Chapter 6:Electron Theory and Band Theory of Metals

Example 6.1 , Page no:146

In [1]:
import math
from __future__ import division

#given
n=8.5E28; #in 1/m^3 (density of electron)
m_e=9.11E-31; #in Kg (mass of electron)
k=1.38E-23; #in J/K (Boltzmann's constant)
e=1.6E-19; #in C (charge of electron)
T=300; #in K (temperature)
p=1.69E-8; #in ohm-m (resistivity)

#calculate
lambda1=math.sqrt(3*k*m_e*T)/(n*e**2*p); #calculation of mean free path
lambda2=lambda1*1E9; #changing unit from m to nanometer

#result
print"The mean free path of electron is =",'%.3E'%lambda1,"m";
print"\t\t\t\t  =",round(lambda2,2),"nm";
#Note: answer in the book is wrong due to printing mistake

The mean free path of electron is = 2.893E-09 m
				  = 2.89 nm

Example 6.2 , Page no:146

In [2]:
import math
from __future__ import division

#given
k=1.38E-23; #in J/K (Boltzmann's constant)
e=1.6E-19; #in C (charge of electron)
P_E=1; #in percentage (probability that a state with an energy 0.5 eV above Fermi energy will be occupied)
E=0.5; #in eV (energy above Fermi level)

#calculate
P_E=1/100; #changing percentage into ratio
E=E*e; #changing unit from eV to J
#P_E=1/(1+exp((E-E_F)/k*T))
#Rearranging this equation, we get
#T=(E-E_F)/k*log((1/P_E)-1)
#Since E-E_F has been denoted by E  therefore
T=E/(k*math.log((1/P_E)-1));

#result
print"The temperature is T=",round(T),"K";
# Note: There is slight variation in the answer due to logarithm function

The temperature is T= 1262.0 K

Example 6.3 , Page no:147

In [3]:
import math
from __future__ import division

#given
n=5.8E28; #in 1/m^3 (density of electron)
m=9.1E-31; #in Kg (mass of electron)
e=1.6E-19; #in C (charge of electron)
p=1.54E-8; #in ohm-m (resistivity)

#calculate
t=m/(n*e**2*p); #calculation of relaxation time

#result
print"The relaxation time of conduction electrons is =",'%.3E'%t,"sec";

The relaxation time of conduction electrons is = 3.980E-14 sec

Example 6.4 , Page no:147

In [4]:
import math
from __future__ import division

#given
n=8.5E28; #in 1/m^3 (density of electron)
m=9.1E-31; #in Kg (mass of electron)
e=1.6E-19; #in C (charge of electron)
sigma=6E7; #in 1/ohm-m (conductivity)
E_F=7; #in E=eV (Fermi energy of Copper)

#calculate
E_F=E_F*e; #changing unit from eV to J
v_F=math.sqrt(2*E_F/m); #calculation of velocity of electrons
#Since sigma=n*e^2*lambda/(2*m*v_F)
#Therefore we have
lambda1=2*m*v_F*sigma/(n*e**2); #calculation of mean free path
lambda2=lambda1*1E10; #changing unit from m to Angstrom

#result
print"The velocity of the electrons is v_F=",'%.3E'%v_F,"m/s   (roundoff error)";
print"The mean free path traveled by the electrons is=",round(lambda2),"Angstrom";

The velocity of the electrons is v_F= 1.569E+06 m/s   (roundoff error)
The mean free path traveled by the electrons is= 787.0 Angstrom

Example 6.5 , Page no:147

In [5]:
import math
from __future__ import division

#given
n=6.5E28; #in 1/m^3 (density of electron)
m=9.1E-31; #in Kg (mass of electron)
e=1.6E-19; #in C (charge of electron)
p=1.43E-8; #in ohm-m (resistivity)

#calculate
t=m/(n*e**2*p); #calculation of relaxation time

#result
print"The relaxation time of conduction electrons is =",'%.3E'%t,"sec";

The relaxation time of conduction electrons is = 3.824E-14 sec

Example 6.6 , Page no:148

In [6]:
import math
from __future__ import division

#given
T=30; #in Celcius (temperature)
k=1.38E-23; #in J/K (Boltzmann's constant)
m_p=1.67E-27; #in Kg (mass of proton)
e=1.6E-19; #in C (charge of electron)

#calculate
T=T+273; #changing temperature from Celcius to Kelvin
KE=(3/2)*k*T; #calculation of average kinetic energy
KE1=KE/e; #changing unit from J to eV
m=1.008*2*m_p; #calculating mass of hydrogen gas molecule
c=math.sqrt(3*k*T/m); #calculation of velocity

#result
print"The average kinetic energy of gas ,molecules is KE=",'%.3E'%KE,"J";
print"\t\t\t\t\t\t  =",round(KE1,4),"eV   (Answer is wrong in textbook)";
print"The velocity of molecules is c=",round(c,2),"m/s";

The average kinetic energy of gas ,molecules is KE= 6.272E-21 J
						  = 0.0392 eV   (Answer is wrong in textbook)
The velocity of molecules is c= 1930.27 m/s

