Chapter 4: Free Electron Theory of Metals

Example 4.1,Page number 112

In [1]:
import math

#Given Data
m = 9.1*10**-31;   # Mass of an electron, kg
e = 1.6*10**-19;   # Charge on an electron, C
n = 8.5*10**28;   # Concentration of electron in Cu, per metre cube
rho = 1.7*10**-8;  # Resistivity of Cu, ohm-m
t = m/(n*e**2*rho);  # Collision time for an electron in monovalent Cu, s
print"The collision time for an electron in monovalent Cu =","{0:.3e}".format(t),"s";
The collision time for an electron in monovalent Cu = 2.460e-14 s

Example 4.2,Page number 112

In [2]:
import math
#Given Data
m = 9.1*10**-31;   # Mass of an electron, kg
e = 1.6*10**-19;   # Charge on an electron, C
n = 10**29;   # Concentration of electron in material, per metre cube
rho = 27*10**-8;  # Resistivity of the material, ohm-m
tau = m/(n*e**2*rho);  # Collision time for an electron in the material, s
v_F = 1*10**8;    # Velocity of free electron, cm/s
lamda = v_F*tau;   # Mean free path of electron in the material, cm
print"The collision time for an electron in monovalent Cu =","{0:.3e}".format(tau),"s";
print"The mean free path of electron at 0K =","{0:.3e}".format(lamda),"cm";
The collision time for an electron in monovalent Cu = 1.317e-15 s
The mean free path of electron at 0K = 1.317e-07 cm

Example 4.3,Page number 112

In [3]:
import math

#Given Data
m = 9.1*10**-31;   # Mass of an electron, kg
e = 1.6*10**-19;   # Charge on an electron, C
r = 1.28*10**-10;  # Atomic radius of cupper, m
a = 4*r/math.sqrt(2);    # Lattice parameter of fcc structure of Cu, m
V = a**3;    # Volume of unit cell of Cu, metre cube
n = 4/V;    # Number of atoms per unit volume of Cu, per metre cube
tau = 2.7*10**-4;  # Relaxation time for an electron in monovalent Cu, s
sigma = n*e**2*tau/m;    # Electrical conductivity of Cu, mho per cm
print"The free electron density in monovalent Cu =","{0:.3e}".format(n),"per metre cube";
print"The electrical conductivity of monovalent Cu =","{0:.3e}".format(sigma),"mho per cm";
The free electron density in monovalent Cu = 8.429e+28 per metre cube
The electrical conductivity of monovalent Cu = 6.403e+17 mho per cm

Example 4.4,Page number 118

In [4]:
import math

#Given Data
m = 9.1*10**-31;   # Mass of an electron, kg
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
h = 6.625*10**-34;     # Planck's constant, Js
L = 10*10**-3;     # Length of side of the cube, m
# For nth level
nx = 1; ny = 1; nz = 1;     # Positive integers along three axis
En = h**2/(8*m*L**2)*(nx**2+ny**2+nz**2)/e;      # Energy of nth level for electrons, eV
# For (n+1)th level
nx = 2; ny = 1; nz = 1;     # Positive integers along three axis
En_plus_1 = h**2/(8*m*L**2)*(nx**2+ny**2+nz**2)/e;      # Energy of (n+1)th level for electrons, eV
delta_E = En_plus_1 - En;       # Energy difference between two levels for the free electrons
print"The energy difference between two levels for the free electrons =","{0:.3e}".format( delta_E),"eV";
The energy difference between two levels for the free electrons = 1.130e-14 eV

Example 4.5,Page number 119

In [5]:
import math

#Given Data
T = 300.0;    # Room temperature of tungsten, K
k = 1.38*10**-23;  # Boltzmann constant, J/mol/K
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
E_F = 4.5*e;    # Fermi energy of tungsten, J
E = E_F-0.1*E_F; # 10% energy below Fermi energy, J
f_T = 1.0/(1+math.exp((E-E_F)/(k*T)));     # Probability of the electron in tungsten at room temperature at an nergy 10% below the Fermi energy
print"The probability of the electron at an energy 10 percent below the Fermi energy in tungsten at 300 K =",round(f_T,3);
E = 2*k*T+E_F; # For energy equal to 2kT + E_F
f_T = 1.0/(1+math.exp((E-E_F)/(k*T)));     # Probability of the electron in tungsten at an energy 2kT above the Fermi energy
print"The probability of the electron at an energy 2kT above the Fermi energy =",round(f_T,4);
The probability of the electron at an energy 10 percent below the Fermi energy in tungsten at 300 K = 1.0
The probability of the electron at an energy 2kT above the Fermi energy = 0.1192

