9.7 The Model System with Potential Energy
421
Solution
a.
u(σ) 4ε
σ
σ
12 −
σ
σ
6
4ε(1 − 1) 0
b. At an extremum, the derivative du/dr is equal to zero:
du
dr
4ε
−12σ 12 1
r 13 + 6σ 6 1
r 7
0
0 4ε
6σ 6 1
r 7
−
2σ 6
r 6 + 1
This expression equals zero if r → ∞ or if
r 2 1/6 σ 1.12246σ
The minimum is at r r min 2 1/6 σ. There is a relative maximum at r → ∞.
Exercise 9.21
a. Show that
u LJ (r min ) −ε
(9.7-8)
b. The force in the r direction is given by −du/dr. Show that the force on one particle due to
another particle at distance r is
F r 4ε
12
σ 12
r 13 − 6
σ 6
r 7
(9.7-9)
c. Show that F r 0 if r 2 1/6 σ.
In a liquid or solid the average separation of the molecules from their nearest neighbors is approximately equal to the intermolecular distance at the minimum in u. Energy
is required either to compress a liquid or solid or to pull its molecules away from each
other, which results in a nearly constant volume. In a gas at ordinary pressures, the
average nearest-neighbor distance is roughly 10 times as great as in a liquid, and the
average intermolecular forces are relatively small.
A simpler but less realistic representation of the pair potential function is the square
well potential
u(r)
⎧
⎨
⎩
∞
(0 < r < d)
−u 0 (d < r < c)
0
(r > c)
(9.7-10)
This representation is shown in Figure 9.16. If optimum values of d and c are chosen,
this function can approximate the actual intermolecular potential for some purposes.
u
r = d r = c
r
0
2u 0
`
Figure 9.16 The Square-Well Representation of the Intermolecular Potential of a Pair of Atoms.