MATERIAL SCIENCE &
ENGINEERING
Homework: 03
Name: Serajee MD Toriqul Arman
Student ID: 191861317
3.1 Difference between atomic structure and crystal structure:
 Any matter in this universe is collection of atom on the other hand
crystal is regularly bounded solid structure.
Atomic structure
is the arrangement of sub
atomic particles like electron
proton & neutron.
Atomic structure is complex
& the structure is not
stable.it can be define as
structure of positive nucleus
& the negatively charged
electrons.
Atomic structure is
composed of sub atomic
particles: electron proton &
neutron.
Overall, Atomic structure is
same in all atoms.
Crystal structure
is the arrangement of atoms or
molecules. Like crystalline materials
diamond.
Crystal structure is well-organized
lattice structure. Where in this
arrangement, atoms are
interconnected making a uniform
structure.
Crystal structure mainly composed
with anion, cation, free electrons &
metal ion.
Crystal structure are different,
depending on the component
present in the crystal.
3.2 If the atomic radius of lead is 0.175 nm, the volume of its unit cell in
cubic meters:
Given radius of lead atom R=0.175 nm & we have asked to calculate 𝑉𝑐
(volume of unit cell crystal structure) in cubic meters.
Solution:
To calculate 𝑉𝑐 of unit cell we know that, crystal structure unit cell shape of
lead has a FCC crystal structure, thus
VC=16𝑅3 ξ2
Now we can plug the value of R=0.175 in the equation to compute & the
unit of the output will be calculate in cubic meters.
VC=16𝑅3 ξ2
=16ሺ0.175 ∗ 10−9 𝑚ሻ3 ξ2
=1.21 ∗ 10−9 𝑚3
3.3 Showing for the body-centered cubic crystal structure that the unit cell
edge length (a)and the atomic radius R are related through 𝑎 = 4𝑅/ ξ3.
Solution:
At first we can consider a unite cell cube for body-centered cubic crystal,
Every side of the cube is equal to a. & diagonal between two point of cube is
𝑙.
Now we can draw a right tringle in the bottom face of the cube
⊿𝑁𝑂𝑃 & according to Pythagorean Theorem
ሺlሻ2 = 𝑎 2 + 𝑎 2
= 2𝑎 2
now then we can draw next hyphens for triangle "𝑁𝑃𝑄" , hance
തതതതሻ2 = ሺQP
തതതതሻ2 + ሺNQ
തതതതሻ2
ሺNQ
ሺdሻ2 = ሺaሻ2 + ሺ2aሻ2
We know that longest diagonal of the cube d = 4R, R is the atomic radius.
Therefore,
ሺ4Rሻ2 = 𝑎 2 + 2𝑎 2
𝑎=
4𝑅
ξ3
3.4 For the HCP crystal structure, showing that the ideal c/a ratio is 1.633:
Here we can consider the hexagonal structure where bottom layer is a
hexagon & it has 7 particle, we can get 6 tringle from it. If we take one
tringle then we can draw a line from A that intersect BC & also we can draw
a line from C that intersect AB. Now we can get the tringle AED. if we draw a
straight line from D to the middle layer of particle B. now we can draw a
tetrahedron from that. Hence,
From ⊿𝐴𝐸𝐷
𝐴𝐸
cos 30= 𝐴𝐷
ξ3
2
AD=
From the Pythagorean Theorem
𝑎
= 𝐴𝐷
2
𝑎∗2
2∗ξ3
=
𝑎
ξ3
𝐴𝐹 2 = 𝐴𝐷2 + 𝐹𝐷2
𝑎 2
𝑐 2
𝑎2 = ൬ ൰ + ቀ ቁ
2
ξ3
2
2
𝑎
𝑐
𝑎2 =
+
3
4
2
𝑎
𝑐2
2
𝑎 −
=
3
4
2
3𝑎2 − 𝑎
𝑐2
=
3
4
2𝑎2 𝑐 2
=
3
4
𝑐
ξ2
𝑎=
3
2
𝑐 2ξ2
=
= 1.633
𝑎
ξ3
3.5 Molybdenum (Mo) has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight
of 95.94 g/mol. Compute and compare its theoretical density with the experimental value found inside
the front cover of the book.