Example 4.6,Page number 121

In [7]:
import math

#Given Data
h = 6.625*10**-34; # Planck's constant, Js
h_cross = h/(2*math.pi);    # Reduced Planck's constant, Js
m = 9.1*10**-31;   # Mass of an electron, kg
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
a = 5.34*10**-10;      # Lattice constant of monovalent bcc lattice, m
V = a**3;    # Volume of bcc unit cell, metre cube
n = 2.0/V;    # Number of atoms per metre cube
E_F = h_cross**2.0/(2*m*e)*(3*math.pi**2*n)**(2.0/3);    # Fermi energy of monovalent bcc solid, eV

print"The Fermi energy of a monovalent bcc solid =",round(E_F,4),"eV";
The Fermi energy of a monovalent bcc solid = 2.0341 eV

Example 4.7,Page number 121

In [8]:
import math

#Given Data
h = 6.625*10**-34; # Planck's constant, Js
h_cross = h/(2*math.pi);    # Reduced Planck's constant, Js
m = 9.11*10**-31;   # Mass of an electron, kg
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
V = 1*10**-5;    # Volume of cubical box, metre cube
E_F = 5*e;  # Fermi energy, J 
D_EF = V/(2*math.pi**2)*(2*m/h_cross**2)**(3.0/2)*E_F**(1.0/2)*e;     # Density of states at Fermi energy, states/eV
print"The density of states at Fermi energy =","{0:.3e}".format( D_EF),"states/eV";
The density of states at Fermi energy = 1.521e+23 states/eV

Example 4.8,Page number 121

In [9]:
import math

#Given Data
h = 6.626*10**-34; # Planck's constant, Js
h_cross = h/(2*math.pi);    # Reduced Planck's constant, Js
m = 9.1*10**-31;   # Mass of an electron, kg
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
V = 1*10**-6;    # Volume of cubical box, metre cube
E_F = 7.13*e;  # Fermi energy for Mg, J 
D_EF = V/(2*math.pi**2)*(2*m/h_cross**2)**(3.0/2)*E_F**(1.0/2);     # Density of states at Fermi energy for Cs, states/eV
E_Mg = 1.0/D_EF;  # The energy separation between adjacent energy levels of Mg, J
print"The energy separation between adjacent energy levels of Mg =","{0:.3e}".format(E_Mg/e),"eV";
E_F = 1.58*e;  # Fermi energy for Cs, J 
D_EF = V/(2*math.pi**2)*(2*m/h_cross**2)**(3.0/2)*E_F**(1.0/2);     # Density of states at Fermi energy for Mg, states/eV
E_Mg = 1.0/D_EF;  # The energy separation between adjacent energy levels of Cs, J
print"The energy separation between adjacent energy levels of Cs =","{0:.3e}".format(E_Mg/e),"eV";
The energy separation between adjacent energy levels of Mg = 5.517e-23 eV
The energy separation between adjacent energy levels of Cs = 1.172e-22 eV

Example 4.9,Page number 122

In [10]:
import math 

#Given Data
m = 9.1*10**-31;   # Mass of an electron, kg
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
E_F = 3.2*e;    # Fermi energy of sodium, J
P_F = math.sqrt(E_F*2*m);    # Fermi momentum of sodium, kg-m/s
print"The Fermi momentum of sodium =","{0:.3e}".format(P_F),"kg-m/sec";
The Fermi momentum of sodium = 9.653e-25 kg-m/sec

Example 4.10,Page number 122

In [11]:
import math

#Given Data
k = 1.38*10**-23;  # Boltzmann constant, J/mol/K
T = 500.0;    # Rise in temperature of Al, K
EF_0 = 11.63;    # Fermi energy of Al, eV
EF_T = EF_0*(1-math.pi**2.0/12*(k*T/EF_0)**2);  # Change in Fermi energy of Al with temperature, eV
print"The change in Fermi energy of Al with tempertaure rise of 500 degree celsius =",round(EF_T,3),"eV";
The change in Fermi energy of Al with tempertaure rise of 500 degree celsius = 11.63 eV

Example 4.11,Page number 122

In [12]:
import math

#Given Data
m = 9.18*10**-31;   # Mass of an electron, kg
e = 1.6*10**-19;   # Charge on an electron, C
lamda = 1.0*10**-9;   # Mean free path of electron in metal, m
v = 1.11*10**5;    # Average velocity of the electron in metal, m/s

