5.3 Solutions 303
5.11 Force,F=−
dV
dr=
a
r^2−
7 b
r^8
The particles will separate most easily when the force between them is a min-
imum, that is whendF
dr=0. This gives:
dF
dr=−
2 a
r^3+
56 b
r^9= 0
r=(
28 b
a) 1 / 6
5.12 The inter-nuclear distance is found from
dE
dr= 0
2 A
ro^3−
8 B
ro^9= 0 →ro^6 =4 B
A
(1)
The dissociation energyDis formed from−D=E(ro)−D=−A
r^2 o+
B
ro^8=−
A
ro^2+
A
4 ro^2=−
3 A
4 ro^2where we have used (1).A=
4 Dr^2 o
3
=4
3
× 3 × 1. 6 × 10 −^19 ×(0. 4 × 10 −^9 )^2 = 1. 02 × 10 −^37
B=
A
4
r^60 =1. 02 × 10 −^37
4
×(0. 4 × 10 −^9 )^6
= 1. 04 × 10 −^91
5.13=
(
2 kT
K) 1 / 2
The force constantK =Ya 0 = 1. 6 × 1010 × 4. 94 × 10 −^10 = 7 .9N/m^2T=K
2 k〈A〉^2 =
7. 9 ×(0. 46 × 10 −^10 )^2
2 × 1. 38 × 10 −^23
=606 K= 333 ◦C
5.3.3 Metals ...........................................
5.14 (a)vF=(
2 EF
m) 1 / 2
=
[
2 × 5. 52 × 1. 6 × 10 −^19
9. 11 × 10 −^31
] 1 / 2
= 1. 39 × 106 m/s(b)τ=m
ne^2 ρ=
9. 11 × 10 −^31
(5. 86 × 1028 )(1. 6 × 10 −^19 )^2 (1. 62 × 10 −^8 )
= 3. 7 × 10 −^14 s