Download Problem # 1

Problem # 1
Select the correct answer by putting [9] in the appropriate box.
(a)
Materials' properties are strongly related to:
[
(b)
]
[x]
atomic structure
[
]
atomic weight
[
]
none of the above
Which primary bonding is directional:
[
]
(d)
(e)
ionic
[x]
covalent
[
metallic
]
[]
(c)
atomic number
all of the above
Which material has the highest range of electrical conductivities:
[x ]
metals
[
]
ceramics
[
]
polymers
[
]
all of the above
Which one of the following statements in not correct for HCP crystals:
[
]
coordination number is 12
[
]
{0001} is a close packed plane
[
]
closed-packed planes are arranged as ABABABAB …
[x ]
there are 4 atoms per unit cell
(e)
The APF for simple cubic structure is:
]
π/8
[
]
π/4
[ x]
π/6
[
]
π/2
[
(20 Points)
2
(f)
The relationship between the unit cell size and the atomic radius R in FCC is:
[
(g)
(h)
]
a = 2R/√2
[x]
a = 4R/√2
[
]
a = R√2
[
]
a = 4R√2
The linear density along [110] in BCC is:
[
]
√3/(√2R), where R is the atomic radius
[
]
√2/(4√3R)
[x]
√3/(4√2R)
[
√2/(√2R)
]
The planar density in (100) plane in simple cubic is:
[
]
1/(2R2)
[
]
2/(R2)
[
]
1/(R2)
[x]
(i)
Vacancies are created when:
[
]
electrons are missing from lattice sites
[x]
atoms are missing from lattice sites
[
atoms are missing from interstitial sites
]
[ ]
(j)
1/(4R2)
none of the above
The number of vacancies in metals:
[
]
decreases with increasing temperature
[
]
do not depend on the total number of atomic sites
[
]
is independent of temperature
[x]
none of the above
3
(k)
(l)
(m)
(n)
(o)
The energy required to form a vacancy:
[
]
decreases with decreasing temperature
[
]
increases with increasing temperature
[x]
is independent of temperature
[ ]
none of the above
An atom occupying a position that is not a normal lattice site is:
[
]
a vacancy
[
]
a substitution atom
[x]
an interstitial atom
[
a dislocation
]
The crystallographic direction shown below is:
[ ]
[101]
[
]
[111]
[x]
[111]
[
[111]
]
−−
−−
The crystallographic plane shown below is:
[ ]
⎡ −⎤
⎢⎣3 21⎥⎦
[ ]
(2 31)
[ ]
< 3 21 >
[x]
none of the above
−
−
A grain boundary:
[ ]
is a point defect
[ ]
is a linear defect
[x]
separates two grains or crystals having deferent crystallographic orientation
[ ]
none of the above
4
(p)
(q)
(r)
(s)
(t)
The dislocation line in screw dislocation is:
[ ]
not related to Burger’s vector
[ ]
normal (perpendicular) to Burger’s vector
[x]
parallel to Burger’s vector
[ ]
none of the above
In a substitutional solid solution:
[ ]
impurity atoms fill interstitial positions
[ ]
a new phase with different crystal structure will form
[ ]
the two elements are not soluble in each other
[x]
none of the above
Two metallic elements will form a solid solution with substantial solubility if they have:
[ ]
different crystal structures
[ ]
large difference in electronegativities
[ ]
a difference in atomic radii much more than 15 %
[x]
none of the above
Impurities often segregate along grain boundaries because:
[ ]
grain boundaries are less chemically reactive compared to grains
[x]
grain boundaries have higher energy state compared to grains
[ ]
grain boundaries have lower energy state compared to grains
[ ]
none of the above
Single crystal materials have:
[
]
small number of grain boundaries
[ ]
large number of grain boundaries
[ ]
random distribution of grains
[x]
none of the above
5
Problem # 2
(10 Points)
Calcium has a cubic crystal structure, the atomic radius is 0.197 nm, the density is 1.55 g/cm3,
and the atomic weight is 40.08 g/mol. Determine whether it has simple, FCC, or BCC structure.
(Navg = 6.023 x 1023 atoms/mol)
Solution
R = 0.197 nm, ρ = 1.55 g/cm3, A = 40.08 g/mol
ρ=
nA
nA
= 3
Vc N a N
If SC structure: n = 1 atom and a=2R
ρ=
1atom × 40.08 g / mol
(
2 × 0.197 ×10−7 cm
)
3
6.023 × 10−7 atom / mol
40.08 g
g
=
1.09
cm 3
36.84cm 3
=
If BCC structure: n = 2 atoms and a=4R/√3
ρ=
2atoms × 40.08 g / mol
3
⎛ 4
⎞
−7
−7
×
0.197
×
10
cm
⎜
⎟ 6.023 × 10 atom / mol
⎝ 3
⎠
=
80.16 g
g
= 1.41 3
3
56.72cm
cm
If FCC structure: n = 4 atoms and a=2√2xR
ρ=
(
4atoms × 40.08 g / mol
2 2 × 0.197 × 10 −7 cm
)
3
6.023 × 10 −7 atom / mol
Therefore, FCC is the appropriate structure of Calcium.
=
160.32 g
g
=
1.54
104.20cm 3
cm 3
6
Problem # 3
(10 Points)
a) (8 points): Calculate the relative volume change of a piece of iron (Fe) when its crystal
structure changes from BCC to FCC upon heating through 912°C. The BCC unit cell is
0.293 nm and the FCC unit cell is 0.363 nm at the transformation temperature.
V
−V Initial
)
(The relative volume change = Final
V Initial
Solution
There are different ways for soving this problem. One of them is to assume a number of atoms in
the material. For example 100 atoms.
1) Number of unit cells of BCC= 100/2 = 50
The volume of the sample with BCC structure = 50*(0.293nm)^3 = 1.258nm3
2) Number of unit cells of FCC= 100/4 = 25
The volume of the sample with FCC structure = 25*(0.363nm)^3 = 1.196nm3
V Final −V Initial 1.196 − 1.258
=
= −0.0493 or
1.258
V Initial
- 4.93%
An alternative way is shown below.
The following equations will be used for solving this problem.
V Final −V Initial
M
nA
(1),
ρ=
(2),
ρ=
V
V Initial
V c N avg
(3)
At the transformation temperature the mass of the material is not changing; therefore,
M = ρ BCCV Initial = ρ FCCV Final
Using equation (3):
Therefore,
ρ BCC
ρ FCC
⇒ V Final =
ρ BCC
V
ρ FCC Initial
n BCC A
V BCC N avg n BCCV FCC
=
=
n FCC A
n FCCV BCC
V FCC N avg
ρ BCC
V
−V Initial
ρ FCC Initial
V Final −V Initial
n BCCV FCC
2 × 0.3633
=
=
−1 =
− 1 = −0.0492 or
4 × 0.2933
V Initial
V Initial
n FCCV BCC
The advantage here is that only one caculation which means that less error.
b) (2 points): Is the change in volume considered as expansion or contraction?
The negative sign indicates that the volume is contracting.
− 4.92%
7
Problem # 4
(10 Points)
The numbers of vacancies present in silver at 500°C and 900°C are 3.59 x1021 m−3 and 1024 m−3,
respectively. Calculate the energy for vacancy formation in silver. Assume that the density at both
temperatures is the same. (k = 1.38 x 10-23 J/atom-K or 8.62 x 10-5 eV/atom-K).
Solution
At T1 = 500 + 273 = 773 K
NV 1 = N exp(
− QV
) …………1
KT 1
At T2 = 900 + 273 = 1173 K
NV 2 = N exp(
− QV
) …………2
KT 2
Taking the ratio and ln of both sides and rearranging gives:
N v1
Nv2
⎛ Q ⎞
N s exp ⎜ − v ⎟
⎝ kT 1 ⎠ = exp ⎛ − Q v
=
⎜⎜
⎛ Qv ⎞
⎝ k1
N s exp ⎜ −
⎟
⎝ kT 2 ⎠
⎛ 1 1 ⎞⎞
⎜ − ⎟ ⎟⎟
⎝ T1 T 2 ⎠ ⎠
⎛ T ×T ⎞ ⎛ N ⎞
Qv = − k ⎜ 1 2 ⎟ ln ⎜ v 1 ⎟ = − 8.62 × 10 −5 eV / atom .K
⎝ T 2 −T 1 ⎠ ⎝ N v 2 ⎠
QV = 1.10 eV/atom.
21
⎛ 773 × 1173 ⎞ ⎛ 3.59 × 10 ⎞
⎟
⎜
⎟ ln ⎜
24
⎝ 1173 − 773 ⎠ ⎝ 10
⎠
8
Problem # 5
(10 Points)
Two metallic elements A and B having densities of 0.534 g/cm3 and 2.71 g/cm3 respectively
are available. Compute the concentration of element B (in wt.%) that is required to develop
an alloy with a density of 2.47 g/cm3.
Solution
To compute the concentration of elemnt A (in wt%) that, when added to element B, will yield a
density of 2.47 g/cm3 we use the average density:
V T otal = V A +V B
ρ ave =
mass
M
M
=
=
total volume V Total V A +V B
⇒
Assume 100g of material.
MA + MB = 100g
ρ AV E =
MB
ρB
+
100 g
100 g − M B
ρA
100 ρ B ( ρ A − ρ AV E )
ρ AV E ( ρ A − ρ B )
100 g × 2.71g / cm 3 × (0.534 g / cm 3 − 2.47 g / cm 3 )
MB =
= 97.62 g
3
3
3
2.47 g / cm (0.534 g / cm − 2.71g / cm )
MB =
CB =
97.62 g
= 0.9762 or
100 g
97.62wt %