# For Lead
n = 13.2*10**28;  # Electronic concentration of Pb, per metre cube
sigma = n*e**2*lamda/(m*v);     # Electrical conductivity of lead, mho per metre
print"The electrical conductivity of lead =","{0:.3e}".format(sigma),"mho per metre";

# For Silver
n = 5.85*10**28;  # Electronic concentration of Ag, per metre cube
sigma = n*e**2*lamda/(m*v);     # Electrical conductivity of Ag, mho per metre
print"The electrical conductivity of silver =","{0:.3e}".format(sigma),"mho per metre";
The electrical conductivity of lead = 3.316e+07 mho per metre
The electrical conductivity of silver = 1.470e+07 mho per metre

Example 4.12,Page number 125

In [13]:
import math

#Given Data
k = 1.38*10**-23;  # Boltzmann constant, J/mol/K
e = 1.6*10**-19;   # Charge on an electron, C
L = math.pi**2.0/3*(k/e)**2;    # Lorentz number, watt-ohm/degree-square
print"The Lorentz number =","{0:.3e}".format(L),"watt-ohm/degree-square";
The Lorentz number = 2.447e-08 watt-ohm/degree-square

Example 4.13,Page number 125

In [14]:
import math

#Given Data
A =[[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]];      # Declare a 4X4 cell
A[0][0] = 'Mg';  
A[0][1] = 2.54*10**-5;  
A[0][2] = 1.5;  
A[0][3] = 2.32*10**2;  
A[1][0] = 'Cu';  
A[1][1] = 6.45*10**-5;  
A[1][2] = 3.85;  
A[1][3] = 2.30*10**2;  
A[2][0] = 'Al';  
A[2][1] = 4.0*10**-5;  
A[2][2] = 2.38;
A[2][3] = 2.57*10**2;  
A[3][0] = 'Pt';  
A[3][1] = 1.02*10**-5;  
A[3][2] = 0.69;
A[3][3] = 2.56*10**2;  
T1 = 273;   # First temperature, K
T2 = 373;   # Second temperature, K
print"_________________________________________________________________";
print"Metal     sigma x 10**-05   K(W/cm-K)   Lorentz number             ";
print"          (mho per cm)                (watt-ohm/deg-square)x10**-2";
print"_________________________________________________________________";
for i in range (0,4) :
    L1 = A[i][2]/(A[i][1]*T1); 
    L2 = A[i][3];
    print"",A[i][0],"        ",A[i][1]/10**-5,"         ",A[i][2],"      ",L2/10**2,"         ",L2/10**2;
print"_________________________________________________________________";
_________________________________________________________________
Metal     sigma x 10**-05   K(W/cm-K)   Lorentz number             
          (mho per cm)                (watt-ohm/deg-square)x10**-2
_________________________________________________________________
 Mg          2.54           1.5        2.32           2.32
 Cu          6.45           3.85        2.3           2.3
 Al          4.0           2.38        2.57           2.57
 Pt          1.02           0.69        2.56           2.56
_________________________________________________________________

Example 4.14,Page number 125

In [15]:
import math

#Given Data
A = [[1,2],[3,4]];      # Declare a 2X3 cell
A[0][0] = 1.6*10**8;  # Electrcal conductivity of Au at 100 K, mho per metre
A[0][1] = 2.0*10**-8;   # Lorentz number of Au at 100 K, volt/K-square
A[1][0] = 5.0*10**8;  # Electrcal conductivity of Au at 273  K, mho per metre
A[1][1] = 2.4*10**-8;   # Lorentz number of Au at 273 K, volt/K-square
T1 = 100;       # First temperature, K
T2 = 273;       # Second temperature, K

print"___________________________________________________________________________";
print"          T = 100 K                               T = 273 K                ";
print"_________________________________       ___________________________________";
print"Electrical conductivity)  L              Electrical conductivity)  L       ";
print"    mho per metre       V/K-square           mho per metre       V/K-square";
print"___________________________________________________________________________";
K1 = A[0][0]*T1*A[0][1]; 
K2 = A[1][0]*T2*A[1][1];
print"{0:.3e}".format(A[0][0]),"            ","{0:.3e}".format(A[0][1]),"         ","{0:.3e}".format(A[1][0]),"           ","{0:.3e}".format(A[1][1]) 
print"K =",K1,"W/cm-K                           K =",K2,"W/cm-K";
print"___________________________________________________________________________";
___________________________________________________________________________
          T = 100 K                               T = 273 K                
_________________________________       ___________________________________
Electrical conductivity)  L              Electrical conductivity)  L       
    mho per metre       V/K-square           mho per metre       V/K-square
___________________________________________________________________________
1.600e+08              2.000e-08           5.000e+08             2.400e-08
K = 320.0 W/cm-K                           K = 3276.0 W/cm-K
___________________________________________________________________________

