18.22 Yes, compound semiconductors can exhibit intrinsic behavior. Provided are that there are no dopants, the
electronic structure of a compound semiconductor can be described as having an almost full valence band and
almost empty conduction band with a small number of electrons thermally excited across the forbidden energy
gap.
18-31
18.27 (a) No hole is generated by an electron excitation involving a donor impurity atom because the
excitation comes from a level within the band gap, and thus, no missing electron is created within the normally
filled valence band.
(b) No free electron is generated by an electron excitation involving an acceptor impurity atom because the
electron is excited from the valence band into the impurity level within the band gap; no free electron is introduced
into the conduction band.
18.28 Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table (Figure 2.6),
and an N atom has one more valence electron than an Si atom.
Boron will act as an acceptor in Ge. Since it (B) is from group IIIA of the periodic table, a B atom has one
less valence electron than a Ge atom.
Sulfur will act as a donor in InSb. Since S is from group VIA of the periodic table, it will substitute for Sb;
also, an S atom has one more valence electron than an Sb atom.
Indium will act as a donor in CdS. Since In is from group IIIA of the periodic table, it will substitute for
Cd; and, an In atom has one more valence electron than a Cd atom.
Arsenic will act as an acceptor in ZnTe. Since As is from group VA of the periodic table, it will substitute
for Te; furthermore, an As atom has one less valence electron than a Te atom.
1The atomic weight of vanadium (V) is 50.94 g/mole. Its density is =5.96 g/cm3. It
has the BCC crystal structure. Calculate the atomic radius of vanadium. Confirm that
this matches the value given in Appendix A.
3a = 4 R
a
2a
V has the BCC crystal structure,
3
1
4R then Vv= , and there are 1+ 8 =2 atoms in one unit cell
8
3
(R is the Radius of V atomVv= volume of the BCC vanadium unit cell)
=
nA
nA
=
3
Vv N A
4R NA
3
1
nA 3
3
R = 4
NA 1
3
2 50.94 g / mole
3
= 3
23
4
5.96 g / cm 6.023 10 / mole = 1.32 108 cm
=0.132 nm
In good agreement with the tabulated values
Core 3.20
For FCC
BCC
Vc = a 3 = ( 2 R 2 ) 3
Vc' = a'3 = (
percentage =
4R
3
)3
No. of unit cell N
No. of unit cell 2N
2 NVc '− NVc
= 8.87%
NVc
7.A
8.C
9.B
3.1 Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as
the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to
the arrangement of atoms in the crystalline solid material.
3.14 For each of these three alloys we need, by trial and error, to calculate the density using Equation 3.5,
and compare it to the value cited in the problem. For SC, BCC, and FCC crystal structures, the respective values of
4R
.
n are 1, 2, and 4, whereas the expressions for a (since VC = a3) are 2R, 2 R 2 , and
3
For alloy A, let us calculate ρ assuming a BCC crystal structure.
ρ =
=
=
nAA
VC N A
nAA
⎛ 4 R ⎞3
⎜ ⎟ NA
⎝ 3⎠
(2 atoms/unit cell)(43.1 g/mol)
⎧⎡
⎫
3
⎪ (4)(1.22 × 10-8 cm) ⎤
⎪
⎥ /(unit cell) ⎬ (6.023 × 1023 atoms/mol)
⎨⎢
3
⎥⎦
⎪⎢⎣
⎪
⎩
⎭
= 6.40 g/cm3
Therefore, its crystal structure is BCC.
For alloy B, let us calculate ρ assuming a simple cubic crystal structure.
ρ =
=
nAB
(2a) 3 N A
(1 atom/unit cell)(184.4 g/mol)
⎫
⎧
3
⎨ (2)(1.46 × 10-8 cm) /(unit cell) ⎬ (6.023 × 1023 atoms/mol)
⎭
⎩
[
]
= 12.3 g/cm3
Therefore, its crystal structure is simple cubic.
For alloy C, let us calculate ρ assuming a BCC crystal structure.