SME
Materials Science & Material Properties
Crystal structures
Use this for “Copper is FCC with an atomic radius of 0.128 nm and a molar mass of 63.55 g/mol. Find its cell edge and density.”
Diagram
Atoms on corners count ⅛, on edges ¼, on faces ½ and inside 1: n = 4 atoms per cell.
They
touch
along
a
face
diagonal:
4r
=
√
2
a.
This is an example. Type your own values to start a new problem.
ρ (g/cm³)
=
×
A (g/mol)
(
a (nm)
)
3
×
6.022
×
10
23
R
Atomic radius
example
Formulas
a
=
k
×
R
,
k
=
2
(SC),
4
÷
√
3
(BCC)
or
2
√
2
(FCC)
for
n
atoms
per
cell
0.362
=
k
×
0.128,
k
=
2
(SC),
4
÷
√
3
(BCC)
or
2
√
2
(FCC)
for
4
atoms
per
cell
ρ
=
n
×
A
÷
(6.022
×
10
23
×
(
a
×
10
-7
)
3
)
8.8955
=
4
×
63.55
÷
(6.022
×
10
23
×
(0.362 ×
10
-7
)
3
)
Assumptions
•
The atoms are hard spheres that touch along the cell’s closest line.
•
A corner atom is shared by 8 cells and a face atom by 2: n = 1, 2 or 4.
•
Avogadro’s number NA = 6.022 × 10²³ per mole; 1 nm = 10⁻⁷ cm.
Step-by-step
WE KNOW
n = 4, R = 0.128 nm, A = 63.55 g/mol
FIND
cell edge (a), density (ρ)
Step 1 · Find cell edge (a)
Use a = k × R, k = 2 (SC), 4 ÷ √3 (BCC) or 2√2 (FCC) for n atoms per cell
Face-centred:
they
touch
along
a
face
diagonal,
so
√
2
a
=
4
R
.
a
=
2
×
√
2
×
R
a
=
2
×
√
2
×
0.128
a
=
2
×
1.414
×
0.128
a
=
2.828
×
0.128
a
=
0.362039
nm
Step 2 · Find density (ρ)
Use ρ = n × A ÷ (6.022 × 10²³ × (a × 10⁻⁷)³)
The
mass
of
the
cell’s
n
atoms
over
its
volume,
with
a
in
cm.
ρ
=
4
×
63.55
÷
(6.022
×
10
23
×
(0.362039 ×
10
-7
)
3
)
ρ
=
4
×
63.55
(6.022 ×
10
23
) × (4.7453 ×
10
-23
)
ρ
=
4
×
63.55
28.58
ρ
=
254.2
28.58
ρ
=
8.8955
g/cm³
Check
Put the numbers back in. Both sides should match.
0.362
=
2
×
√
2
×
0.128
✓
8.8955
=
4
×
63.55
÷
(6.022
×
10
23
×
(0.362039 ×
10
-7
)
3
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.