Example 4.15,Page number 131

In [16]:
import math

#Given Data
e = 1.6*10**-19;   # Electronic charge, C
a = 0.428*10**-9;  # Lattice constant of Na, m
V = a**3;    # Volume of unit cell, metre cube
N = 2;  # No. of atoms per unit cell of Na
n = N/V;    # No. of electrons per metre cube, per metre cube
R_H = -1.0/(n*e);  # Hall coeffcient of Na, metre cube per coulomb
print"The Hall coefficient of sodium =","{0:.3e}".format(R_H),"metre cube per coulomb";
The Hall coefficient of sodium = -2.450e-10 metre cube per coulomb

Example 4.16,Page number 131

In [17]:
import math

#Given Data
e = 1.6*10**-19;   # Electronic charge, C
n = 24.2*10**28;    # No. of electrons per metre cube, per metre cube
R_H = -1.0/(n*e);  # Hall coeffcient of Be, metre cube per coulomb
print"The Hall coefficient of beryllium =","{0:.3e}".format(R_H),"metre cube per coulomb";
The Hall coefficient of beryllium = -2.583e-11 metre cube per coulomb

Example 4.17,Page number 131

In [18]:
import math

#Given Data
e = 1.6*10**-19;   # Electronic charge, C
R_H = -8.4*10**-11;  # Hall coeffcient of Ag, metre cube per coulomb
n = -3*math.pi/(8*R_H*e);    # Electronic concentration of Ag, per metre cube
print"The electronic concentration of Ag =","{0:.3e}".format(n),"per metre cube";
The electronic concentration of Ag = 8.766e+28 per metre cube

Example 4.18,Page number 134

In [19]:
import math

#Given Data
# We have from Mattheissen rule, rho = rho_0 + alpha*T1
T1 = 300.0;   # Initial temperature, K
T2 = 1000.0;  # Final temperature, K
rho = 1*10**-6;    # Resistivity of the metal, ohm-m
delta_rho = 0.07*rho;      # Increase in resistivity of metal, ohm-m
alpha = delta_rho/(T2-T1);  # A constant, ohm-m/K
rho_0 = rho - alpha*T1;     # Resistivity at room temperature, ohm-m
print"The resistivity at room temperature =","{0:.3e}".format(rho),"ohm-m";
The resistivity at room temperature = 1.000e-06 ohm-m

Example 4.19,Page number 134

In [20]:
import math

#Given Data
# We have from Mattheissen rule, rho = rho_0 + alpha*T1
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
k = 1.38*10**-23;  # Boltzmann constant, J/mol/K
rho_40 = 0.2;   # Resistivity of Ge at 40 degree celsius, ohm-m
E_g = 0.7;  # Bandgap for Ge, eV
T1 = 20+273;    # Second temperature, K
T2 = 40 + 273;  # First temperature, K
rho_20 = rho_40*math.exp(E_g*e/(2*k)*(1.0/T1-1.0/T2));    # Resistivity of Ge at 20 degree celsius, ohm-m
print"The resistivity of Ge at 20 degree celsius =",round(rho_20,1),"ohm-m";
The resistivity of Ge at 20 degree celsius = 0.5 ohm-m

Example 4.20,Page number 135

In [21]:
import math

#Given Data
rs_a0_ratio = 3.25;     # Ratio of solid radius to the lattice parameter
E_F = 50.1*(rs_a0_ratio)**(-2);  # Fermi level energy of Li, eV
T_F = 58.2e+04*(rs_a0_ratio)**(-2);  # Fermi level temperature of Li, K
V_F = 4.20e+08*(rs_a0_ratio)**(-1);  # Fermi level velocity of electron in Li, cm/sec
K_F = 3.63e+08*(rs_a0_ratio)**(-1);   
print"E_F =",round(E_F,2),"eV";
print"T_F =","{0:.3e}".format(T_F),"K";
print"V_F =","{0:.3e}".format(V_F),"cm/sec";
print"K_F =","{0:.3e}".format(K_F),"per cm";
E_F = 4.74 eV
T_F = 5.510e+04 K
V_F = 1.292e+08 cm/sec
K_F = 1.117e+08 per cm

