3.13 In order to determine whether Nb has an FCC or a BCC crystal structure, we need to compute its
density for each of the crystal structures.
For FCC,
n
= 4, and
a
=
2
R
2
(Equation 3.1). Also, from Figure 2.6, its
atomic weight is 92.91 g/mol.
Thus, for FCC (employing Equation 3.5)
ρ
=
nA
Nb
a
3
N
A
=
nA
Nb
(
2
R
2
)
3
N
A
=
(4 atoms/unit cell)(92.91 g/mol)
(2)
(
1.43
×
10
-8
cm
)
(
2
)
[
]
3
/(unitcell)
⎧
⎨
⎩
⎫
⎬
⎭
(
6.023
×
10
23
atoms /mol
)
= 9.33 g/cm
3
For BCC,
n
= 2, and
a
=
4
R
3
(Equation 3.3), thus
ρ
=
nA
Nb
4
R
3
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
3
N
A
ρ
=
(2 atoms/unit cell)(92.91 g/mol)
(4)
(
1.43
×
10
-8
cm
)
3
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
3
/(unitcell)
⎧
⎨
⎪
⎩
⎪
⎫
⎬
⎪
⎭
⎪
(
6.023
×
10
23
atoms/mol
)
= 8.57 g/cm
3
which is the value provided in the problem statement.
Therefore, Nb has a BCC crystal structure.

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3. 17
(a)
We are asked to calculate the unit cell volume for Be.
For HCP, from the solution to Problem
3.6
V
C
= 6
R
2
c
3
But,
c
= 1.568
a
, and
a
= 2
R
, or
c
= 3.14
R
, and
V
C
= (6)(3.14)
R
3
3
= (6)(3.14)
(
3
)
0.1143
×
10
-7
cm
[
]
3
= 4.87
×
10
−
23
cm
3
/unit cell
(b)
The theoretical density of Be is determined, using Equation 3.5, as follows:
ρ
=
nA
Be
V
C
N
A
For HCP,
n
= 6 atoms/unit cell, and for Be,
A
Be
= 9.01 g/mol (as noted inside the front cover).
Thus,
ρ
=
(6 atoms/unit cell)(9.01 g/mol)
(
4.87
×
10
-23
cm
3
/unit cell
)(
6.023
×
10
23
atoms/mol
)
= 1.84 g/cm
3
The value given in the literature is 1.85 g/cm
3
.