9: Energy bands in solids

Example number 9.1, Page number 240

In [29]:
#importing modules
import math
from __future__ import division

#Variable declaration
#E-EF=KT
#K=KB is the boltzmann constant in m^2 Kg s^-2 k^-1

#Calculation
f=1/(1+math.exp(1));    #The fermi function for an energy kt above fermi energy

#Result
print "The fermi function for an energy kt above fermi energy is",round(f,3)

The fermi function for an energy kt above fermi energy is 0.269

Example number 9.2, Page number 240

In [32]:
#importing modules
import math
from __future__ import division

#Variable declaration
X=0.01*1.6*10**-19;    #difference between energy and fermi energy(J)
T=200;    #temperature(K)
KB=1.38*10**-23;    #Boltzmann's Constant(J/K)

#Calculation
f=1/(1+math.exp(X/(KB*T)));     #The fermi function

#Result
print "The fermi function is",round(f,6)
print "answer given in the book is wrong"

The fermi function is 0.358999
answer given in the book is wrong

Example number 9.3, Page number 241

In [37]:
#importing modules
import math
from __future__ import division

#Variable declaration
EF=11.63*1.6*10**-19;    #fermi energy of conducting electron in aluminium(J)
t=7.3*10**-15;       #relaxation time for electron(sec)
m=9.11*10**-31;      #mass of electon(Kg)

#Calculation
Vf=math.sqrt(2*EF/m);     #The fermi velocity fo conducting electron in aluminium(ms^-1)
x=t*Vf*10**9;     #mean free path for conducting electron of aluminium(nm)

#Result
print "The fermi velocity fo conducting electron in aluminium is",round(Vf/10**6,5),"*10**6 ms^-1"
print "The mean free path for conducting electron of aluminium is",round(x,4),"nm"

The fermi velocity fo conducting electron in aluminium is 2.02118 *10**6 ms^-1
The mean free path for conducting electron of aluminium is 14.7546 nm

Example number 9.4, Page number 241

In [46]:
#importing modules
import math
from __future__ import division

#Variable declaration
Vf=0.86*10**6;    #The femi energy of electons in the metal(m/sec)
m=9.11*10**-31;    #mass of electon(Kg)
KB=1.38*10**-23;   #Boltzmann's Constant(m^2 Kg s^-2 k^-1)

#Calculation
Ef=(1/2)*m*Vf**2;   #The fermi energy in a metal(J)
Tf=Ef/KB;    #The fermi temperature in a metal(K)

#Result
print "The fermi energy in a metal is",round(Ef/10**-19,5),"*10**-19 J or",round(Ef/(1.6*10**-19),4),"eV"
print "The fermi temperature in a metal is",round(Tf/10**3,2),"*10**3 K"

The fermi energy in a metal is 3.36888 *10**-19 J or 2.1055 eV
The fermi temperature in a metal is 24.41 *10**3 K

Example number 9.5, Page number 242

In [3]:
#importing modules
import math
from __future__ import division

#Variable declaration
Ef=3.2*1.6*10**-19;    #The fermi energy in a metal(J)
m=9.11*10**-31;   #mass of electon(Kg)
KB=1.38*10**-23;    #Boltzmann's Constant(m^2 Kg s^-2 k^-1)

#Calculation
Tf=Ef/KB;     #The fermi temparature for sodium(K)
Vf=math.sqrt(2*Ef/m);     #The fermi velocity fo conducting electron in aluminium(ms^-1)

#Result
print "The fermi temparature for sodium is",round(Tf/10**3,2),"*10**3 K"
print "The fermi velocity fo conducting electron in aluminium is",round(Vf/10**6,4),"*10**6 ms^-1"

The fermi temparature for sodium is 37.1 *10**3 K
The fermi velocity fo conducting electron in aluminium is 1.0602 *10**6 ms^-1

Example number 9.6, Page number 242

In [7]:
#importing modules
import math
from __future__ import division

#Variable declaration
E=5.5*1.6*10**-19;     #energy level(J)
Ef=5*1.6*10**-19;     #fermi energy level(J)
x=0.5*1.6*10**-19;    #Difference between energy and fermi energy(J)
f=0.01;   #fermi function at which there is 1% probability that an electron in a solid
KB=1.38*10**-23;    #Boltzmann's Constant(m^2 Kg s^-2 k^-1)

#Calculation
T=x/(KB*(math.log(1-f)-math.log(f)));      #The temperature at which there is 1% probability that an electron in a solid(K)

#Result
print "The temperature at which there is 1% probability that an electron in a solid is",round(T/10**3,5),"*10**3 K"

The temperature at which there is 1% probability that an electron in a solid is 1.26158 *10**3 K

Example number 9.7, Page number 244

In [9]:
#importing modules
import math
from __future__ import division

#Variable declaration
Ef=2.1*1.6*10**-19;    #fermi energy level in potassium(J)
f1=0.99;    #fermi factor for 1st
f2=0.01;    #fermi factor for 2nd
f3=0.5;     #fermi factor for 3rd
T=300;     #temperature(K)
e=1.6*10**-19;    #charge of electron(C)
KB=1.38*10**-23;     #Boltzmann's Constant(m^2 Kg s^-2 k^-1)

#Calculation
E1=(Ef+((KB*T)*(math.log(1-f1)-math.log(f1))))/e;    #The energy for probability of occupancy at 1st(eV)
E2=(Ef+((KB*T)*(math.log(1-f2)-math.log(f2))))/e;    #The energy for 1st at which the probability of occupancy(eV)
E3=(Ef+((KB*T)*(math.log(1-f3)-math.log(f3))))/e;    #The energy for 1st at which the probability of occupancy(eV)

