Chapter 3: Introduction to the Quantum Theory of Solids¶

Example 3.1, Page 59¶

In [1]:
#Variable declaration
v1=10**5 #m per sec
deltav=0.01 #m per sec
m=9.11*10**-31# kg*m/s
e=1.6*10**-9 #Coulombs

#calculations&Results
#deltaE=0.5*m*(v2**2-v1**2)
#deltav=v2-v1.............deltav<<v1
print "The change in kinetic energy"
deltaE=m*v1*deltav
print "in J is %.2e"%deltaE
deltaE=deltaE/e
print "in eV is %.1e"%deltaE

The change in kinetic energy
in J is 9.11e-28
in eV is 5.7e-19


Example 3.2, Page 68¶

In [2]:
import math

#Variable declaration
P=10
a=5*10**-10 # a=5 Armstrong
h=1.054*10**-34
m=9.11*10**-31 #kg

#Calculations&Results
#alpha*a=%pi
#sqrt((2*m*E2)/h**2)*a=pi
E2=(math.pi**2*h**2)/(2*m*a**2)
print 'E2= %.3e J'%E2
E2=E2*6.2415*10**18
print 'E2= %.2f eV'%E2
E1=1.053 #eV
AE=E2-E1
print 'AE= %.3f eV'%AE

E2= 2.407e-19 J
E2= 1.50 eV
AE= 0.449 eV


Example 3.3, Page 84¶

In [3]:
import math

#Variable declaration
m=9.11*10**-31 #kg
E=1.6*10**-19 #C
h=6.625*10**-34 #J sec

#Calculations
N=(4*math.pi*(2*m)**(3./2)*2*E**(3./2))/(h**3*3)

#Result
print 'E2= %.1e per meter**3'%N

E2= 4.5e+27 per meter**3


Example 3.4, Page 90¶

In [8]:
import math

#Variable declaration
y=10
Ni=10
gi=10

#Calculations
#(gi-Ni)!=1
z=(math.factorial(gi)/math.factorial(Ni)*math.factorial(gi-Ni))

#Result
print 'the possible number of ways of realizing a particular distribution = %.f'%z

the possible number of ways of realizing a particular distribution = 1


Example 3.5, Page 90¶

In [11]:
import math

#Variable declaration
gi=10
Ni=9

#Calculations
x=gi-Ni
#factorial(gi-Ni)=1
z=(math.factorial(gi)/math.factorial(Ni)*math.factorial(gi-Ni))

#Result
print 'possible arrangements = %.f'%z

possible arrangements = 10


Example 3.6, Page 93¶

In [12]:
import math

#Variable declaration
T=300 #K

#Calculations
#a=E-Ef
#a=3*k*T
#fF(E)=1/(1+exp(E-Ef/kT))
z=math.exp(3)
y=1./(1+z)                    #y=fF(E)

#Result
print 'fF(E)=%.2f %%'%(y*100)

fF(E)=4.74 %


Example 3.7, Page 94¶

In [13]:
import math

#Variable declaration
Ef=6.25 #eV
E=5.95  #eV
p=0.01
k= 8.617 *10**-5 #eV K^-1

#Calculations
#p=1-fF(E)
#fF(E)=1./(1+exp(E-Ef/kT))
#p=1-1./(1+exp(E-Ef/kT))               equation 1
x=E-Ef
y=(1./(1-p))-1     # solving the above equtaion 1
T=x/(k*math.log(y))

#Result
print 'T=%.1f K\n'%T

T=757.6 K