Example 4.21,Page number 135

In [22]:
import math

#Given Data
n = 6.04*10**22;  # Concentration of electrons in yittrium, per metre cube
r_s = (3/(4*math.pi*n))**(1.0/3)/10**-8;    # Radius of the solid, angstrom
a0 = 0.529; # Lattice parameter of yittrium, angstrom
rs_a0_ratio = r_s/a0;   # Solid radius to lattice parameter ratio
E_F = 50.1*(rs_a0_ratio)**(-2);  # Fermi level energy of Y, eV
print"The Fermi energy of yittrium =",round(E_F,4),"eV";
Ryd = 13.6;     # Rydberg energy constant, eV
E_bs = 0.396*Ryd;   # Band structure energy value of Y, eV
print"The band structure value of E_F =",round(E_bs,3),"eV is in close agreement with the calculated value of",round(E_F,4),"eV";
The Fermi energy of yittrium = 5.6083 eV
The band structure value of E_F = 5.386 eV is in close agreement with the calculated value of 5.6083 eV

Example 4.22,Page number 137

In [23]:
import math

#Given Data
rs_a0_ratio = 2.07;   # Solid radius to lattice parameter ratio for Al
E_F = 50.1*(rs_a0_ratio)**(-2);  # Fermi level energy of Y, eV
# According to Jellium model, h_cross*omega_P = E = 47.1 eV *(rs_a0_ratio)**(-3/2)
E = 47.1*(rs_a0_ratio)**(-3.0/2);      # Plasmon energy of Al, eV
print"The plasmon energy of Al =",round(E,4),"eV";
print"The experimental value is 15 eV";
The plasmon energy of Al = 15.8149 eV
The experimental value is 15 eV

Example 4.1a,Page number 137

In [24]:
import math

#Given Data
E_F = 1;    # For simplicity assume Fermi energy to be unity, eV
k = 1.38*10**-23;      # Boltzmann constant, J/mol/K
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
dE = 0.1;        # Exces energy above Fermi level, eV
T = 300;        # Room temperature, K
E = E_F + dE;    # Energy of the level above Fermi level, eV
f_E = 1.0/(math.exp((E-E_F)*e/(k*T))+1);    # Occupation probability of the electron at 0.1 eV above E_F
print"At 300 K:";
print"=========";
print"The occupation probability of electron at",round(dE,3),"eV above Fermi energy =",round(f_E,3);
E = E_F - dE;    # Energy of the level below Fermi level, eV
f_E = 1.0/(math.exp((E-E_F)*e/(k*T))+1);    # Occupation probability of the electron at 0.1 eV below E_F
print"The occupation probability of electron at",round(dE,3),"below Fermi energy =",round(f_E,3);

T = 1000;        # New temperature, K
print"At 1000 K:";
print"=========";
E = E_F + dE;    # Energy of the level above Fermi level, eV
f_E = 1.0/(math.exp((E-E_F)*e/(k*T))+1);    # Occupation probability of the electron at 0.1 eV above E_F
print"The occupation probability of electron at",round(dE,3),"eV above Fermi energy =",round(f_E,3);
E = E_F - dE;    # Energy of the level below Fermi level, eV
f_E = 1.0/(math.exp((E-E_F)*e/(k*T))+1);    # Occupation probability of the electron at 0.1 eV below E_F
print"The occupation probability of electron at",round(dE,3),"eV below Fermi energy =",round(f_E,3);
At 300 K:
=========
The occupation probability of electron at 0.1 eV above Fermi energy = 0.021
The occupation probability of electron at 0.1 below Fermi energy = 0.979
At 1000 K:
=========
The occupation probability of electron at 0.1 eV above Fermi energy = 0.239
The occupation probability of electron at 0.1 eV below Fermi energy = 0.761

Example 4.2a,Page number 138

In [25]:
import math

#Given Data
f_E = 0.01;     # Occupation probability of electron
E_F = 1;    # For simplicity assume Fermi energy to be unity, eV
k = 1.38*10**-23;      # Boltzmann constant, J/mol/K
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
dE = 0.5;        # Exces energy above Fermi level, eV
E = E_F + dE;    # Energy of the level above Fermi level, eV
# We have, f_E = 1/(exp((E-E_F)*e/(k*T))+1), solving for T
T = (E-E_F)*e/k*1.0/math.log(1.0/f_E-1);   # Temperature at which the electron will have energy 0.1 eV above the Fermi energy, K
print"The temperature at which the electron will have energy",round(dE,3),"eV above the Fermi energy =",round(T,3),"K";
The temperature at which the electron will have energy 0.5 eV above the Fermi energy = 1261.578 K