Solution:
At first we have to calculate the density of molybdenum (Mo):
Hence Given 𝑟 = 0.1363𝑛𝑚
𝑀 = 95.94𝑔/𝑚𝑜𝑙𝑒
Avogadro 𝑁𝑎 = 6.022 ∗ 1023
For BCC we know that number of atom in the unit cell is
𝑁=2
Then we can calculate number of mole:
𝑁
2
𝑛= =
= 3.3211 ∗ 10−24 𝑚𝑜𝑙𝑒
23
𝑁𝑎
6.022∗10
Then we can calculate mass of atom
𝑚 = 𝑛𝑀 = ሺ3.3211 ∗ 10−24 𝑚𝑜𝑙𝑒ሻሺ95.94𝑔/𝑚𝑜𝑙𝑒ሻ
= 3.1863 ∗ 10−22 𝑚𝑜𝑙𝑒
4𝑅 3
Volume of the unit cell 𝑉𝑐 = ቀ ቁ =
ξ3
64
3 ξ3
ሺ0.1363𝑛𝑚ሻ 3
= 3.1187 ∗ 10−2 𝑛𝑚3
Density 𝜌 =
𝑚𝑎𝑠𝑠 𝑜𝑓 2 𝑎𝑡𝑜𝑚
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
=
3.1863∗10−22 𝑔
3.1187∗10−2 𝑛𝑚3
= 1.02168 ∗ 10−20 𝑔/𝑛𝑚3
3.6 Determining the indices for the directions shown in the following cubic
unit cell:
Solution:
Here we can see 4 pointed direction vector we have to find miller indices for
that.
Direction
vectors
Vector origin head-tail
A
(0,1,1)-(1,0,1)
1
1
2
2
B
( ,1, )-(0,0,0)
C
(1,0,0)-(1, ,1)
D
(1,0, )-( ,1,0)
1
2
1
1
2
2
Integer
Miller indices
(-1,0,0)
ഥ𝟎𝟎ሿ
ሾ𝟏
(1,2,1)
ሾ𝟏𝟐𝟏ሿ
(0,-1,-2)
ഥഥ
ሾ𝟎𝟏
𝟐ሿ
(1,-2,1)
ഥ𝟏ሿ
ሾ𝟏𝟐
3.7 Determining the Miller indices for the planes shown in the following unit
cell:
Miller indices for the plane (A)
Unit Cell
Intercept
x
Reciprocal
1
Miller Indices: [111ത]
y
1
1
1
z
-1
-1
Miller indices for the plane (B)
Unit Cell
Intercept
x
Reciprocal
2
1
2
Y
1
2
z
O
2
0
Miller Indices: [2 2 0]
3.8 (a) linear density expressions for BCC [110] and [111] directions in terms
of the atomic radius R.
In the figure below shown a [110] directions with in a BCC unit cell.
[110]
in the [110]-direction vector we can see there is only one atom/ particle
therefore, we can get the result above by 1 to get the expression for the
linear density which yield
=
ξ3
4𝑅 ξ2
Thus, here 1 atom that is considered on the direction vector the length of
the direction vector is denoted by x in this figure which is equal to
𝑥 = ඥ𝑧 2 − 𝑦 2
same as we found in the 3.3 ሺlሻ2 = 𝑎 2 + 𝑎 2
thus
𝑥=
2
32𝑅2
2
൰ =ඨ
= 4𝑅ඨ
3
3
ξ3
4𝑅
ඨሺ4𝑅ሻ2 ൬
Therefore, the expression for the linear density of this direction is
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚 𝑐𝑒𝑛𝑡𝑒𝑟𝑒𝑑 𝑜𝑛 ሾ110ሿ𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ሾ110ሿ 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟
1𝑎𝑡𝑜𝑚
ξ3
=
=
2 4𝑅ξ2
4𝑅ට
3
A BCC unit cell within which is drawn a [111] direction is shown below.
𝐿𝐷110 =
[111]
Vector length in the [111] directing is is the longest diagonal length of the
cube, in this direction we can get 2mole/ particle. Furthermore, the length
of the vector shown is equal to 4r,
𝐿𝐷111 =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚 𝑒𝑛𝑡𝑒𝑟 𝑜𝑛 ሾ111ሿ𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ሾ111ሿ𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟
=
2𝑎𝑡𝑜𝑚𝑠
4𝑅
=
1
2𝑅
(b) Compute and compare linear density values for these same two
directions for iron (Fe)
Iron has an atomic radius of 𝑟 = 0.124𝑛𝑚. to find the density we can use
that formula:
ξ3
ቆ
ቇ 𝑓𝑜𝑟 ሾ110ሿ
4𝑅ξ2
and
1
൬ ൰ 𝑓𝑜𝑟 ሾ111ሿ
2𝑅
Hence now we can plug the value to calculate:
[110]
ξ3
ξ3
=
4𝑅ξ2 4ሺ0.124ሻξ2
= 2.4 ∗ 10−9
[111]
1
=
1
2𝑅 2ሺ0.124ሻ
= 4 ∗ 10−9