SME
Materials Science & Material Properties · Problem type
Atomic packing factor
Use this for “Iron is BCC with an atomic radius of 0.124 nm. What fraction of its cell do the atoms fill?”
Diagram
Atoms on corners count ⅛, on edges ¼, on faces ½ and inside 1: n = 2 atoms per cell.
They
touch
along
the
body
diagonal:
4r
=
√
3
a.
This is an example. Type your own values to start a new problem.
APF
=
×
4
3
×
π
×
(
R (nm)
)
3
(
a (nm)
)
3
Formulas
a
=
k
×
R
,
k
=
2
(SC),
4
÷
√
3
(BCC)
or
2
√
2
(FCC)
for
n
atoms
per
cell
0.2864
=
k
×
0.124,
k
=
2
(SC),
4
÷
√
3
(BCC)
or
2
√
2
(FCC)
for
2
atoms
per
cell
APF
=
n
×
(
4
3
)
×
π
×
R
3
a
3
0.6802
=
2
×
(
4
3
)
×
π
×
0.124
3
0.2864
3
Assumptions
•
The atoms are hard spheres that touch along the cell’s closest line.
•
n atoms per cell: SC 1, BCC 2, FCC 4.
•
The factor is the same for every radius: SC 0.52, BCC 0.68, FCC 0.74.
Step-by-step
WE KNOW
n = 2, R = 0.124 nm
FIND
cell edge (a), atomic packing factor (APF)
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
Body-centred:
they
touch
along
the
body
diagonal,
so
√
3
a
=
4
R
.
a
=
4
×
R
÷
√
3
a
=
4
×
0.124
÷
√
3
a
=
4
×
0.124
1.732
a
=
0.496
1.732
a
=
0.286366
nm
Step 2 · Find atomic packing factor (APF)
Use APF = n × (4 ÷ 3) × π × R³ ÷ a³
The
volume
of
the
cell’s
n
spheres
over
the
cell’s
volume
a
3
.
APF
=
2
×
(
4
3
)
×
π
×
0.124
3
0.286366
3
APF = 0.6802
Check
Put the numbers back in. Both sides should match.
0.2864
=
4
÷
√
3
×
0.124
✓
0.6802
=
2
×
(
4
3
)
×
π
×
0.124
3
0.286366
3
✓
Topic check
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