Example 4.3a,Page number 139

In [26]:
import math

#Given Data
E_F = 10;   # Fermi energy of electron in metal, eV
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
m = 9.1*10**-31;   # Mass of an electron, kg
E_av = 3.0/5*E_F;     # Average energy of free electron in metal at 0 K, eV
V_F = math.sqrt(2*E_av*e/m);   # Speed of free electron in metal at 0 K, eV
print"The average energy of free electron in metal at 0 K =",round(E_av,3),"eV";
print"The speed of free electron in metal at 0 K =","{0:.3e}".format(V_F),"m/s";
The average energy of free electron in metal at 0 K = 6.0 eV
The speed of free electron in metal at 0 K = 1.453e+06 m/s

Example 4.4a,Page number 139

In [27]:
import math

#Given Data
f_E = 0.1;     # Occupation probability of electron
E_F = 5.5;    # Fermi energy of Cu, eV
k = 1.38*10**-23;      # Boltzmann constant, J/mol/K
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
dE = 0.05*E_F;        # Exces energy above Fermi level, eV
E = E_F + dE;    # Energy of the level above Fermi level, eV
# We have, f_E = 1/(exp((E-E_F)*e/(k*T))+1), solving for T
T = (E-E_F)*e/k*1.0/math.log(1.0/f_E-1);   # Temperature at which the electron will have energy 0.1 eV above the Fermi energy, K
print"The temperature at which the electron will have energy",round(dE/E_F*100,3),"percent above the Fermi energy",round(T,3),"K";

#(The answer given in the textbook is wrong) 
The temperature at which the electron will have energy 5.0 percent above the Fermi energy 1451.106 K

Example 4.5a,Page number 139

In [28]:
import math

#Given Data
T_F = 24600;   # Fermi temperature of potassium, K
k = 1.38*10**-23;  # Boltzmann constant, J/mol/K
m = 9.1*10**-31;   # Mass of an electron, kg
E_F = k*T_F;    # Fermi energy of potassium, eV
v_F = math.sqrt(2*k*T_F/m);      # Fermi velocity of potassium, m/s
print"The Fermi velocity of potassium =","{0:.3e}".format(v_F),"m/s";
The Fermi velocity of potassium = 8.638e+05 m/s

Example 4.6a,Page number 139

In [29]:
import math

#Given Data
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
E_F = 7.0;  # Fermi energy of Cu, eV
f_E = 0.9;  # Occupation probability of Cu
k = 1.38*10**-23;  # Boltzmann constant, J/mol/K
T = 1000;   # Given temperature, K
# We have, f_E = 1/(exp((E-E_F)*e/(k*T))+1), solving for E
E = k*T*math.log(1.0/f_E-1) + E_F*e;     # Energy level of Cu for 10% occupation probability at 1000 K, J
print"The energy level of Cu for 10 percent occupation probability at 1000 K =",round(E/e,3),"eV";
The energy level of Cu for 10 percent occupation probability at 1000 K = 6.81 eV

Example 4.7a,Page number 140

In [30]:
import math

#Given Data
m = 9.1*10**-31;   # Mass of an electron, kg
e = 1.6*10**-19;   # Electronic charge, C
h = 6.626*10**-34; # Planck's constant, Js
E_F = 1.55;  # Fermi energy of Cu, eV
n = (math.pi/3)*(8*m/h**2)**(3.0/2)*(E_F*e)**(3.0/2);    # Electronic concentration in cesium, electrons/cc
print"The electronic concentration in cesium =","{0:.3e}".format(n),"electrons/cc";
The electronic concentration in cesium = 8.733e+27 electrons/cc

Example 4.8a,Page number 141

In [31]:
import math

#Given Data
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
E_F = 7;  # Fermi energy, eV
k = 1.38*10**-23;  # Boltzmann constant, J/mol/K
T_F = E_F*e/k;    # Fermi temperature, K
print"The Fermi temperature corresponding to Fermi energy =","{0:.3e}".format(T_F),"K";
The Fermi temperature corresponding to Fermi energy = 8.116e+04 K

Example 4.9a,Page number 141

In [33]:
import math

#Given Data
m = 9.1*10**-31;   # Mass of the electron, kg
h = 6.626*10**-34; # Planck's constant, Js
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
h_cross = h/(2*math.pi);    # Reduced Planck's constant, Js
s = 0.01;   # Side of the box, m
E = 2;  # Energy range of the electron in the box, eV
V = s**3;    # Volume of the box, metre cube
I = I = 2*E**(3.0/2)/3;    # Definite integral over E : I = 2*E**(3/2)/3
D_E = V/(2*math.pi**2)*(2*m/h_cross**2)**(3.0/2)*I*e**(3.0/2);  # Density of states for the electron in a cubical box, states
print"The density of states for the electron in a cubical box =","{0:.3e}".format(D_E),"states";
The density of states for the electron in a cubical box = 1.280e+22 states

