3.091 Fall Term 2004
Homework #6
Solutions
From the text:
9-50.
n
a3
=
N Av
, Vmolar
Vmolar
=
M atomic
ρ
+ solve for n, the number of atoms/unit cell.
N Av ρ
6.02 × 1023 × 0.856
× a3 =
× (5.247 × 10−8 )3
39.0893
M atomic
thus, K must be BCC with 2 atoms/unit cell
∴ n =
9-51.
= 1.90 ≅ 2
⎛
⎞
1
mol
51.996
g
Cr
⎜
⎟
×
⎜
6.022 × 1023 ⎟
1 mol
⎝
⎠
Vunit cell = 2
g
7.20 3
cm
Vunit cell = a3 = 2.40 × 10–23 cm3
a=
3
2.40 × 10−23 cm3 = 2.88 × 10-8 cm
d2body = a2 + a2 + a2 = 3a2
dbody = 3a = 3 × (2.88 × 10-8 cm) = 5.00 × 10-8 cm
dbody = 4 rCr
rCr =
9-59.
5.00 × 10−8 cm
= 1.25 × 10-8 cm
4
From the geometry of the body-centered cubic structure, we find the shortest distance
between barium atoms will occur down the body diagonal where
dbody = 3 a = 4rBa
rBa =
3
1 m
100 cm
× 0.5025 nm × 9
×
= 2.178 × 10–8 cm
4
10 nm
1 m
The shortest distance between barium atoms will be 2 rBa.
The Ba–Ba distance = 4.356 × 10–8 cm.
3.091 Fall Term 2004 Homework #6 Solutions
page 2
Additional questions:
0.
in BCC there are 2 atoms / unit cell, so
2
a3
=
2
a3
=
N Av ⋅ ρ
A
N Av
, where Vmolar = A/ρ
Vmolar
(A is atomic mass of iron)
1
⎛ 2A ⎞ 3
4
∴ a = ⎜
=
r
⎟
N
ρ
⋅
3
⎝ Av
⎠
−8
∴ r = 1.24 ×10 cm
if we assume that change of phase does not change the radius of the iron atom, then we
repeat the calculation in the context of an FCC crystal structure, i.e., 4 atoms per unit cell
and a = 2 2 r
ρ =
N Av
4A
(2 2 r )3
= 8.60 g cm −3
FCC iron is more closely packed than BCC suggesting that iron contracts upon changing
from BCC to FCC. This is consistent with the packing density calculations reported in
lecture that give FCC as being 74% dense and BCC 68% dense. The ratio of the densities
calculated here is precisely the same:
7.86
0.68
=
8.60
0.74
1.
First determine the P.D. (packing density) for Au (FCC); then relate it to the molar
volume given in the PT.
16π r3
Vol.Atom UC
16π r3
= 33 =
P.D. =
3
Vol.UC
a
3a
16πr 3
=
3 × 16 2r 3
π
= 0.74 =74%
3 2
Void Volume = 1 – P.D. = 1 – 0.74 = 100% – 74%
P.D. =
From the Packing Density (74%) we recognize the void volume to be 26%. Given the
molar volume at 10.3 cm3/mole, the void volume is:
0.26 × 10.3 cm3 = 2.68 cm3/mole
3.091 Fall Term 2004 Homework #6 Solutions
page 3
(a) The answer can be found by looking at a unit cell of Cu (FCC).
2.
z
Nearest neighbor distance is observed
along <110>; second nearest along
<100>. The second nearest neighbor
distance is found to be “a” (Another way
of finding it is looking at LN4, page 12).
y
x
a
Cu: at. Vol. = 7.1 × 10-6 m3/mole =
N 3
a (Cu: FCC; 4 at/UC)
4
7.1 × 10 −6 × 4
a = 3
= 3.61 × 10 −10 m
23
6.02 × 10
(b) d
hkl
a
=
h2 + k 2 + l 2
3.61 × 10 −10
d
=
= 2.55 × 10 −10 m
110
2
Let's look at the unit cell.
3.
z
[101]
[011]
[110]
y
[110]
x
[011]
[101]
There are six [110]-type directions in the (111) plane. Their indices are:
3.091 Fall Term 2004 Homework #6 Solutions
page 4
[10 1 ], [1 01], [1 10], [1 1 0 ], [0 1 1], [011 ]
Determine the lattice parameter and look at the Unit Cell occupation.
