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diff --git a/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/Chapter2.ipynb b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/Chapter2.ipynb new file mode 100644 index 00000000..5bd122ad --- /dev/null +++ b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/Chapter2.ipynb @@ -0,0 +1,337 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:74a00fabf3de3a229499fd336c46d9a546ea42ad7cb4fbe98a92a6ea72f21fa8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2: Bonding in Solids" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1,Page number 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "epsilon_0 = 8.854*10**-12; # Absolute electrical permittivity of free space, F/m\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, eV/J\n", + "r = 3.147*10**-10; # Nearest neighbour distance for KCl, m\n", + "n = 9.1; # Repulsive exponent of KCl\n", + "A = 1.748; # Madelung constant for lattice binding energy\n", + "E = A*e**2/(4*math.pi*epsilon_0*r)*(n-1)/n/e; # Binding energy of KCl, eV\n", + "print\"The binding energy of KCl = \",round(E,4),\"eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The binding energy of KCl = 7.10982502818 eV\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2,Page number 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "epsilon_0 = 8.854*10**-12; # Absolute electrical permittivity of free space, F/m\n", + "N = 6.023*10**23; # Avogadro's number\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, eV/J\n", + "a0 = 5.63*10**-10; # Lattice parameter of NaCl, m\n", + "r0 = a0/2; # Nearest neighbour distance for NaCl, m\n", + "n = 8.4; # Repulsive exponent of NaCl\n", + "A = 1.748; # Madelung constant for lattice binding energy\n", + "E = A*e**2/(4*pi*epsilon_0*r0)*(n-1)/n/e; # Binding energy of NaCl, eV\n", + "print\"The binding energy of NaCl = \",round(E*N*e/(4.186*1000),4),\"kcal/mol\" ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The binding energy of NaCl = 181.1005 kcal/mol\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3,Page number 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "epsilon_0 = 8.854*10**-12; # Absolute electrical permittivity of free space, F/m\n", + "N = 6.023*10**23; # Avogadro's number\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, eV/J\n", + "E = 162.9*10**3; # Binding energy of KCl, cal/mol\n", + "n = 8.6; # Repulsive exponent of KCl\n", + "A = 1.747; # Madelung constant for lattice binding energy\n", + "# As lattice binding energy, E = A*e**2/(4*%pi*epsilon_0*r0)*(n-1)/n, solving for r0\n", + "r0 = A*N*e**2/(4*pi*epsilon_0*E*4.186)*(n-1)/n; # Nearest neighbour distance of KCl, m\n", + "print\"The nearest neighbour distance of KCl = \",round(r0*10**10,4),\"angstorm\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The nearest neighbour distance of KCl = 3.1376 angstorm\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.4,Page number 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "epsilon_0 = 8.854*10**-12; # Absolute electrical permittivity of free space, F/m\n", + "N = 6.023*10**23; # Avogadro's number\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, eV/J\n", + "E = 152*10**3; # Binding energy of CsCl, cal/mol\n", + "n = 10.6; # Repulsive exponent of CsCl\n", + "A = 1.763; # Madelung constant for lattice binding energy\n", + "\n", + "# As lattice binding energy, E = A*e**2/(4*pi*epsilon_0*r0)*(n-1)/n, solving for r0\n", + "r0 = A*N*e**2/(4*pi*epsilon_0*E*4.186)*(n-1)/n; # Nearest neighbour distance of CsCl, m\n", + "print\"The nearest neighbour distance of CsCl = \",round(r0*10**10,4),\"angstrom\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The nearest neighbour distance of CsCl = 3.4776 angstrom\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5,Page number 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "epsilon_0 = 8.854*10**-12; # Absolute electrical permittivity of free space, F/m\n", + "N = 6.023*10**23; # Avogadro's number\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, eV/J\n", + "r0 = 6.46*10**-10; # Nearest neighbour distance of NaI\n", + "E = 157.1*10**3; # Binding energy of NaI, cal/mol\n", + "A = 1.747; # Madelung constant for lattice binding energy\n", + "\n", + "# As lattice binding energy, E = -A*e**2/(4*pi*epsilon_0*r0)*(n-1)/n, solving for n\n", + "n = 1/(1+(4.186*E*4*pi*epsilon_0*r0)/(N*A*e**2)); # Repulsive exponent of NaI\n", + "print\"\\nThe repulsive exponent of NaI = \",round(n,4);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The repulsive exponent of NaI = 0.363\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.6,Page number 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, eV/J\n", + "a0 = 2.8158*10**-10; # Nearest neighbour distance of solid\n", + "A = 1.747; # Madelung constant for lattice binding energy\n", + "n = 8.6; # The repulsive exponent of solid\n", + "c = 2; # Structural factor for rocksalt\n", + "# As n = 1 + (9*c*a0**4)/(K0*e**2*A), solving for K0\n", + "K0 = 9*c*a0**4/((n-1)*e**2*A); # Compressibility of solid, metre square per newton\n", + "print\"The compressibility of the solid = \", \"{0:.3e}\".format(K0),\"metre square per newton\";" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The compressibility of the solid = 3.329e-01 metre square per newton\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7,Page number 69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "chi_diff = 1; # Electronegativity difference between the constituent of elements of solid\n", + "percent_ion = 100*(1-math.e**(-(0.25*chi_diff**2))); # Percentage ionic character present in solid given by Pauling\n", + "print\"The percentage ionic character present in solid = \",round(percent_ion,2),\"percent \";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The percentage ionic character present in solid = 22.12 percent \n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.8,Page number 69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "Eh_GaAs = 4.3; # Homopolar gap of GaAs compound, eV\n", + "C_GaAs = 2.90; # Ionic gap of GaAs compound, eV\n", + "Eh_CdTe = 3.08; # Homopolar gap of CdTe compound, eV\n", + "C_CdTe = 4.90; # Ionic gap of CdTe compound, eV\n", + "\n", + "fi_GaAs = C_GaAs**2/(Eh_GaAs**2 + C_GaAs**2);\n", + "fi_CdTe = C_CdTe**2/(Eh_CdTe**2 + C_CdTe**2);\n", + "print\"The fractional ionicity of GaAs = \",round(fi_GaAs,4);\n", + "print\"The fractional ionicity of CdTe = \",round(fi_CdTe,4);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The fractional ionicity of GaAs = 0.3126\n", + "The fractional ionicity of CdTe = 0.7168\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter1.ipynb b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter1.ipynb new file mode 100644 index 00000000..55c0317e --- /dev/null +++ b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter1.ipynb @@ -0,0 +1,1136 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:09ec47f79034ef2298c5ba335d328b021b3f5f7d5d50c5362b4013942de55a1c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1: Structure of Solids" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.1,Page number 13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "r = 1.278*10**-10; # Atomic radius of fcc structure, m\n", + "a = 4*r/sqrt(2); # Lattice parameter of fcc strucure, m\n", + "V = a**3; # Volume of fcc unit cell, metre, cube\n", + "print\"The lattice parameter of fcc strucure =\",\"{0:.3e}\".format(a),\"m\";\n", + "print\"The volume of fcc unit cell =\",\"{0:.3e}\".format(V),\"metre cube\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The lattice parameter of fcc strucure = 3.615e-10 m\n", + "The volume of fcc unit cell = 4.723e-29 metre cube\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.2,Page number 14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "r = 0.143*10**-9; # Radius of Nb unit cell, m\n", + "d = 8.57*10**3; # Density of Nb unit cell, kg/metre-cube\n", + "M = 92.91*10**-3; # Atomic weight of Nb, kg per mole\n", + "N = 6.023*10**23; # Avogadro's No.\n", + "\n", + "# For fcc\n", + "a = 4*r/sqrt(2); # Lattice parameter for fcc structure of Nb, m\n", + "n = a**3*d*N/M; # Number of lattice points per unit cell\n", + "if ((n%int(n))< 0.001) :\n", + " print\"The number of atoms associated with the cell is\",int(n),\", Nb should have fcc structure\";\n", + "\n", + "\n", + "# For bcc\n", + "a = 4*r/sqrt(3); # Lattice parameter for bcc structure of Nb, m\n", + "n = a**3*d*N/M; # Number of lattice points per unit cell\n", + "if ((n%int(n)) < 0.001) :\n", + " print\"The number of atoms associated with the cell is\",int(n),\", Nb should have bcc structure\";\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of atoms associated with the cell is 2 , Nb should have bcc structure\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3,Page number 17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "V = 10.58*10**-29; # Volume of the unit cell, metre cube\n", + "#3*sqrt(3)/2*1.58*a**3-V) =0; # First lattice parameter, m\n", + "a= (V/(3*sqrt(3)/2*1.58))**(1.0/3); #Solving for a\n", + "c = 1.58*a; # Third lattice parameter, m\n", + "print\"The lattice parameters of hcp structure of Ti are:\";\n", + "print\"a =\",round(a/10**-10,3),\"angstorm, c =\",round(c/10**-10,3),\"angstorm\";\n", + "\n", + "# Result \n", + "# The lattice parameters of hcp structure of Ti are:\n", + "# a = 2.95 angstorm, c = 4.67 angstorm \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The lattice parameters of hcp structure of Ti are:\n", + "a = 2.954 angstorm, c = 4.667 angstorm\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.4,Page number 17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "c_by_a_ratio = 1.633; # Ideal c/a ratio\n", + "A = [[1,2,3,4],[5,6,7,8]] # Declare a cell\n", + "# Assign values to the elements of the cell from the table\n", + "A[0][0] = 'Mg';\n", + "A[1][0] = 'Cd';\n", + "A[0][1] = 5.21;\n", + "A[1][1] = 5.62;\n", + "A[0][2] = 3.21;\n", + "A[1][2]= 2.98;\n", + "A[0][3] = A[0][1]/A[0][2];\n", + "A[1][3] = A[1][1]/A[1][2];\n", + "if (A[0][3] - c_by_a_ratio) < 0.01:\n", + " print\"\",A[0][0],\"satisfies ideal c/a ratio and\",A[1][0],\"has large deviation from this value.