Example 4.10a,Page number 141

In [34]:
import math

#Given Data
E_F = 1;    # For simplicity assume Fermi energy to be unity, eV
k = 1.38*10**-23;      # Boltzmann constant, J/mol/K
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
dE = 0.5;        # Exces energy above Fermi level, eV
T = 300;        # Room temperature, K
E = E_F + dE;    # Energy of the level above Fermi level, eV
f_E = 1./(math.exp((E-E_F)*e/(k*T))+1);    # Occupation probability of the electron at 0.1 eV above E_F
print"At 300 K:";
print"=========";
print"The occupation probability of electron at",dE,"eV above Fermi energy =","{0:.3e}".format(f_E);
E = E_F - dE;    # Energy of the level below Fermi level, eV
f_E = 1.0/(math.exp((E-E_F)*e/(k*T))+1);    # Occupation probability of the electron at 0.1 eV below E_F
print"The occupation probability of electron at",dE,"eV above Fermi energy =","{0:.3e}".format(f_E);
At 300 K:
=========
The occupation probability of electron at 0.5 eV above Fermi energy = 4.054e-09
The occupation probability of electron at 0.5 eV above Fermi energy = 1.000e+00

Example 4.11a,Page number 141

In [35]:
import math

#Given Data
E_F = 1;    # For simplicity assume Fermi energy to be unity, eV
k = 1.38*10**-23;      # Boltzmann constant, J/mol/K
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
dE = 0.2;        # Exces energy above Fermi level, eV
T = 0+273;        # Room temperature, K
E = E_F + dE;    # Energy of the level above Fermi level, eV
f_E = 1.0/(math.exp((E-E_F)*e/(k*T))+1);    # Occupation probability of the electron at 0.1 eV above E_F
print"At 273 K:";
print"=========";
print"The occupation probability of electron at",dE,"eV above Fermi energy =","{0:.3e}".format(f_E);
T = 100+273;    # Given temperature of 100 degree celsius, K
f_E = 1.0/(math.exp((E-E_F)*e/(k*T))+1);    # Occupation probability of the electron at 0.1 eV below E_F
print"At 373 K:";
print"=========";
print"The occupation probability of electron at",dE,"eV above Fermi energy =","{0:.3e}".format(f_E);
At 273 K:
=========
The occupation probability of electron at 0.2 eV above Fermi energy = 2.047e-04
At 373 K:
=========
The occupation probability of electron at 0.2 eV above Fermi energy = 1.992e-03

Example 4.12a,Page number 142

In [36]:
import math

#Given Data
m = 9.1*10**-31;   # Mass of the electron, kg
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
r = 1.28*10**-10;   # Atomic radius of Cu, m
a = 4*r/math.sqrt(2);    # Lattice constant of Cu, m
tau = 2.7*10**-14;      # Relaxation time for the electron in Cu, s
V = a**3;    # Volume of the cell, metre cube
n = 4.0/V;    # Concentration of free electrons in monovalent copper, 
sigma = n*e**2*tau/m;    # Electrical conductivity of monovalent copper, mho per m
print"The electrical conductivity of monovalent copper =","{0:.3e}".format( sigma/100),"mho per cm";
The electrical conductivity of monovalent copper = 6.403e+05 mho per cm

Example 4.13a,Page number 142

In [37]:
import math 

#Given Data
n = 18.1*10**22;  # Number of electrons per unit volume, per cm cube
N = n/2;    # Pauli's principle for number of energy levels, per cm cube
E_F = 11.58;     # Fermi energy of Al, eV
E = E_F/N;     # Interelectronic energy separation between bands of Al, eV
print"The interelectronic energy separation between bands of Al =","{0:.3e}".format(E),"eV";
The interelectronic energy separation between bands of Al = 1.280e-22 eV

Example 4.14a,Page number 142

In [38]:
import math

#Given Data
m = 9.1*10**-31;   # Mass of the electron, kg
h = 6.626*10**-34; # Planck's constant, Js
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
h_cross = h/(2*math.pi);    # Reduced Planck's constant, Js
E_F = 7;    # Fermi energy of Cu, eV
V = 10**-6;  # Volume of the cubic metal, metre cube
D_EF = V/(2*math.pi**2)*(2*m/h_cross**2)**(3.0/2)*(E_F)**(1.0/2)*e**(3.0/2);   # Density of states in Cu contained in cubic metal, states/eV
print"The density of states in Cu contained in cubic metal =","{0:.3e}".format(D_EF),"states/eV";
The density of states in Cu contained in cubic metal = 1.796e+22 states/eV