#Result
print "The energy for probability of occupancy at 1st is",round(E1,2),"eV"
print "The energy for probability of occupancy at 2nd is",round(E2,3),"eV"
print "The energy for probability of occupancy at 3rd is",E3,"eV"

The energy for probability of occupancy at 1st is 1.98 eV
The energy for probability of occupancy at 2nd is 2.219 eV
The energy for probability of occupancy at 3rd is 2.1 eV

Example number 9.8, Page number 245

In [11]:
#importing modules
import math
from __future__ import division

#Variable declaration
X=0.1*1.6*10**-19;     #difference between energy and fermi energy(J)
T=300;    #temperature(K)
KB=1.38*10**-23;    #Boltzmann's Constant(m^2 Kg s^-2 k^-1)

#Calculation
f=1-1/(1+math.exp(X/(KB*T)));     #The probability of unoccupancy by an electron at room temperature 

#Result
print "The probability of unoccupancy by an electron at room temperature is",round(f,5)

The probability of unoccupancy by an electron at room temperature is 0.97946

Example number 9.9, Page number 252

In [18]:
#importing modules
import math
from __future__ import division

#Variable declaration
n=4;     #number of atoms/unit cell in Al
a=4.05*10**-10;    #lattice constant of Aluminium which is FCC crystal(m)
nf=3;   #number of free electrons per atom in Al
T=300;   #ambient temperature(K)
x=0.1*1.6*10**-19;   #The same difference energy and fermi energy(J)
m=9.11*10**-31;    #mass of electon(kg)
h=6.625*10**-34;   #plank's constant(m^2 Kg/sec)
KB=1.38*10**-23;   #Boltzmann's Constant(m^2 Kg s^-2 k^-1)

#Calculation
nc=n*nf/(a**3);   #number of electrons per unit volume
Ef=h**2/(8*m)*((3*nc)/math.pi)**(2/3);   #The fermi energy for the metal(eV)
f=1/(1+math.exp(x/(KB*T)));    #he fermi factor

#Result
print "The fermi energy for the metal is",round(Ef/(1.6*10**-19),2),"eV"
print "The fermi factor is",round(f,4)

The fermi energy for the metal is 11.66 eV
The fermi factor is 0.0205

Example number 9.10, Page number 253

In [20]:
#importing modules
import math
from __future__ import division

#Variable declaration
n=2;   #number of atoms/unit cell in cesium which is Bcc
a=6.14*10**-10;   #lattice constant of cesium which is BCC crystal(m)
nf=1;   #number of free electrons per atom in cesium
m=9.11*10**-31;   #mass of electon(kg)
h=6.625*10**-34;    #plank's constant(m^2 Kg/sec)
KB=1.38*10**-23;   #Boltzmann's Constant(m^2 Kg s^-2 k^-1)
e=1.6*10**-19;    #charge of electron(C)

#Calculation
nc=n*nf/(a**3);   #number of electrons per unit volume
Ef=(h**2/(8*m)*((3*nc)/math.pi)**(2/3))/e;    #The fermi energy for the metal(eV)

#Result
print "The fermi energy for cesium is",round(Ef,3),"eV"

The fermi energy for cesium is 1.537 eV

Example number 9.11, Page number 254

In [22]:
#importing modules
import math
from __future__ import division

#Variable declaration
Ef=2.1*1.6*10**-19;    #The fermi energy level in potassium at a particular temperature(J)
m=9.11*10**-31;     #mass of electron(kg)
h=6.625*10**-34;    #plank's constant(m^2 Kg/sec)

#Calculation
nc=(8*m/(h**2)*Ef)**(3/2)*(math.pi/3);   #ThE Number of free electrons per unit volume in potassium(electrons/m^3)

#Result
print "The number of free electrons per unit volume in potassium is",round(nc/10**28,2),"*10**28 electrons/m^3"

The number of free electrons per unit volume in potassium is 1.38 *10**28 electrons/m^3

Example number 9.12, Page number 254

In [24]:
#importing modules
import math
from __future__ import division

#Variable declaration
AW=23;    #atomic weight of sodium(gm/mole)
d=0.971*10**6;   #density of sodium(gm/m^3)
m=9.11*10**-31;   #mass of electon(kg)
h=6.625*10**-34;  #plank's constant(m^2 Kg/sec)
AV=6.02*10**23;   #Avagadro number(mole^-1)
e=1.6*10**-19;    #charge of electron(C)

#Calculation
nc=AV*d/AW;    #number of electrons per unit volume
Ef=(h**2/(8*m)*((3*nc)/math.pi)**(2/3))/e;    #The fermi energy for the sodium(eV)

#Result
print "The fermi energy for the sodium is",round(Ef,3),"eV"

The fermi energy for the sodium is 3.155 eV

Example number 9.13, Page number 255

In [26]:
#importing modules
import math
from __future__ import division

#Variable declaration
AW=63.5;    #atomic weight of copper(u)
M=63.5*1.66*10**-27;   #mass of one copper atom(kg)
d=8.94*10**3;    #density of sodium(Kg/m^3)
m=9.11*10**-31;  #mass of electon(Kg)
h=6.625*10**-34;   #plank's constant(m^2 Kg/sec)
e=1.6*10**-19;   #charge of electron(C)

#Calculation
nc=d/M;   #number of electrons per unit volume(electrons/m^3)
Ef=h**2/(8*m)*((3*nc)/math.pi)**(2/3)/e;   #The fermi energy for the sodium(eV)

#Result
print "The fermi energy for the sodium is",round(Ef,3),"eV"
print "answer varies due to rounding off errors"

The fermi energy for the sodium is 7.046 eV
answer varies due to rounding off errors