MatE271
Answer HW # 1
9/14/2001
[1] See page 39-40
Though the straight bonding between Ethylene-molecule’s C-C and C-H is strong covalent bonds,
the bonding occurs between adjacent sections of the long Ethylene-molecule chains is a weak
secondary bonding (van der Waals bonding), which leads to the low strength and low melting
points for the Ethylene-molecule. By contrast, diamond has covalent bonding between each
adjacent pair of the C atoms, which leads to a exceptional high hardness and a high melting point.
[2] See page 65 figure 3-5
Aluminum is an FCC structure
R = 0.143nm
a = 4R
2 = 0.4045 nm
d
Vol = a 3 = 0.4045 x10-9
i
3
= 6.6184 x 10-29 m3
[3] see page 61 fig-3.3
No of atoms / cell = 8 x
1
= 1 atom
8
R = 0.126nm
a = 2R = 0.126 * 2 = 0.252 nm
Vol / unit cell = a 3 = 0(.252 x10-9 ) 3 = 1.6003 x 10-29
m3
Na = 6.023 x1023 atoms / mol
atomic mass = Weight / Na =
Density =
70.4 x10-3 (kg / mol)
= 1.1689 x10-25 kg
6.023 x1023 (atoms / mol)
No. of atoms / cell x atomic mass 1 x1.1689 x10-25
=
= 7,304.3 kg m3
Unit cell volume
1.6003 x 10-29
[4] See page 63 table-3.2
If a cubic system will have a face atom on its
base and top only, it will lose the cubic
b
g
symmetry a a = b ≠ c and becomes a simple
tetragonal structure
[5] Noting the footnote for table 3.1, we see that, when β=90º, the monoclinic system becomes
orthorhombic.