Example 4.15a,Page number 143

In [39]:
import math

#Given Data
m = 9.1*10**-31;   # Mass of the electron, kg
h = 6.626*10**-34; # Planck's constant, Js
e = 1.6*10**-19;   # Energy equivalent of 1 eV, J/eV
h_cross = h/(2*math.pi);    # Reduced Planck's constant, Js
E_F = 7;    # Fermi energy of Cu, eV
V = 10**-6;  # Volume of the cubic metal, metre cube
D_EF = V/(2*math.pi**2)*(2*m/h_cross**2)**(3.0/2)*(E_F)**(1.0/2)*e**(3.0/2);   # Density of states in Cu contained in cubic metal, states/eV
d = 1.0/(D_EF);     # Electronic energy level spacing between successive levels of Cu, eV
print"The electronic energy level spacing between successive levels of Cu =","{0:.3e}".format(d),"eV";
The electronic energy level spacing between successive levels of Cu = 5.568e-23 eV

Example 4.16a,Page number 143

In [40]:
import math

#Given Data
A = [[1,2],[3,4],[5,6],[7,8]];      # Declare a 4X2 matrix 
A[0][0] = 'Li';       
A[0][1] = -0.4039;   # Energy of outermost atomic orbital of Li, Rydberg unit
A[1][0] = 'Na';      # 
A[1][1] = -0.3777;   # Energy of outermost atomic orbital of Na, Rydberg unit
A[2][0] = 'F';      # 
A[2][1] = -1.2502;   # Energy of outermost atomic orbital of F, Rydberg unit
A[3][0] = 'Cl';      # 
A[3][1] = -0.9067;   # Energy of outermost atomic orbital of Cl, Rydberg unit
cf = 13.6;  # Conversion factor for Rydberg to eV
print"________________________________________";
print" Atom                          Energy gap";
print"",A[1][0],A[3][0],"                         ",(A[1][1]-A[3][1])*cf,"eV";
print"",A[1][0],A[2][0],"                          ",(A[1][1]-A[2][1])*cf,"eV";
print"",A[0][0],A[2][0],"                          ",(A[0][1]-A[2][1])*cf,"eV";

print"________________________________________";
________________________________________
 Atom                          Energy gap
 Na Cl                           7.1944 eV
 Na F                            11.866 eV
 Li F                            11.50968 eV
________________________________________

Example 4.18a,Page number 144

In [41]:
import math

#Given Data
# For Cu
rs_a0_ratio = 2.67;     # Ratio of solid radius to the lattice parameter
E_F = 50.1*(rs_a0_ratio)**(-2);  # Fermi level energy of Cu, eV
T_F = 58.2*10**4*(rs_a0_ratio)**(-2);  # Fermi level temperature of Cu, K
V_F = 4.20*10**8*(rs_a0_ratio)**(-1);  # Fermi level velocity of electron in Cu, cm/sec
K_F = 3.63*10**8*(rs_a0_ratio)**(-1);   
print"For Cu :";
print"========";
print"E_F =",round(E_F,3),"eV";
print"T_F =","{0:.3e}".format(T_F),"K";
print"V_F =","{0:.3e}".format(V_F),"cm/sec";
print"K_F =","{0:.3e}".format(K_F),"per cm";
rs_a0_ratio = 3.07;     # Ratio of solid radius to the lattice parameter
E_F = 50.1*(rs_a0_ratio)**(-2);  # Fermi level energy of Nb, eV
T_F = 58.2*10**4*(rs_a0_ratio)**(-2);  # Fermi level temperature of Nb, K
V_F = 4.20*10**8*(rs_a0_ratio)**(-1);  # Fermi level velocity of electron in Nb, cm/sec
K_F = 3.63*10**8*(rs_a0_ratio)**(-1);   
print"For Nb :";
print"========";
print"E_F =",round(E_F,3),"eV";
print"T_F =","{0:.3e}".format(T_F),"K";
print"V_F =","{0:.3e}".format(V_F),"cm/sec";
print"K_F =","{0:.3e}".format(K_F),"per cm";
For Cu :
========
E_F = 7.028 eV
T_F = 8.164e+04 K
V_F = 1.573e+08 cm/sec
K_F = 1.360e+08 per cm
For Nb :
========
E_F = 5.316 eV
T_F = 6.175e+04 K
V_F = 1.368e+08 cm/sec
K_F = 1.182e+08 per cm
In [ ]: