to 900oC :
α + liquid
900oC
to 1000oC :
only liquid
(b)
Liquidus: 962oC at 100% Ag, to 779oC at 71.9% Ag, to 1084oC
at 100% Cu.
Solidus: 962oC
at 100% Ag to 779oC at 91.2% Ag, remaining at 779oC to
92% Cu, to 1084oC at 100% Cu
(c) With a
single component (only Ag or only Cu), two phase are present (solid + liquid) where the liquidus and solidus meet.
With two
components, three phases present in the temperature range between the liquidus
and solidus of a two component, or binary, phase diagram.
Q: Consider the Cu-Ni system at 1300oC
(a)
What is the solubility limit of
copper in solid α?
(b)
What is the solubility limit of
nickel in the liquid?
(c)
What are the chemical
composition of the phases in a 45 Cu-58 Ni alloy at 1300oC?
Soln: (a) 42% Cu (b) 45% Ni (c) α
(42 Cu-58 Ni), L (55Cu-45Ni)
Q: An alloy of 40Ag-60Cu is cooled slowly
from 1000oC to room temperature. (a) What phases will be present as cooling
progresses? (b) Indicate their compositions. (c) Write the eutectic reaction?
Soln: (a) & (b)
Temperature,oC
α Liquid β
1000 -- 40Ag-60Cu --
800 -- 66Ag-34Cu 8Ag-92Cu
779.4 91.2Ag-8.8Cu 71.9Ag-28.1Cu 8Ag-92Cu
600 96.5Ag-3.5Cu -- 2Ag-98Cu
400 99Ag-1Cu -- Near 100Cu
20 Near 100 Ag -- Near 100Cu
780oC
(c) Liquid (71.9%Ag) → α (91.2%Ag) + β (8% Ag)
Q: Consider a phase transformation that
occurs at 1025oC. The boundary energy, γ, is 0.50 J/m2.
And the value for energy per unit volume that is releases when the
supersaturated phase reacts to give the equilibrium phases is -2.1 X 109
J/m3 at 900oC (it is 0 at 1025oC). Determine
the critical radius for homogeneous nucleation at 900oC.
Soln: The critical radius is at the maximum
energy for nucleation. Therefore,
dΔFn/dr=0
dΔFn/dr=8πrγ
+4πr2ΔFv=0
rc=-2γ/ΔFv=0.5 X 10-9 m
