Chapter 16 Bonding in solids

Example 16.1 Page no 75

In [12]:
#Given
p1=6.0
p2=12

#Calculation
R06=(p2/p1)
Umin=1/R06*(1-(1/R06))

#Result
print"(a) The inner molecular distance R0=(",R06,"B/A)**1/6"
print"(b) The minimum potential energy is",-Umin,"A**2/B"

(a) The inner molecular distance R0=( 2.0 B/A)**1/6
(b) The minimum potential energy is -0.25 A**2/B

Example 16.2 Page no 76

In [21]:
#Given
m=2.0
n=10

#Calculation
R02=(n/m)**0.25
R10=(n/m)**1.25
U=-((1/R02)+(1/R10))

#Result
print"Potential energy is", round(U,1),"A/R0**2"

Potential energy is -0.8 A/R0**2