Ba: BCC; at.vol. = 39.24 cm3/mole; n = 2 atoms/UC
4.
3.924 × 10−5 ⎛⎝ m 3 / mole⎞⎠ =
z
3
a=
y
[110]
x
N
A a3
2
2 × 3.924 × 10 −5
23
6.02 × 10
a = 5.08 × 10-10 m
1 atom
linear density = a 2
=
1
5.08 × 10 −10 ×
2
= 1. 39 × 10 9 atoms/m
Aluminum at 300 K has FCC structure:
5.
[001]
(110) plane
[010]
[100]
Volume unit of a cell:
V =
10 cm 3
1 mole
4 atoms
×
×
mole
1 unit cell
6.02 × 10 23 atoms
= 6.64 × 10 −23 cm3 / unit cell
3.091 Fall Term 2004 Homework #6 Solutions
V=a
3
1
∴ a = ⎛⎝ 6.64 × 10 − 23 cm3 ⎞⎠ 3
2a = 4r ; ∴atomic radius r =
For FCC:
= 1.43 × 10-8 cm
page 5
= 4.05 × 10 −8 cm
2
a =
4
2 ⎛
4.05 × 10 −8 cm⎞⎠
⎝
4
Planar Packing Fraction of the (110) plane:
2a 2
⎛ 1⎞
⎛ 1⎞
number of lattice points in the shaded area = 2⎜ ⎟ + 4 ⎜ ⎟ = 2
⎝ 2⎠
⎝ 4⎠
area occupied by 1 atom = π r 2
2 ⋅ π r2
area occupied by atoms
Fraction =
=
total area
2a 2
2
−8
⎛
⎞
2 ⋅ π ⎝ 1.43 × 10
cm⎠
=
2 = 0.554
2 ⎛⎝ 4.05 × 10 −8 cm⎞⎠
area of shadowed plane in above unit cell =
Linear Packing Density of the [100] direction:
Density =
6.
1 atom
1 atom
=
= 2.47 × 10 7 atoms / cm
−8
a
4.05 × 10 cm
(111) inverse =
z
1 1 1
1 1 1
x = 1, y = 1, z = 1
This plane intersects x axis at x = 1
y axis at y = 1
z axis at z=1
y
x
(210) inverse =
1 1 1
2 1 0
3.091 Fall Term 2004 Homework #6 Solutions
z
page 6
x = 1/2, y = 1, z = infinity
This plane intersects x axis at x = 1/2
y axis at y = 1
This plane does not intersect the z axis.
y
x
(003) inverse =
z
1 1 1
0 0 3
x = infinity, y = infinity, z = 1/3
This plane does not intersect either the x or y axis.
This plane intersects the z axis at z = 1/3.
y
x
(011) looks like this:
8.
4×
a
1
atoms = 1 atom
4
area = 2 a 2
2 a
1/ 3
1
N Av
22.23 ⎞
⇒ a = ⎛⎝
= 3.33 × 10−8 cm
3 =
23 ⎠
a
Vmolar
6.02 × 10
1
14
2
∴ atomic density =
cm
2 = 6.376 × 10
2a
3.091 Fall Term 2004 Homework #6 Solutions
9.
page 7
A characteristic x-ray spectrum ofCr will show λSWL, Kβ, Kα and the continuous
spectrum or Bremsstrahlung. We may quantify λKα and λSWL.
24Cr:
ν K = 3 4 R ( Z − 1) = 3 4 × 1.097 ×107 ( 23)
= 4.35 × 109 m–1
λK = 2.3 × 10–10 m
2
2
α
α
λ SWL = hc
=
1.24 ×10−6 m =
V
1.24 ×10−6 m
6 ×104
eV
= 2.07 × 10–11 m
Kα
Kβ
characteristic lines
intensity
continuous spectrum
λSWL
10.
(a) λSWL =
λ
12400 12400
=
= 0.188 Å
v
66 ×103
(b) see sketch above in answer to problem 9. Lα and Lβ line will appear to the right of
the analogous K lines (at higher values of λ), the Lα to the right of the Lβ.
(c) - incident electrons are deflected by negative charge if the electrons of the target
- change in velocity (speed or direction or both) is an acceleration.
- accelerating charge emits radiation.
- extent of acceleration is NOT QUANTIZED.
- spectrum is continuous.