Example 6.7 , Page no:148

In [7]:
import math
from __future__ import division

#given
E=10; #in eV (kinetic energy for each electron and proton)
m_e=9.1E-31; #in Kg (mass of electron)
m_p=1.67E-27; #in Kg (mass of proton)
e=1.6E-19; #in C (charge of electron)

#calculate
E=E*e; #changing unit from eV to J
#since E=m*v^2/2
#therefore  v=sqrt(2E/m)
v_e=math.sqrt(2*E/m_e); #calculation of kinetic energy of electron
v_p=math.sqrt(2*E/m_p); #calculation of kinetic energy of proton

#result
print"The kinetic energy of electron is v_e=",'%.3E'%v_e,"m/s";
print"The kinetic energy of proton is v_p=",'%.3E'%v_p,"m/s";
print "Note: Answer is wrong in textbook"

The kinetic energy of electron is v_e= 1.875E+06 m/s
The kinetic energy of proton is v_p= 4.377E+04 m/s
Note: Answer is wrong in textbook

Example 6.8 , Page no:149

In [8]:
import math
from __future__ import division

#given
I=100; #in A (current in the wire)
e=1.6E-19; #in C (charge of electron)
A=10; #in mm^2 (cross-sectional area)
n=8.5E28; #in 1/m^3 (density of electron)

#calculate
A=A*1E-6; #changing unit from mm^2 to m^2
v_d=I/(n*A*e);

#result
print"The drift velocity of free electrons is v_d=",'%.3E'%v_d,"m/s";

The drift velocity of free electrons is v_d= 7.353E-04 m/s

Example 6.9 , Page no:149

In [9]:
import math
from __future__ import division

#given
I=4;#in A (current in the conductor)
e=1.6E-19; #in C (charge of electron)
A=1E-6; #in m^2 (cross-sectional area)
N_A=6.02E23; #in atoms/gram-atom (Avogadro's number)
p=8.9; #in g/cm^3 (density)
M=63.6; #atomic mass of copper

#calculate
n=N_A*p/M; #Calculation of density of electrons in g/cm^3
n1=n*1E6; #changing unit from g/cm^3 to g/m^3
v_d=I/(n1*A*e);

#result
print"The density of copper atoms is n=",'%.3E'%n,"atoms/m^3";
print"\t\t\t        =",'%.3E'%n1,"atoms/m^3";
print"The average drift velocity of free electrons is v_d=",'%.3E'%v_d,"m/s";

The density of copper atoms is n= 8.424E+22 atoms/m^3
			        = 8.424E+28 atoms/m^3
The average drift velocity of free electrons is v_d= 2.968E-04 m/s

Example 6.10 , Page no:149

In [10]:
import math
from __future__ import division

#given
n=9E28; #in 1/m^3 (density of valence electrons)
sigma=6E7; #in mho/m (conductivity of copper)
e=1.6E-19; #in C (charge of electron)

#calculate
#Since sigma=n*e*mu  therefore
mu=sigma/(n*e); #calculation of mobility of electron

#result
print"The mobility of electrons is =",'%.3E'%mu,"m^2/V-s";

The mobility of electrons is = 4.167E-03 m^2/V-s

Example 6.11 , Page no:150

In [11]:
import math
from __future__ import division

#given
E_F=5.51; #in eV (Fermi energy in Silver)
k=1.38E-23; #in J/K (Boltzmann's constant)
e=1.6E-19; #in C (charge of electron)

#calculate
#part-(a)
Eo=(3/5)*E_F; #calculation of average energy of free electron at 0K
#part-(b)
Eo1=Eo*e; #changing unit from eV to J
#Since for a classical particle E=(3/2)*k*T
#therefroe we have
T=(2/3)*Eo1/k; #calculation of temperature for a classical particle (an ideal gas)

#result
print"The average energy of free electron at 0K is Eo=",Eo,"eV";
print"The temperature at which a classical particle have this much energy is T=",'%.3E'%T,"K";

The average energy of free electron at 0K is Eo= 3.306 eV
The temperature at which a classical particle have this much energy is T= 2.555E+04 K

Example 6.12 , Page no:150

In [12]:
import math
from __future__ import division

#given
E_F_L=4.7; #in eV (Fermi energy in Lithium)
E_F_M=2.35; #in eV (Fermi energy in a metal)
n_L=4.6E28; #in 1/m^3 (density of electron in Lithium)

#calculate
#Since n=((2*m/h)^3/2)*E_F^(3/2)*(8*pi/3) and all things except E_F are constant
# Therefore we have n=C*E_F^(3/2)   where C is proportionality constant
#n1/n2=(E_F_1/E_F_2)^(3/2)
#Therefore we have
n_M=n_L*(E_F_M/E_F_L); #calculation of electron density for a metal

#result
print"The lectron density for a metal is =",n_M,"1/m^3";
print "Note: Answer in the book is wrong "

The lectron density for a metal is = 2.3e+28 1/m^3
Note: Answer in the book is wrong