5.3 Solutions
303
5.11 Force, F = −
dV
dr
=
a
r 2 −
7b
r 8
The particles will separate most easily when the force between them is a minimum, that is when
dF
dr
= 0. This gives:
dF
dr
= −
2a
r 3 +
56b
r 9 = 0
r =
28b
a
1/6
5.12 The inter-nuclear distance is found from
dE
dr
= 0
2A
r 3
o
−
8B
r 9
o
= 0 → r
6
o =
4B
A
(1)
The dissociation energy D is formed from −D = E(r o )
−D = −
A
r 2
o
+
B
r 8
o
= −
A
r 2
o
+
A
4r 2
o
= −
3A
4r 2
o
where we have used (1).
A =
4Dr
2
o
3
=
4
3
× 3 × 1.6 × 10
−19
× (0.4 × 10
−9 )
2
= 1.02 × 10
−37
B =
A
4
r
6
0 =
1.02 × 10
−37
4
× (0.4 × 10
−9 )
6
= 1.04 × 10
−91
5.13 =
2kT
K
1/2
The force constant K = Ya 0 = 1.6 × 10
10
× 4.94 × 10
−10
= 7.9 N/m
2
T =
K
2k
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) v F =
2E F
m
1/2
=
2 × 5.52 × 1.6 × 10
−19
9.11 × 10 −31
1/2
= 1.39 × 10
6 m/s
(b) τ =
m
ne 2 ρ
=
9.11 × 10
−31
(5.86 × 10 28 )(1.6 × 10 −19 ) 2 (1.62 × 10 −8 )
= 3.7 × 10
−14 s