\";\n", + "else:\n", + " if (A[0][3] - c_by_a_ratio) < 0.01:\n", + " print\"\",A[1][0],\"satisfies ideal c/a ratio and\",A[0][0],\"has large deviation from this value.\"; \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Mg satisfies ideal c/a ratio and Cd has large deviation from this value.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5,Page number 18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "M_Na = 23; # Atomic weight of Na, gram per mole\n", + "M_Cl = 35.5; # Atomic weight of Cl, gram per mole\n", + "d = 2.18*10**6; # Density of Nacl salt, g per metre cube\n", + "n = 4; # No. of atoms per unit cell for an fcc lattice of NaCl crystal\n", + "N = 6.023*10**23; # Avogadro's No.\n", + "# Volume of the unit cell is given by\n", + "# a**3 = M*n/(N*d)\n", + "# Solving for a\n", + "a = (n*(M_Na + M_Cl)/(d*N))**(1.0/3); # Lattice constant of unit cell of NaCl\n", + "print\"Lattice constant for the NaCl crystal =\",round(a/10**-10,3),\"angstorm\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lattice constant for the NaCl crystal = 5.627 angstorm\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6,Page number 18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "r = 1.33; # Ionic radii of K+ ion, angstrom\n", + "R = 1.81; # Ionic radii of Cl- ion, angstrom\n", + "n = 4; # No. of atoms per unit cell for an fcc lattice of NaCl crystal\n", + "APF = (n*(4*pi*r**3/3)+n*(4*pi*R**3/3))/(2*r+2*R)**3; # Atomic packing factor of fcc KCl\n", + "print\"The ionic packing factor of fcc KCl =\",round(APF,3);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ionic packing factor of fcc KCl = 0.56\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.7,Page number 20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "N = 6.023*10**23; # Avogadro's number\n", + "M = 12.01*10**-3; # Atomic weight of diamond/graphite, kg\n", + "\n", + "# For diamond\n", + "a = 3.568*10**-10; # Lattice parameter of diamond, m\n", + "rho = 3.518*10**3; # Density of diamond, kg per metre cube\n", + "n = a**3*rho*N/M; # Number of atoms in the unit cell of diamond structure\n", + "print\"The number of atoms in the unit cell of diamond structure =\",int(n);\n", + "\n", + "# For graphite\n", + "a = 2.451*10**-10; # First lattice parameter of graphite, m\n", + "c = 6.701*10**-10; # Third lattice parameter of graphite, m\n", + "rho = 2.2589*10**3; # Density of graphite, kg per metre cube\n", + "V = 3*sqrt(3)*a**2*c/2; # Volume of hexagonal unit cell of graphite, metre cube\n", + "n = V*rho*N/M; # Number of atoms in the unit cell of graphite structure\n", + "print\"The number of atoms in the unit cell of graphite structure =\",round(n);\n", + "\n", + "# Result \n", + "# The number of atoms in the unit cell of diamond structure = 8\n", + "# The number of atoms in the unit cell of graphite structure = 12 \n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of atoms in the unit cell of diamond structure = 8\n", + "The number of atoms in the unit cell of graphite structure = 12.0\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.8,Page number 21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "N = 6.023*10**23; # Avogadro's number\n", + "\n", + "# For silicon crystallized into diamond structure\n", + "a = 5.43*10**-8; # Lattice parameter of Si, cm\n", + "M = 28.1; # Atomic mass of Si, g/mol\n", + "n = 8/a**3; # Number of atoms per unit volume, atoms per cm cube\n", + "d = n*M/N; # Density of Si crytal, g/cm\n", + "print\"The density of crystallized Si =\",round(d,3),\"gram per cm cube\";\n", + "\n", + "# For GaAs crystallized into Zinc Blende structure\n", + "a = 5.65*10**-8; # Lattice parameter of GaAs, cm\n", + "M_Ga = 69.7; # Atomic weight of Ga, g/mol\n", + "M_As = 74.9; # Atomic weight of As, g/mol\n", + "M = M_Ga + M_As; # Atomic weight of GaAs, g/mol\n", + "n = 4.0/a**3; # Number of atoms per unit volume, atoms per cm cube\n", + "d = n*M/N; # Density of Si crytal, g/cm\n", + "print\"The density of crystallized GaAs =\",round(d,3),\"gram per cm cube\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The density of crystallized Si = 2.331 gram per cm cube\n", + "The density of crystallized GaAs = 5.324 gram per cm cube\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.9,Page number 21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "N = 6.023*10**23; # Avogadro's number\n", + "\n", + "r1 = 0.122*10**-9; # Ionic radii of Ga, m\n", + "r2 = 0.125*10**-9; # Ionic radii of As, m\n", + "r3 = 0.11*10**-9; # Ionic radii of P, m\n", + "\n", + "# For GaP\n", + "r = r1 + r3; # Interatomic separation between Ga and P atoms, m\n", + "a = 4*r/3**(1.0/2); # Lattice parameter of GaP structure, m\n", + "print\"The lattice parameter of GaP structure =\",round(a/10**-10,3),\"angstrom\";\n", + "\n", + "# For GaAs\n", + "r = r1 + r2; # Interatomic separation between Ga and As atoms, m\n", + "a = 4*r/3**(1.0/2); # Lattice parameter of GaP structure, m\n", + "print\"The lattice parameter of GaAs structure =\",round(a/10**-10,3),\"angstrom\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The lattice parameter of GaP structure = 5.358 angstrom\n", + "The lattice parameter of GaAs structure = 5.704 angstrom\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.10,Page number 24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "def string(r_ratio):\n", + " if(r_ratio > 0.732):\n", + " st = 'Caesium Chloride';\n", + " else :\n", + " if(r_ratio < 0.732):\n", + " st = 'Rock Salt';\n", + " else :\n", + " if(r_ratio < 0.414):\n", + " st = 'Rutile';\n", + " return st\n", + "\n", + "crystal = [[1,2],[3,4],[5,6],[7,8],[9,10],[11,12]]; # Declare cells of 6 rows and 2 columns\n", + "crystal[0][0] = 'I';\n", + "crystal[0][1] = 2.19; # Ionic radius of I, angstrom\n", + "crystal[1][0] = 'Cl';\n", + "crystal[1][1]= 1.81; # Ionic radius of Cl, angstrom\n", + "crystal[2][0] = 'Na';\n", + "crystal[2][1] = 0.95; # Ionic radius of Na, angstrom\n", + "crystal[3][0] = 'Cs';\n", + "crystal[3][1] = 1.69; # Ionic radius of Cs, angstrom\n", + "crystal[4][0] = 'Mg';\n", + "crystal[4][1] = 0.99; # Ionic radius of Mg2+, angstrom\n", + "crystal[5][0] = 'O';\n", + "crystal[5][1] = 1.40; # Ionic radius of O2-, angstrom\n", + "\n", + "print\"The crystal structure of\",crystal[2][0],crystal[0][0],\"with radius ratio =\",round(crystal[2][1]/crystal[0][1],4),\"is\",string(crystal[2][1]/crystal[0][1]);\n", + "\n", + "print\"The crystal structure of\",crystal[2][0],crystal[1][0],\"with radius ratio =\",round(crystal[2][1]/crystal[1][1],4),\"is\",string(crystal[2][1]/crystal[1][1]);\n", + "\n", + "print\"The crystal structure of\",crystal[3][0],crystal[1][0],\"with radius ratio =\",round(crystal[3][1]/crystal[1][1],4),\"is\",string(crystal[3][1]/crystal[1][1]);\n", + "\n", + "print\"The crystal structure of\",crystal[3][0],crystal[0][0],\"with radius ratio =\",round(crystal[3][1]/crystal[0][1],4),\"is\",string(crystal[3][1]/crystal[0][1]);\n", + "\n", + "print\"The crystal structure of\",crystal[4][0],crystal[5][0],\"with radius ratio =\",round(crystal[4][1]/crystal[5][1],4),\"is\",string(crystal[4][1]/crystal[5][1]);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The crystal structure of Na I with radius ratio = 0.4338 is Rock Salt\n", + "The crystal structure of Na Cl with radius ratio = 0.5249 is Rock Salt\n", + "The crystal structure of Cs Cl with radius ratio = 0.9337 is Caesium Chloride\n", + "The crystal structure of Cs I with radius ratio = 0.7717 is Caesium Chloride\n", + "The crystal structure of Mg O with radius ratio = 0.7071 is Rock Salt\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.11,Page number 25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "R = 1; # For simplicity we assume radius of atom to be unity, m\n", + "# For bcc Structure,\n", + "a = 4*R/sqrt(3); # Lattice parameter of bcc crystal, m\n", + "# We have R+r = a/2, solving for r\n", + "r = a/2-R # Relation between radius of the void and radius of the atom, m \n", + "print\"The maxiumum radius of the sphere that can fit into void between two bcc unit cells =\",round(r,3),\"R\"; \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maxiumum radius of the sphere that can fit into void between two bcc unit cells = 0.155 R\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.12,Page number 25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "R = 1; # For simplicity we assume radius of atom to be unity, m\n", + "# For fcc Structure,\n", + "a = 4*R/sqrt(2); # Lattice parameter of fcc crystal, m\n", + "# We have R+r = a/2, solving for r\n", + "r = a/2-R # Relation between radius of the void and radius of the atom, m \n", + "print\"The maxiumum radius of the sphere that can fit into void between two fcc unit cells =\",round(r,3),\"R\"; \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maxiumum radius of the sphere that can fit into void between two fcc unit cells = 0.414 R\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.13,Page number 26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "R = 1; # For simplicity we assume radius of atom to be unity, m\n", + "# For bcc Structure,\n", + "a = 4*R/sqrt(3); # Lattice parameter of bcc crystal, m\n", + "# We have (R+r)**2 = (a/2)**2+(a/4)**2, solving for r\n", + "r = sqrt(5)*a/4-R # Relation between radius of the void and radius of the atom, m \n", + "print\"The radius of largest void in the bcc lattice =\",round(r,3),\"R\"; \n", + "\n", + "# For fcc Structure,\n", + "a = 4*R/sqrt(2); # Lattice parameter of fcc crystal, m\n", + "# We have (R+r)**2 = (a/2)**2+(a/4)**2, solving for r\n", + "r_fcc = a/2-R # Relation between radius of the void and radius of the atom, m \n", + "print\"The radius of largest void in the fcc lattice is\",round(r_fcc/r,3),\"times larger than that in the bcc lattice\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The radius of largest void in the bcc lattice = 0.291 R\n", + "The radius of largest void in the fcc lattice is 1.423 times larger than that in the bcc lattice\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14,Page number 26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "R = 1; # For simplicity we assume radius of atom to be unity, m\n", + "\n", + "# For bcc Structure,\n", + "a = 4*R/sqrt(3); # Lattice parameter of bcc crystal, m\n", + "# We have (R+r)**2 = (a/2)**2+(a/4)**2, solving for r\n", + "r = a/2-R # Relation between radius of the void and radius of the atom, m \n", + "print\"The radius of void for carbon atoms in iron =\",round(r,3),\"R\"; \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The radius of void for carbon atoms in iron = 0.155 R\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15,Page number 27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "R = 1; # For simplicity we assume radius of atom to be unity, m\n", + "# From the right triangle LMO, LM/LO = R/(R + r) = cosd(30), solving for r\n", + "r =R/cos(radians(30))-R;\n", + "print\"The radius of triangular void =\",round(r,3),\"R\"; \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The radius of triangular void = 0.155 R\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.16,Page number 27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "R = 1; # For simplicity we assume radius of atom to be unity, m\n", + "# From the right triangle LMN similar to trinagle LPO, LM/LO = R/(R + r) = LP/LO = sqrt(2/3), solving for r\n", + "r = R/sqrt(2./3)-R;\n", + "print\"The radius ratio of tetragonal void =\",round(r/R,3); \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The radius ratio of tetragonal void = 0.225\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.17,Page number 28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "R = 1; # For simplicity we assume radius of atom to be unity, m\n", + "# From the isosceles right triangle LMN, LM/LO = (R + r)/R = sqrt(2)/1, solving for r\n", + "r =R*sqrt(2)-R;\n", + "print\"The radius ratio of octahedral void =\",round(r/R,3); \n", + "s" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The radius ratio of octahedral void = 0.414\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.18,Page number 32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "p = 3; q = -3; r = 3.0/2; # Coefficients of intercepts along three axes\n", + "h = 1.0/p; # Reciprocate the first coefficient\n", + "k = 1.0/q; # Reciprocate the second coefficient\n", + "l = 1.0/r; # Reciprocate the third coefficient\n", + "# Find l.c.m. of m,n and p is 3\n", + "mul_fact = 3;\n", + "h = h*mul_fact; # Clear the first fraction\n", + "k = k*mul_fact; # Clear the second fraction\n", + "l = l*mul_fact; # Clear the third fraction\n", + "print\"The required miller indices are : (\",int(h),int(k),int(l),\")\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required miller indices are : ( 1 -1 2 )\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.19,Page number 32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "p = 2; q = 3; r = 4; # Coefficients of intercepts along three axes\n", + "h = 1.0/p; # Reciprocate the first coefficient\n", + "k = 1.0/q; # Reciprocate the second coefficient\n", + "l = 1.0/r; # Reciprocate the third coefficient\n", + "# Find l.c.m. of m,n and p\n", + "# l.c.m. of 2 and 4 is 4 and l.c.m. of 4 and 3 is 12\n", + "#hence l.c.m = 12\n", + "mul_fact =12 ; \n", + "h = h*mul_fact; # Clear the first fraction\n", + "k = k*mul_fact; # Clear the second fraction\n", + "l = l*mul_fact; # Clear the third fraction\n", + "print\"The required miller indices are : (\",int(h),int(k),int(l),\")\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required miller indices are : ( 6 4 3 )\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.20,Page number 32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "p = 4; q = 4; r = float('inf'); # Coefficients of intercepts along three axes\n", + "h = 1.0/p; # Reciprocate the first coefficient\n", + "k = 1.0/q; # Reciprocate the second coefficient\n", + "l = 1.0/r; # Reciprocate the third coefficient\n", + "# Find l.c.m. of m,n and p \n", + "#l.c.m of p and q is 4\n", + "mul_fact =4;\n", + "h = h*mul_fact; # Clear the first fraction\n", + "k = k*mul_fact; # Clear the second fraction\n", + "l = l*mul_fact; # Clear the third fraction\n", + "print\"The required miller indices are : (\",int(h),int(k),int(l),\")\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required miller indices are : ( 1 1 0 )\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.21,Page number 32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "a = 0.424; b = 2; c = 0.367; # Intercepts on planes along three axes, m\n", + "# Here pa = 0.424; qb = 2; rc = 0.183, solving for p, q and r, we have\n", + "p = 0.424/a; q = 2/b; r = 0.183/c; # Coefficients of intercepts along three axes\n", + "h = 1.0/p; # Reciprocate the first coefficient\n", + "k = 1.0/q; # Reciprocate the second coefficient\n", + "l = 1.0/r; # Reciprocate the third coefficient\n", + "print\"The required miller indices are :(\",int(h),int(k),int(l),\")\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required miller indices are :( 1 1 2 )\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.22,Page number 33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "r = 1.746*10**-10; # Atomic radius of lead atom, angstrom\n", + "a = 4*r/sqrt(2); # Interatomic spacing, m\n", + "h = 1; k = 0; l = 0; # Miller Indices for planes in a cubic crystal\n", + "d_100 = a/(h**2+k**2+l**2)**(1.0/2); # The interplanar spacing for cubic crystals, m\n", + "print\"The interplanar spacing between consecutive (100) planes =\",round(d_100/1e-010,3),\"angstrom\";\n", + "\n", + "h = 1; k = 1; l = 0; # Miller Indices for planes in a cubic crystal\n", + "d_110 = a/(h**2+k**2+l**2)**(1.0/2); # The interplanar spacing for cubic crystals, m\n", + "print\"The interplanar spacing between consecutive (110) planes =\",round(d_110/1e-010,3),\"angstrom\";\n", + "\n", + "h = 1; k = 1; l = 1; # Miller Indices for planes in a cubic crystal\n", + "d_111 = a/(h**2+k**2+l**2)**(1.0/2); # The interplanar spacing for cubic crystals, m\n", + "print\"The interplanar spacing between consecutive (111) planes =\",round(d_111/1e-010,3),\"angstrom\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The interplanar spacing between consecutive (100) planes = 4.938 angstrom\n", + "The interplanar spacing between consecutive (110) planes = 3.492 angstrom\n", + "The interplanar spacing between consecutive (111) planes = 2.851 angstrom\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.23,Page number 34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "c = 3.0*10**8; # Speed of light, m/s\n", + "E_K = 13.6*29**2; # Energy of electron in the K-shell\n", + "E_L = 13.6*29**2/4; # Energy of electron in the L-shell\n", + "# As E_K - E_L = h*c/lambda, solving for lambda\n", + "lamda = h*c/((E_K - E_L)*e); # Wavelength of K_alpha radiation of tungsten, m\n", + "print\"The wavelength of K_alpha radiation of Cu =\",round(lamda/1e-010,3),\"angstrom\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The wavelength of K_alpha radiation of Cu = 1.448 angstrom\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.24,Page number 35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "c = 3.0*10**8; # Speed of light, m/s\n", + "E_K = 13.6*74**2; # Energy of electron in the K-shell\n", + "E_L = 13.6*74**2/4; # Energy of electron in the L-shell\n", + "# As E_K - E_L = h*c/lambda, solving for lambda\n", + "lamda = h*c/((E_K - E_L)*e); # Wavelength of K_alpha radiation of tungsten, m\n", + "print\"The wavelength of K_alpha radiation of tungsten =\",\"{0:.3e}\".format(lamda/1e-010),\"angstrom\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The wavelength of K_alpha radiation of tungsten = 2.224e-01 angstrom\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.25,Page number 35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "a_Cu = 3.61; # Lattice constant of Cu, angstrom\n", + "a_Pd = 3.89; # Lattice constant of Pd, angstrom\n", + "\n", + "# For x = 20% of Pd\n", + "x = 0.20; # Percentage of Pd in Cu-Pd alloy\n", + "a_Cu_Pd = ((1-x)*a_Cu + x*a_Pd);\n", + "print\"For\",x*100,\"percent of Pd in Cu-Pd alloy, a =\",a_Cu_Pd,\"angstrom\";\n", + "\n", + "# For x = 40% of Pd\n", + "x = 0.40; # Percentage of Pd in Cu-Pd alloy\n", + "a_Cu_Pd = ((1-x)*a_Cu + x*a_Pd);\n", + "print\"For\",x*100,\"percent of Pd in Cu-Pd alloy, a =\",a_Cu_Pd,\"angstrom\";\n", + "\n", + "# For x = 60% of Pd\n", + "x = 0.60; # Percentage of Pd in Cu-Pd alloy\n", + "a_Cu_Pd = ((1-x)*a_Cu + x*a_Pd);\n", + "print\"For\",x*100,\"percent of Pd in Cu-Pd alloy, a =\",a_Cu_Pd,\"angstrom\";\n", + "\n", + "# For x = 80% of Pd\n", + "x = 0.80; # Percentage of Pd in Cu-Pd alloy\n", + "a_Cu_Pd = ((1-x)*a_Cu + x*a_Pd);\n", + "print\"For\",x*100,\"percent of Pd in Cu-Pd alloy, a =\",a_Cu_Pd,\"angstrom\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For 20.0 percent of Pd in Cu-Pd alloy, a = 3.666 angstrom\n", + "For 40.0 percent of Pd in Cu-Pd alloy, a = 3.722 angstrom\n", + "For 60.0 percent of Pd in Cu-Pd alloy, a = 3.778 angstrom\n", + "For 80.0 percent of Pd in Cu-Pd alloy, a = 3.834 angstrom\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.26,Page number 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "a_Rh = 3.80; # Lattice constant of Rh, angstrom\n", + "a_Pt = 3.92; # Lattice constant of Pt, angstrom\n", + "a_Pt_Rh = 3.78; # Lattice constant of unit cell of Pt-Rh alloy, angstrom\n", + "V = (a_Pt*1e-08)**3; # Volume of unit cell of Pt, metre cube\n", + "V_90 = 0.9*V; # 90 percent of the cell volume of Pt, metre cube\n", + "\n", + "# For x = 20% of Rh in Pt-Rh alloy, we have\n", + "# a_Pt_Rh = ((1-x)*a_Pt + x*a_Rh), solving for x\n", + "x = (a_Pt_Rh - a_Pt)/(a_Rh-a_Pt); # Amount of required Rh in Pt to change the unit cell volume\n", + "print\"The amount of Rh required in Pt to change the unit cell volume =\",round(x,2),\"percent\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The amount of Rh required in Pt to change the unit cell volume = 1.17 percent\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.27,Page number 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "r_bcc = 0.126; # Atomic radius of the iron atoms in the bcc structure, nm\n", + "r_fcc = 0.129; # Atomic radius of the iron atoms in the fcc structure, nm\n", + "a_bcc = 4*r_bcc/sqrt(3);\n", + "a_fcc = 4*r_fcc/sqrt(2);\n", + "V_bcc = 2*a_bcc**3; # Volume of bcc unit cell, nm cube\n", + "V_fcc = a_fcc**3; # Volume of fcc unit cell, nm cube\n", + "delta_V = V_fcc - V_bcc; # Change in volume from bcc to fcc structure, nm cube\n", + "V = V_bcc;\n", + "V_frac = delta_V/V; # Fractional change in volume from bcc to fcc structure\n", + "\n", + "print\"The percentage change in volume from bcc to fcc structure =\",round(V_frac*100,2),\"percent\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The percentage change in volume from bcc to fcc structure = -1.43 percent\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter3.ipynb b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter3.ipynb new file mode 100644 index 00000000..46b18892 --- /dev/null +++ b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter3.ipynb @@ -0,0 +1,825 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:ab6a8819ae55a050f752b7af592620b211f459f05ec8dc687c071a119643f8a0" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3: Specific Heat of Solids and\n", + "Lattice Vibrations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1,Page number 79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "V0 = 9.1*10**-5; # Atomic volume of Pb, metre cube per kg\n", + "K = 2.3*10**-11; # Compressibility of Pb, metre square per newton\n", + "alpha = 86*10**-6; # Coefficient of thermal expansion, per K\n", + "Cv = 1.4*10**2; # Specific heat at constant volume, J/kg\n", + "gama = alpha*V0/(K*Cv); # Grunesien parameter for Pb\n", + "print\"The Grunesien parameter for Pb = \",round(gama,3);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Grunesien parameter for Pb = 2.43\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2,Page number 79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "V0 = 11*10**-5; # Atomic volume of Cu, metre cube per kg\n", + "K = 0.75*10**-11; # Compressibility of Cu, metre square per newton\n", + "alpha = 49*10**-6; # Coefficient of thermal expansion, per K\n", + "gama = 1.9; # The Grunesien parameter for Cu = 2.4 \n", + "Cv = alpha*V0/(K*gama); # Specific heat of Cu at constant volume, J/kg\n", + "print\"The specific heat capacity of Cu = \",round(Cv,3),\"J/kg\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The specific heat capacity of Cu = 378.246 J/kg\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3,Page number 88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "N = 6.02*10**26; # Avogadro's number, per kmole\n", + "C_t = 6.32*10**3; # Velocity of transverse wave, m/s\n", + "C_l = 3.1*10**3; # Velocity of longitudinal wave, m/s\n", + "rho = 2.7*10**3; # Density of Al, kg per metre cube\n", + "M = 26.97; # Atomic weight of Al, gram per mol\n", + "V = M/rho; # Atomic volume of Al, metre cube\n", + "f_c = (9*N/(4*pi*V*(1.0/C_t**3+2.0/C_l**3)))**(1.0/3);\n", + "print\"The Debye cut-off frequency of Al = \",\"{0:.3e}\".format(f_c),\"per sec\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Debye cut-off frequency of Al = 8.468e+12 per sec\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4,Page number 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "N = 6.02*10**23; # Avogadro's number, per mole\n", + "k = 1.38*10**-23; # Boltzmann constant, J/K\n", + "R = N*k; # Molar gas constant, J/mol/K\n", + "theta_D = 2230; # Debye temperature for diamond, K\n", + "T = 300.0; # Room temperature, K\n", + "C_v = 12.0/5*(pi**4*R)*(T/theta_D)**3; # Specific heat capacity per unit volume of diamond, J/mol-K\n", + "print\"The heat capacity per unit volume of diamond = \",round(C_v,3),\"J/mol-K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The heat capacity per unit volume of diamond = 4.729 J/mol-K\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5,Page number 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "k = 1.38*10**-23; # Boltzmann constant, J/K\n", + "theta_D = 1440.0; # Debye temperature for Be, K\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "f_D = k*theta_D/h; # Debye cut off frequency of Be, Hz\n", + "print\"The Debye cut off frequency of Be = \",\"{0:.3e}\".format(f_D),\"sec\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Debye cut off frequency of Be = 2.999e+13 sec\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6,Page number 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "N = 6.023*10**23; # Avogadro's number, per kmol\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "k = 1.38*10**-23; # Boltzmann constant, J/K\n", + "R = N*k; # Molar gas constant, J/kmol/K\n", + "E_F = 7; # Fermi energy of Cu, eV\n", + "theta_D = 348.0; # Debye temperature of Cu, K\n", + "T = 300.0; # Room temperature, K\n", + "T_F = E_F/k; # Fermi temperature of Cu, K\n", + "C_e = pi**2/2*R*10**3*(T/(T_F*e)); # Electronic heat capacity of Cu, J/kmol/K\n", + "C_l = 12.0/5*(pi**4*R)*(T/theta_D)**3; # Lattice heat capacity of Cu, J/kmol/K\n", + "print\"The electronic heat capacity of Cu = \",round(C_e,3),\"J/kmol/K\";\n", + "print\"The lattice heat capacity of Cu = \",round(C_l,3),\"J/mol/K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electronic heat capacity of Cu = 151.616 J/kmol/K\n", + "The lattice heat capacity of Cu = 1244.884 J/mol/K\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7,Page number 90" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "N = 6.023*10**23; # Avogadro's number, per kmol\n", + "e = 1.602*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "k = 1.38*10**-23; # Boltzmann constant, J/K\n", + "R = N*k; # Molar gas constant, J/kmol/K\n", + "E_F = 7.0; # Fermi energy of Cu, eV\n", + "theta_D = 348.0; # Debye temperature of Cu, K\n", + "T = 0.01; # Room temperature, K\n", + "T_F = E_F/k; # Fermi temperature of Cu, K\n", + "C_e = pi**2/2*R*(T/(T_F*e)); # Electronic heat capacity of Cu, J/mol/K\n", + "C_l = 12.0/5*(pi**4*R)*(T/theta_D)**3; # Lattice heat capacity of Cu, J/kmol/K\n", + "print\"The electronic heat capacity of Cu = \",\"{0:.3e}\".format(C_e),\"J/mol/K\";\n", + "print\"The lattice heat capacity of Cu = \",\"{0:.3e}\".format(C_l),\"J/mol/K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electronic heat capacity of Cu = 5.048e-06 J/mol/K\n", + "The lattice heat capacity of Cu = 4.611e-11 J/mol/K\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8,Page number 90" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "N = 6.023*10**23; # Avogadro's number, per kmol\n", + "e = 1.602*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "k = 1.38*10**-23; # Boltzmann constant, J/K\n", + "R = N*k; # Molar gas constant, J/kmol/K\n", + "E_F = 3.2; # Fermi energy of Cu, eV\n", + "theta_D = 150.0; # Debye temperature of Cu, K\n", + "T = 20.0; # Given temperature, K\n", + "T_F = E_F/k; # Fermi temperature of Cu, K\n", + "C_e = pi**2/2*R*(T/(T_F*e)); # Electronic heat capacity of Cu, J/mol/K\n", + "C_l = 12.0/5*(pi**4*R)*(T/theta_D)**3; # Lattice heat capacity of Cu, J/kmol/K\n", + "print\"The electronic heat capacity of Na = \",\"{0:.3e}\".format(C_e),\"J/mol/K\";\n", + "print\"The lattice heat capacity of Na = \",round(C_l,4),\"J/mol/K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electronic heat capacity of Na = 2.208e-02 J/mol/K\n", + "The lattice heat capacity of Na = 4.6059 J/mol/K\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9,Page number 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "N = 6.023*10**23; # Avogadro's number, per kmol\n", + "e = 1.602*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "k = 1.38*10**-23; # Boltzmann constant, J/K\n", + "R = N*k; # Molar gas constant, J/kmol/K\n", + "E_F = 3.2; # Fermi energy of Hf, eV\n", + "theta_D = 242.0; # Debye temperature of Hf, K\n", + "T_F = E_F/k; # Fermi temperature of Hf, K\n", + "T = [300.0, 200.0, 100.0, 10.0, 5.0]; # Declare a vector of 5 temperature values, K\n", + "print\"________________________\";\n", + "print\"T(K) C_l (J/kmol/K)\";\n", + "print\"________________________\";\n", + "for i in xrange(len(T)):\n", + " C_l = 12.0/5*(pi**4*R)*(T[i]/theta_D)**3; # Lattice heat capacity of Hf, J/kmol/K \n", + " print\"\",T[i],\" \",round(C_l,3);\n", + "print\"________________________\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "________________________\n", + "T(K) C_l (J/kmol/K)\n", + "________________________\n", + " 300.0 3701.863\n", + " 200.0 1096.848\n", + " 100.0 137.106\n", + " 10.0 0.137\n", + " 5.0 0.017\n", + "________________________\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.10,Page number 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "N = 6.023*10**23; # Avogadro's number, per kmol\n", + "e = 1.602*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "k = 1.38*10**-23; # Boltzmann constant, J/K\n", + "R = N*k; # Molar gas constant, J/kmol/K\n", + "E_F = 7.0; # Fermi energy of Hf, eV\n", + "theta_D = 343.0; # Debye temperature of Hf, K\n", + "T_F = E_F/k; # Fermi temperature of Hf, K\n", + "# As C_l = 12/5*(pi**4*R)*(T/theta_D)**3 and C_e = pi**2/2*R*(T/(T_F*e)) so that\n", + "# For C_l = C_e, we have\n", + "T = sqrt((pi**2/2*R*1/(T_F*e))/(12.0/5*pi**4*R)*theta_D**3); # Required temperature when C_l = C_e, K\n", + "print\"The temperature at which lattice specific heat equals electronic specific heat for Cu =\",round(T,3),\"K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The temperature at which lattice specific heat equals electronic specific heat for Cu = 3.238 K\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11,Page number 92" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "C11 = 1.08*10**12; C12 = 0.62*10**12; C44 = 0.28*10**12; # Elastic constants of Al, dynes/cm square\n", + "a = 4.05*10**-8; # Lattice constant for Al cubic structure, cm\n", + "rho = 2.70; # g/cm cube \n", + "k = 1.38*10**-23; # Boltzmann constant, J/K\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "s = 4.0; # Number of atoms in Al unit cell\n", + "Va = a**3; # Volume of unit cell, cm cube\n", + "theta_D =(3.15/(8*pi)*(h/k)**3*s/(rho**(3.0/2)*Va)*(C11-C12)**(1.0/2)*(C11+C12+2*C44)**(1.0/2)*C44**(1.0/2))**(1.0/3);\n", + "print\"The Debye temperature of Al =\",round(theta_D,3),\"K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Debye temperature of Al = 466.605 K\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.12,Page number 93" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "k = 1.38*10**-23; # Boltzmann constant, J/K\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "A =[[1,2,3,4,5,6,7,8],[9,10,11,12,13,14,15,16]]; # Declare a matrix of 2X8\n", + "A[0][0] = 'Cu';\n", + "A[0][1] = 1.684*10**12;\n", + "A[0][2] = 1.214*10**12;\n", + "A[0][3] = 0.754*10**12;\n", + "A[0][4] = 4;\n", + "A[0][5] = 3.61*10**-8;\n", + "A[0][6] = 8.96;\n", + "A[1][0] = 'Na';\n", + "A[1][1] = 0.055*10**12;\n", + "A[1][2] = 0.047*10**12;\n", + "A[1][3] = 0.049*10**12;\n", + "A[1][4] = 2;\n", + "A[1][5] = 4.225*10**-8;\n", + "A[1][6] = 0.971;\n", + "\n", + "# For Cu\n", + "Va = A[0][5]**3; # Volume of unit cell, cm cube\n", + "A[0][7] = (3.15/(8*pi)*(h/k)**3*A[0][4]/(A[0][6]**(3.0/2)*Va)*(A[0][1]-A[0][2])**(1.0/2)*(A[0][1]+A[0][2]+2*A[0][3])**(1.0/2)*A[0][3]**(1.0/2))**(1.0/3);\n", + "\n", + "# For Na\n", + "Va =A[1][5]**3; # Volume of unit cell, cm cube\n", + "A[1][7] = (3.15/(8*pi)*(h/k)**3*A[1][4]/(A[1][6]**(3.0/2)*Va)*(A[1][1]-A[1][2])**(1.0/2)*(A[1][1]+A[1][2]+2*A[1][3])**(1.0/2)*A[1][3]**(1.0/2))**(1.0/3);\n", + "\n", + "print\"________________________________________\";\n", + "print\"Metal C11 C12 C44 thetaD\";\n", + "print\"________________________________________\";\n", + "for i in range (0,2) :\n", + " print\"\",A[i][0],\" \",A[i][1]/10**12,\" \",A[i][2]/10**12,\" \",A[i][3]/10**12,\" \",round(A[i][7],2);\n", + "print\"________________________________________\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "________________________________________\n", + "Metal C11 C12 C44 thetaD\n", + "________________________________________\n", + " Cu 1.684 1.214 0.754 380.2\n", + " Na 0.055 0.047 0.049 150.44\n", + "________________________________________\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.13,Page number 93" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "k = 1.38*10**-23; # Boltzmann constant, J/K\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "A =[[1,2,3,4,5],[6,7,8,9,10],[11,12,13,14,15],[16,17,18,19,20]]; # Declare a matrix of 4X5\n", + "A[0][0] = 300;\n", + "A[0][1] = 0.878*10**10;\n", + "A[0][2] = 0.483*10**10;\n", + "A[0][3] = 0.448*10**10;\n", + "A[1][0] = 200;\n", + "A[1][1] = 0.968*10**10;\n", + "A[1][2] = 0.508*10**10;\n", + "A[1][3] = 0.512*10**10;\n", + "A[2][0] = 100;\n", + "A[2][1] = 1.050*10**10;\n", + "A[2][2] = 0.540*10**10;\n", + "A[2][3] = 0.579*10**10;\n", + "A[3][0] = 20;\n", + "A[3][1] = 1.101*10**10;\n", + "A[3][2] = 0.551*10**10;\n", + "A[3][3] = 0.624*10**10;\n", + "s = 2; # Number of atoms in a unit cell\n", + "a = 4.225*10**-10; # Lattice parameter of Na, m\n", + "rho = 0.971*10**3; # Density of Na, kg/metre-cube\n", + "Va = a**3; # Volume of unit cell, metre cube\n", + "print\"________________________________________\";\n", + "print\"T C11 C12 C44 thetaD\"\n", + "print\"________________________________________\";\n", + "for i in range (0,4) :\n", + " A[i][4] = (3.15/(8*pi)*(h/k)**3*s/(rho**(3.0/2)*Va)*(A[i][1]-A[i][2])**(1.0/2)*(A[i][1]+A[i][2]+2*A[i][3])**(1.0/2)*A[i][3]**(1.0/2))**(1.0/3);\n", + " print\"\",A[i][0],\" \",A[i][1]/10**10,\" \",A[i][2]/10**10,\" \",A[i][3]/10**10,\" \",round(A[i][4],2);\n", + "\n", + "print\"________________________________________\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "________________________________________\n", + "T C11 C12 C44 thetaD\n", + "________________________________________\n", + " 300 0.878 0.483 0.448 197.33\n", + " 200 0.968 0.508 0.512 210.52\n", + " 100 1.05 0.54 0.579 222.08\n", + " 20 1.101 0.551 0.624 229.77\n", + "________________________________________\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.14,Page number 93" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import scipy\n", + "from scipy.integrate import quad\n", + "\n", + "#Given Data\n", + "Lu =[[1,2,3,4,5],[6,7,8,9,10],[11,12,13,14,15],[16,17,18,19,20],[21,22,23,24,25],[26,27,28,29,30]]; # Declare a matrix of 6X5\n", + "Lu[0][0] = 0;\n", + "Lu[0][1] = 5.58;\n", + "Lu[0][2] = 3.517;\n", + "Lu[0][4] = 0.750;\n", + "Lu[1][0] = 36;\n", + "Lu[1][1] = 5.409;\n", + "Lu[1][2] = 3.440;\n", + "Lu[1][4] = 0.560;\n", + "Lu[2][0] = 103;\n", + "Lu[2][1] = 5.213;\n", + "Lu[2][2] = 3.341;\n", + "Lu[2][4] = 0.492;\n", + "Lu[3][0] = 157;\n", + "Lu[3][1] = 5.067;\n", + "Lu[3][2] = 3.259;\n", + "Lu[3][4] = 0.388;\n", + "Lu[4][0] = 191;\n", + "Lu[4][1] = 4.987;\n", + "Lu[4][2] = 3.217;\n", + "Lu[4][4] = 0.357;\n", + "Lu[5][0] = 236;\n", + "Lu[5][1] = 4.921;\n", + "Lu[5][2] = 3.179;\n", + "Lu[5][4] = 0.331;\n", + "V0 = 3*sqrt(3)/2*Lu[0][2]**2*Lu[0][1];\n", + "V = [0,0,0,0,0,0]; # Declare volume array\n", + "print\"______________________________________________________________\";\n", + "print\"P(kbar) c(angstrom) a(angstrom) gamma_G nu_G \";\n", + "print\"______________________________________________________________\";\n", + "for i in range (0,6) :\n", + " V[i] = 3*sqrt(3)/2*Lu[i][2]**2*Lu[i][1];\n", + " Lu[i][3] = Lu[i][4]*V[i]/V0+2.0/3*(1-V[i]/V0)**(1.0/2);\n", + " print\"\",Lu[i][0],\" \",Lu[i][1],\" \",Lu[i][2],\" \",round(Lu[i][3],3),\" \",Lu[i][4];\n", + "\n", + "print\"______________________________________________________________\";\n", + "\n", + "cnt = 0;\n", + "print\"________________________\";\n", + "print\"P(kbar) Theta_D(K)\";\n", + "print\"________________________\";\n", + "for i in range (0,6) :\n", + " def integrand(x, a, b):\n", + " return (-1*Lu[i][4]*(exp(x)/V0)-(2.0/3)*(1-exp(x)/V0)**(1.0/2))\n", + " a=1;\n", + " b=1;\n", + " I = quad(integrand,-0.8+cnt,log(V[i]/1000000), args=(a,b));\n", + " theta_D = exp(I);\n", + " cnt = cnt + 0.01;\n", + " print\"\",Lu[i][0],\" \",round(theta_D[0],0);\n", + "\n", + "print\"________________________\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "______________________________________________________________\n", + "P(kbar) c(angstrom) a(angstrom) gamma_G nu_G \n", + "______________________________________________________________\n", + " 0 5.58 3.517 0.75 0.75\n", + " 36 5.409 3.44 0.699 0.56\n", + " 103 5.213 3.341 0.679 0.492\n", + " 157 5.067 3.259 0.615 0.388\n", + " 191 4.987 3.217 0.602 0.357\n", + " 236 4.921 3.179 0.591 0.331\n", + "______________________________________________________________\n", + "________________________\n", + "P(kbar) Theta_D(K)\n", + "________________________\n", + " 0 185.0\n", + " 36 195.0\n", + " 103 210.0\n", + " 157 222.0\n", + " 191 230.0\n", + " 236 237.0\n", + "________________________\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.15,Page number 94" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#GIven Data\n", + "T_M = 1356.0; # Melting temperature of Cu, K\n", + "V = 7.114; # Atomic volume of Cu, cm cube per g-atom\n", + "M = 63.5; # atomic weight of Cu, g/mole\n", + "K = 138.5; # Lindemann constant\n", + "theta_M = K*(T_M/M)**(1.0/2)*(1/V)**(1.0/3); # Debye temperature by Lindemann method, K\n", + "\n", + "print\"The Debye temperature by Lindemann method =\",round(theta_M,3),\"K\";\n", + "print\"The values obtained from other methods are:\";\n", + "print\"theta_s = 342 K; theta_R = 336 K; theta_E = 345 K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Debye temperature by Lindemann method = 332.778 K\n", + "The values obtained from other methods are:\n", + "theta_s = 342 K; theta_R = 336 K; theta_E = 345 K\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.16,Page number 100" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "N_A = 6.023*10**23; # Avogadro's number\n", + "c = 3.0*10**8; # Speed of light, m/s\n", + "epsilon_0 = 15.0; # Dielectric constant of the medium\n", + "m = 2.0*10**-22; # Mass of ion, g\n", + "e = 4.8*10**-10; # Charge on the ion, C\n", + "rho = 7.0; # Average density of solid, g/cc\n", + "A = 120.0; # Average atomic weight of solid, g\n", + "N = rho/A*N_A; # Number of ions per cc, per cm cube\n", + "f_P = 1/(2*pi)*sqrt(4*pi*N*e**2/(m*epsilon_0)); # Plasma frequency of vibrating ions in the crystal, Hz\n", + "lamda_P = c/f_P; # Plasma wavelength of vibrating ions in the crystal, cm\n", + "print\"The plasma frequency of vibrating ions in InSb crystal = \",\"{0:.3e}\".format(f_P),\"Hz\";\n", + "print\"The plasma wavelength of vibrating ions in InSb crystal =\",round(lamda_P/10**-6,3),\"micron\";\n", + "print\"The calculated frequency lies in the infrared region.\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The plasma frequency of vibrating ions in InSb crystal = 9.268e+11 Hz\n", + "The plasma wavelength of vibrating ions in InSb crystal = 323.706 micron\n", + "The calculated frequency lies in the infrared region.\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.17,Page number 103" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "h = 6.624*10**-34; # Planck's constant, Js\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "q = 1.486*10**11; # Young's modulus of diamond, N/metre-square\n", + "rho = 3500; # Density of diamond, kg/metre-cube\n", + "c = sqrt(q/rho); # Speed of transverse wave through diamond, m/s\n", + "m = 12*1.66*10**-27; # Atomic weight of carbon, kg\n", + "theta_D = (h/k)*c*(3*rho/(4*pi*m))**(1.0/3); # Debye temperature for diamond, K\n", + "print\"The Debye temperature for diamond =\",round(theta_D,3),\"K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Debye temperature for diamond = 1086.709 K\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter4.ipynb b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter4.ipynb new file mode 100644 index 00000000..0b46d733 --- /dev/null +++ b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter4.ipynb @@ -0,0 +1,1614 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:6f64ae81ab14bf773565224cb22f224628ca2a0e167e3fe9ede8614fd4e3f3d5" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4: Free Electron Theory of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1,Page number 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "e = 1.6*10**-19; # Charge on an electron, C\n", + "n = 8.5*10**28; # Concentration of electron in Cu, per metre cube\n", + "rho = 1.7*10**-8; # Resistivity of Cu, ohm-m\n", + "t = m/(n*e**2*rho); # Collision time for an electron in monovalent Cu, s\n", + "print\"The collision time for an electron in monovalent Cu =\",\"{0:.3e}\".format(t),\"s\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The collision time for an electron in monovalent Cu = 2.460e-14 s\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2,Page number 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "e = 1.6*10**-19; # Charge on an electron, C\n", + "n = 10**29; # Concentration of electron in material, per metre cube\n", + "rho = 27*10**-8; # Resistivity of the material, ohm-m\n", + "tau = m/(n*e**2*rho); # Collision time for an electron in the material, s\n", + "v_F = 1*10**8; # Velocity of free electron, cm/s\n", + "lamda = v_F*tau; # Mean free path of electron in the material, cm\n", + "print\"The collision time for an electron in monovalent Cu =\",\"{0:.3e}\".format(tau),\"s\";\n", + "print\"The mean free path of electron at 0K =\",\"{0:.3e}\".format(lamda),\"cm\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The collision time for an electron in monovalent Cu = 1.317e-15 s\n", + "The mean free path of electron at 0K = 1.317e-07 cm\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3,Page number 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "e = 1.6*10**-19; # Charge on an electron, C\n", + "r = 1.28*10**-10; # Atomic radius of cupper, m\n", + "a = 4*r/sqrt(2); # Lattice parameter of fcc structure of Cu, m\n", + "V = a**3; # Volume of unit cell of Cu, metre cube\n", + "n = 4/V; # Number of atoms per unit volume of Cu, per metre cube\n", + "tau = 2.7*10**-4; # Relaxation time for an electron in monovalent Cu, s\n", + "sigma = n*e**2*tau/m; # Electrical conductivity of Cu, mho per cm\n", + "print\"The free electron density in monovalent Cu =\",\"{0:.3e}\".format(n),\"per metre cube\";\n", + "print\"The electrical conductivity of monovalent Cu =\",\"{0:.3e}\".format(sigma),\"mho per cm\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The free electron density in monovalent Cu = 8.429e+28 per metre cube\n", + "The electrical conductivity of monovalent Cu = 6.403e+17 mho per cm\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4,Page number 118" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "h = 6.625*10**-34; # Planck's constant, Js\n", + "L = 10*10**-3; # Length of side of the cube, m\n", + "# For nth level\n", + "nx = 1; ny = 1; nz = 1; # Positive integers along three axis\n", + "En = h**2/(8*m*L**2)*(nx**2+ny**2+nz**2)/e; # Energy of nth level for electrons, eV\n", + "# For (n+1)th level\n", + "nx = 2; ny = 1; nz = 1; # Positive integers along three axis\n", + "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\n", + "delta_E = En_plus_1 - En; # Energy difference between two levels for the free electrons\n", + "print\"The energy difference between two levels for the free electrons =\",\"{0:.3e}\".format( delta_E),\"eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy difference between two levels for the free electrons = 1.130e-14 eV\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5,Page number 119" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "T = 300.0; # Room temperature of tungsten, K\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "E_F = 4.5*e; # Fermi energy of tungsten, J\n", + "E = E_F-0.1*E_F; # 10% energy below Fermi energy, J\n", + "f_T = 1.0/(1+exp((E-E_F)/(k*T))); # Probability of the electron in tungsten at room temperature at an nergy 10% below the Fermi energy\n", + "print\"The probability of the electron at an energy 10 percent below the Fermi energy in tungsten at 300 K =\",round(f_T,3);\n", + "E = 2*k*T+E_F; # For energy equal to 2kT + E_F\n", + "f_T = 1.0/(1+exp((E-E_F)/(k*T))); # Probability of the electron in tungsten at an energy 2kT above the Fermi energy\n", + "print\"The probability of the electron at an energy 2kT above the Fermi energy =\",round(f_T,4);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The probability of the electron at an energy 10 percent below the Fermi energy in tungsten at 300 K = 1.0\n", + "The probability of the electron at an energy 2kT above the Fermi energy = 0.1192\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6,Page number 121" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "h = 6.625*10**-34; # Planck's constant, Js\n", + "h_cross = h/(2*pi); # Reduced Planck's constant, Js\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "a = 5.34*10**-10; # Lattice constant of monovalent bcc lattice, m\n", + "V = a**3; # Volume of bcc unit cell, metre cube\n", + "n = 2.0/V; # Number of atoms per metre cube\n", + "E_F = h_cross**2.0/(2*m*e)*(3*pi**2*n)**(2.0/3); # Fermi energy of monovalent bcc solid, eV\n", + "\n", + "print\"The Fermi energy of a monovalent bcc solid =\",round(E_F,4),\"eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Fermi energy of a monovalent bcc solid = 2.0341 eV\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7,Page number 121" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "h = 6.625*10**-34; # Planck's constant, Js\n", + "h_cross = h/(2*pi); # Reduced Planck's constant, Js\n", + "m = 9.11*10**-31; # Mass of an electron, kg\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "V = 1*10**-5; # Volume of cubical box, metre cube\n", + "E_F = 5*e; # Fermi energy, J \n", + "D_EF = V/(2*pi**2)*(2*m/h_cross**2)**(3.0/2)*E_F**(1.0/2)*e; # Density of states at Fermi energy, states/eV\n", + "print\"The density of states at Fermi energy =\",\"{0:.3e}\".format( D_EF),\"states/eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The density of states at Fermi energy = 1.521e+23 states/eV\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8,Page number 121" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "h_cross = h/(2*pi); # Reduced Planck's constant, Js\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "V = 1*10**-6; # Volume of cubical box, metre cube\n", + "E_F = 7.13*e; # Fermi energy for Mg, J \n", + "D_EF = V/(2*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\n", + "E_Mg = 1.0/D_EF; # The energy separation between adjacent energy levels of Mg, J\n", + "print\"The energy separation between adjacent energy levels of Mg =\",\"{0:.3e}\".format(E_Mg/e),\"eV\";\n", + "E_F = 1.58*e; # Fermi energy for Cs, J \n", + "D_EF = V/(2*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\n", + "E_Mg = 1.0/D_EF; # The energy separation between adjacent energy levels of Cs, J\n", + "print\"The energy separation between adjacent energy levels of Cs =\",\"{0:.3e}\".format(E_Mg/e),\"eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy separation between adjacent energy levels of Mg = 5.517e-23 eV\n", + "The energy separation between adjacent energy levels of Cs = 1.172e-22 eV\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9,Page number 122" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "E_F = 3.2*e; # Fermi energy of sodium, J\n", + "P_F = sqrt(E_F*2*m); # Fermi momentum of sodium, kg-m/s\n", + "print\"The Fermi momentum of sodium =\",\"{0:.3e}\".format(P_F),\"kg-m/sec\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Fermi momentum of sodium = 9.653e-25 kg-m/sec\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.10,Page number 122" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "T = 500.0; # Rise in temperature of Al, K\n", + "EF_0 = 11.63; # Fermi energy of Al, eV\n", + "EF_T = EF_0*(1-pi**2.0/12*(k*T/EF_0)**2); # Change in Fermi energy of Al with temperature, eV\n", + "print\"The change in Fermi energy of Al with tempertaure rise of 500 degree celsius =\",round(EF_T,3),\"eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The change in Fermi energy of Al with tempertaure rise of 500 degree celsius = 11.63 eV\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.11,Page number 122" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.18*10**-31; # Mass of an electron, kg\n", + "e = 1.6*10**-19; # Charge on an electron, C\n", + "lamda = 1.0*10**-9; # Mean free path of electron in metal, m\n", + "v = 1.11*10**5; # Average velocity of the electron in metal, m/s\n", + "\n", + "# For Lead\n", + "n = 13.2*10**28; # Electronic concentration of Pb, per metre cube\n", + "sigma = n*e**2*lamda/(m*v); # Electrical conductivity of lead, mho per metre\n", + "print\"The electrical conductivity of lead =\",\"{0:.3e}\".format(sigma),\"mho per metre\";\n", + "\n", + "# For Silver\n", + "n = 5.85*10**28; # Electronic concentration of Ag, per metre cube\n", + "sigma = n*e**2*lamda/(m*v); # Electrical conductivity of Ag, mho per metre\n", + "print\"The electrical conductivity of silver =\",\"{0:.3e}\".format(sigma),\"mho per metre\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electrical conductivity of lead = 3.316e+07 mho per metre\n", + "The electrical conductivity of silver = 1.470e+07 mho per metre\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.12,Page number 125" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "e = 1.6*10**-19; # Charge on an electron, C\n", + "L = pi**2.0/3*(k/e)**2; # Lorentz number, watt-ohm/degree-square\n", + "print\"The Lorentz number =\",\"{0:.3e}\".format(L),\"watt-ohm/degree-square\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Lorentz number = 2.447e-08 watt-ohm/degree-square\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.13,Page number 125" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "A =[[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]; # Declare a 4X4 cell\n", + "A[0][0] = 'Mg'; \n", + "A[0][1] = 2.54*10**-5; \n", + "A[0][2] = 1.5; \n", + "A[0][3] = 2.32*10**2; \n", + "A[1][0] = 'Cu'; \n", + "A[1][1] = 6.45*10**-5; \n", + "A[1][2] = 3.85; \n", + "A[1][3] = 2.30*10**2; \n", + "A[2][0] = 'Al'; \n", + "A[2][1] = 4.0*10**-5; \n", + "A[2][2] = 2.38;\n", + "A[2][3] = 2.57*10**2; \n", + "A[3][0] = 'Pt'; \n", + "A[3][1] = 1.02*10**-5; \n", + "A[3][2] = 0.69;\n", + "A[3][3] = 2.56*10**2; \n", + "T1 = 273; # First temperature, K\n", + "T2 = 373; # Second temperature, K\n", + "print\"_________________________________________________________________\";\n", + "print\"Metal sigma x 10**-05 K(W/cm-K) Lorentz number \";\n", + "print\" (mho per cm) (watt-ohm/deg-square)x10**-2\";\n", + "print\"_________________________________________________________________\";\n", + "for i in range (0,4) :\n", + " L1 = A[i][2]/(A[i][1]*T1); \n", + " L2 = A[i][3];\n", + " print\"\",A[i][0],\" \",A[i][1]/10**-5,\" \",A[i][2],\" \",L2/10**2,\" \",L2/10**2;\n", + "print\"_________________________________________________________________\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "_________________________________________________________________\n", + "Metal sigma x 10**-05 K(W/cm-K) Lorentz number \n", + " (mho per cm) (watt-ohm/deg-square)x10**-2\n", + "_________________________________________________________________\n", + " Mg 2.54 1.5 2.32 2.32\n", + " Cu 6.45 3.85 2.3 2.3\n", + " Al 4.0 2.38 2.57 2.57\n", + " Pt 1.02 0.69 2.56 2.56\n", + "_________________________________________________________________\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.14,Page number 125" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "A = [[1,2],[3,4]]; # Declare a 2X3 cell\n", + "A[0][0] = 1.6*10**8; # Electrcal conductivity of Au at 100 K, mho per metre\n", + "A[0][1] = 2.0*10**-8; # Lorentz number of Au at 100 K, volt/K-square\n", + "A[1][0] = 5.0*10**8; # Electrcal conductivity of Au at 273 K, mho per metre\n", + "A[1][1] = 2.4*10**-8; # Lorentz number of Au at 273 K, volt/K-square\n", + "T1 = 100; # First temperature, K\n", + "T2 = 273; # Second temperature, K\n", + "\n", + "print\"___________________________________________________________________________\";\n", + "print\" T = 100 K T = 273 K \";\n", + "print\"_________________________________ ___________________________________\";\n", + "print\"Electrical conductivity) L Electrical conductivity) L \";\n", + "print\" mho per metre V/K-square mho per metre V/K-square\";\n", + "print\"___________________________________________________________________________\";\n", + "K1 = A[0][0]*T1*A[0][1]; \n", + "K2 = A[1][0]*T2*A[1][1];\n", + "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]) \n", + "print\"K =\",K1,\"W/cm-K K =\",K2,\"W/cm-K\";\n", + "print\"___________________________________________________________________________\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "___________________________________________________________________________\n", + " T = 100 K T = 273 K \n", + "_________________________________ ___________________________________\n", + "Electrical conductivity) L Electrical conductivity) L \n", + " mho per metre V/K-square mho per metre V/K-square\n", + "___________________________________________________________________________\n", + "1.600e+08 2.000e-08 5.000e+08 2.400e-08\n", + "K = 320.0 W/cm-K K = 3276.0 W/cm-K\n", + "___________________________________________________________________________\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.15,Page number 131" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Electronic charge, C\n", + "a = 0.428*10**-9; # Lattice constant of Na, m\n", + "V = a**3; # Volume of unit cell, metre cube\n", + "N = 2; # No. of atoms per unit cell of Na\n", + "n = N/V; # No. of electrons per metre cube, per metre cube\n", + "R_H = -1.0/(n*e); # Hall coeffcient of Na, metre cube per coulomb\n", + "print\"The Hall coefficient of sodium =\",\"{0:.3e}\".format(R_H),\"metre cube per coulomb\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Hall coefficient of sodium = -2.450e-10 metre cube per coulomb\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.16,Page number 131" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Electronic charge, C\n", + "n = 24.2*10**28; # No. of electrons per metre cube, per metre cube\n", + "R_H = -1.0/(n*e); # Hall coeffcient of Be, metre cube per coulomb\n", + "print\"The Hall coefficient of beryllium =\",\"{0:.3e}\".format(R_H),\"metre cube per coulomb\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Hall coefficient of beryllium = -2.583e-11 metre cube per coulomb\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.17,Page number 131" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Electronic charge, C\n", + "R_H = -8.4*10**-11; # Hall coeffcient of Ag, metre cube per coulomb\n", + "n = -3*pi/(8*R_H*e); # Electronic concentration of Ag, per metre cube\n", + "print\"The electronic concentration of Ag =\",\"{0:.3e}\".format(n),\"per metre cube\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electronic concentration of Ag = 8.766e+28 per metre cube\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.18,Page number 134" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "# We have from Mattheissen rule, rho = rho_0 + alpha*T1\n", + "T1 = 300.0; # Initial temperature, K\n", + "T2 = 1000.0; # Final temperature, K\n", + "rho = 1*10**-6; # Resistivity of the metal, ohm-m\n", + "delta_rho = 0.07*rho; # Increase in resistivity of metal, ohm-m\n", + "alpha = delta_rho/(T2-T1); # A constant, ohm-m/K\n", + "rho_0 = rho - alpha*T1; # Resistivity at room temperature, ohm-m\n", + "print\"The resistivity at room temperature =\",\"{0:.3e}\".format(rho),\"ohm-m\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistivity at room temperature = 1.000e-06 ohm-m\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.19,Page number 134" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "# We have from Mattheissen rule, rho = rho_0 + alpha*T1\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "rho_40 = 0.2; # Resistivity of Ge at 40 degree celsius, ohm-m\n", + "E_g = 0.7; # Bandgap for Ge, eV\n", + "T1 = 20+273; # Second temperature, K\n", + "T2 = 40 + 273; # First temperature, K\n", + "rho_20 = rho_40*exp(E_g*e/(2*k)*(1.0/T1-1.0/T2)); # Resistivity of Ge at 20 degree celsius, ohm-m\n", + "print\"The resistivity of Ge at 20 degree celsius =\",round(rho_20,1),\"ohm-m\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistivity of Ge at 20 degree celsius = 0.5 ohm-m\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.20,Page number 135" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "rs_a0_ratio = 3.25; # Ratio of solid radius to the lattice parameter\n", + "E_F = 50.1*(rs_a0_ratio)**(-2); # Fermi level energy of Li, eV\n", + "T_F = 58.2e+04*(rs_a0_ratio)**(-2); # Fermi level temperature of Li, K\n", + "V_F = 4.20e+08*(rs_a0_ratio)**(-1); # Fermi level velocity of electron in Li, cm/sec\n", + "K_F = 3.63e+08*(rs_a0_ratio)**(-1); \n", + "print\"E_F =\",round(E_F,2),\"eV\";\n", + "print\"T_F =\",\"{0:.3e}\".format(T_F),\"K\";\n", + "print\"V_F =\",\"{0:.3e}\".format(V_F),\"cm/sec\";\n", + "print\"K_F =\",\"{0:.3e}\".format(K_F),\"per cm\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "E_F = 4.74 eV\n", + "T_F = 5.510e+04 K\n", + "V_F = 1.292e+08 cm/sec\n", + "K_F = 1.117e+08 per cm\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.21,Page number 135" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "n = 6.04*10**22; # Concentration of electrons in yittrium, per metre cube\n", + "r_s = (3/(4*pi*n))**(1.0/3)/10**-8; # Radius of the solid, angstrom\n", + "a0 = 0.529; # Lattice parameter of yittrium, angstrom\n", + "rs_a0_ratio = r_s/a0; # Solid radius to lattice parameter ratio\n", + "E_F = 50.1*(rs_a0_ratio)**(-2); # Fermi level energy of Y, eV\n", + "print\"The Fermi energy of yittrium =\",round(E_F,4),\"eV\";\n", + "Ryd = 13.6; # Rydberg energy constant, eV\n", + "E_bs = 0.396*Ryd; # Band structure energy value of Y, eV\n", + "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\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Fermi energy of yittrium = 5.6083 eV\n", + "The band structure value of E_F = 5.386 eV is in close agreement with the calculated value of 5.6083 eV\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.22,Page number 137" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "rs_a0_ratio = 2.07; # Solid radius to lattice parameter ratio for Al\n", + "E_F = 50.1*(rs_a0_ratio)**(-2); # Fermi level energy of Y, eV\n", + "# According to Jellium model, h_cross*omega_P = E = 47.1 eV *(rs_a0_ratio)**(-3/2)\n", + "E = 47.1*(rs_a0_ratio)**(-3.0/2); # Plasmon energy of Al, eV\n", + "print\"The plasmon energy of Al =\",round(E,4),\"eV\";\n", + "print\"The experimental value is 15 eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The plasmon energy of Al = 15.8149 eV\n", + "The experimental value is 15 eV\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1a,Page number 137" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "E_F = 1; # For simplicity assume Fermi energy to be unity, eV\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "dE = 0.1; # Exces energy above Fermi level, eV\n", + "T = 300; # Room temperature, K\n", + "E = E_F + dE; # Energy of the level above Fermi level, eV\n", + "f_E = 1.0/(exp((E-E_F)*e/(k*T))+1); # Occupation probability of the electron at 0.1 eV above E_F\n", + "print\"At 300 K:\";\n", + "print\"=========\";\n", + "print\"The occupation probability of electron at\",round(dE,3),\"eV above Fermi energy =\",round(f_E,3);\n", + "E = E_F - dE; # Energy of the level below Fermi level, eV\n", + "f_E = 1.0/(exp((E-E_F)*e/(k*T))+1); # Occupation probability of the electron at 0.1 eV below E_F\n", + "print\"The occupation probability of electron at\",round(dE,3),\"below Fermi energy =\",round(f_E,3);\n", + "\n", + "T = 1000; # New temperature, K\n", + "print\"At 1000 K:\";\n", + "print\"=========\";\n", + "E = E_F + dE; # Energy of the level above Fermi level, eV\n", + "f_E = 1.0/(exp((E-E_F)*e/(k*T))+1); # Occupation probability of the electron at 0.1 eV above E_F\n", + "print\"The occupation probability of electron at\",round(dE,3),\"eV above Fermi energy =\",round(f_E,3);\n", + "E = E_F - dE; # Energy of the level below Fermi level, eV\n", + "f_E = 1.0/(exp((E-E_F)*e/(k*T))+1); # Occupation probability of the electron at 0.1 eV below E_F\n", + "print\"The occupation probability of electron at\",round(dE,3),\"eV below Fermi energy =\",round(f_E,3);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At 300 K:\n", + "=========\n", + "The occupation probability of electron at 0.1 eV above Fermi energy = 0.021\n", + "The occupation probability of electron at 0.1 below Fermi energy = 0.979\n", + "At 1000 K:\n", + "=========\n", + "The occupation probability of electron at 0.1 eV above Fermi energy = 0.239\n", + "The occupation probability of electron at 0.1 eV below Fermi energy = 0.761\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2a,Page number 138" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "f_E = 0.01; # Occupation probability of electron\n", + "E_F = 1; # For simplicity assume Fermi energy to be unity, eV\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "dE = 0.5; # Exces energy above Fermi level, eV\n", + "E = E_F + dE; # Energy of the level above Fermi level, eV\n", + "# We have, f_E = 1/(exp((E-E_F)*e/(k*T))+1), solving for T\n", + "T = (E-E_F)*e/k*1.0/log(1.0/f_E-1); # Temperature at which the electron will have energy 0.1 eV above the Fermi energy, K\n", + "print\"The temperature at which the electron will have energy\",round(dE,3),\"eV above the Fermi energy =\",round(T,3),\"K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The temperature at which the electron will have energy 0.5 eV above the Fermi energy = 1261.578 K\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3a,Page number 139" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "E_F = 10; # Fermi energy of electron in metal, eV\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "E_av = 3.0/5*E_F; # Average energy of free electron in metal at 0 K, eV\n", + "V_F = sqrt(2*E_av*e/m); # Speed of free electron in metal at 0 K, eV\n", + "print\"The average energy of free electron in metal at 0 K =\",round(E_av,3),\"eV\";\n", + "print\"The speed of free electron in metal at 0 K =\",\"{0:.3e}\".format(V_F),\"m/s\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The average energy of free electron in metal at 0 K = 6.0 eV\n", + "The speed of free electron in metal at 0 K = 1.453e+06 m/s\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4a,Page number 139" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "f_E = 0.1; # Occupation probability of electron\n", + "E_F = 5.5; # Fermi energy of Cu, eV\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "dE = 0.05*E_F; # Exces energy above Fermi level, eV\n", + "E = E_F + dE; # Energy of the level above Fermi level, eV\n", + "# We have, f_E = 1/(exp((E-E_F)*e/(k*T))+1), solving for T\n", + "T = (E-E_F)*e/k*1.0/log(1.0/f_E-1); # Temperature at which the electron will have energy 0.1 eV above the Fermi energy, K\n", + "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\";\n", + "\n", + "#(The answer given in the textbook is wrong) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The temperature at which the electron will have energy 5.0 percent above the Fermi energy 1451.106 K\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5a,Page number 139" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "T_F = 24600; # Fermi temperature of potassium, K\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "E_F = k*T_F; # Fermi energy of potassium, eV\n", + "v_F = sqrt(2*k*T_F/m); # Fermi velocity of potassium, m/s\n", + "print\"The Fermi velocity of potassium =\",\"{0:.3e}\".format(v_F),\"m/s\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Fermi velocity of potassium = 8.638e+05 m/s\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6a,Page number 139" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "E_F = 7.0; # Fermi energy of Cu, eV\n", + "f_E = 0.9; # Occupation probability of Cu\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "T = 1000; # Given temperature, K\n", + "# We have, f_E = 1/(exp((E-E_F)*e/(k*T))+1), solving for E\n", + "E = k*T*log(1.0/f_E-1) + E_F*e; # Energy level of Cu for 10% occupation probability at 1000 K, J\n", + "print\"The energy level of Cu for 10 percent occupation probability at 1000 K =\",round(E/e,3),\"eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy level of Cu for 10 percent occupation probability at 1000 K = 6.81 eV\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7a,Page number 140" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "e = 1.6*10**-19; # Electronic charge, C\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "E_F = 1.55; # Fermi energy of Cu, eV\n", + "n = (pi/3)*(8*m/h**2)**(3.0/2)*(E_F*e)**(3.0/2); # Electronic concentration in cesium, electrons/cc\n", + "print\"The electronic concentration in cesium =\",\"{0:.3e}\".format(n),\"electrons/cc\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electronic concentration in cesium = 8.733e+27 electrons/cc\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8a,Page number 141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "E_F = 7; # Fermi energy, eV\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "T_F = E_F*e/k; # Fermi temperature, K\n", + "print\"The Fermi temperature corresponding to Fermi energy =\",\"{0:.3e}\".format(T_F),\"K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Fermi temperature corresponding to Fermi energy = 8.116e+04 K\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9a,Page number 141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.1*10**-31; # Mass of the electron, kg\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "h_cross = h/(2*pi); # Reduced Planck's constant, Js\n", + "s = 0.01; # Side of the box, m\n", + "E = 2; # Energy range of the electron in the box, eV\n", + "V = s**3; # Volume of the box, metre cube\n", + "I = I = 2*E**(3.0/2)/3; # Definite integral over E : I = 2*E**(3/2)/3\n", + "D_E = V/(2*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\n", + "print\"The density of states for the electron in a cubical box =\",\"{0:.3e}\".format(D_E),\"states\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The density of states for the electron in a cubical box = 1.280e+22 states\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.10a,Page number 141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "E_F = 1; # For simplicity assume Fermi energy to be unity, eV\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "dE = 0.5; # Exces energy above Fermi level, eV\n", + "T = 300; # Room temperature, K\n", + "E = E_F + dE; # Energy of the level above Fermi level, eV\n", + "f_E = 1./(exp((E-E_F)*e/(k*T))+1); # Occupation probability of the electron at 0.1 eV above E_F\n", + "print\"At 300 K:\";\n", + "print\"=========\";\n", + "print\"The occupation probability of electron at\",dE,\"eV above Fermi energy =\",\"{0:.3e}\".format(f_E);\n", + "E = E_F - dE; # Energy of the level below Fermi level, eV\n", + "f_E = 1.0/(exp((E-E_F)*e/(k*T))+1); # Occupation probability of the electron at 0.1 eV below E_F\n", + "print\"The occupation probability of electron at\",dE,\"eV above Fermi energy =\",\"{0:.3e}\".format(f_E);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At 300 K:\n", + "=========\n", + "The occupation probability of electron at 0.5 eV above Fermi energy = 4.054e-09\n", + "The occupation probability of electron at 0.5 eV above Fermi energy = 1.000e+00\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.11a,Page number 141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "E_F = 1; # For simplicity assume Fermi energy to be unity, eV\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "dE = 0.2; # Exces energy above Fermi level, eV\n", + "T = 0+273; # Room temperature, K\n", + "E = E_F + dE; # Energy of the level above Fermi level, eV\n", + "f_E = 1.0/(exp((E-E_F)*e/(k*T))+1); # Occupation probability of the electron at 0.1 eV above E_F\n", + "print\"At 273 K:\";\n", + "print\"=========\";\n", + "print\"The occupation probability of electron at\",dE,\"eV above Fermi energy =\",\"{0:.3e}\".format(f_E);\n", + "T = 100+273; # Given temperature of 100 degree celsius, K\n", + "f_E = 1.0/(exp((E-E_F)*e/(k*T))+1); # Occupation probability of the electron at 0.1 eV below E_F\n", + "print\"At 373 K:\";\n", + "print\"=========\";\n", + "print\"The occupation probability of electron at\",dE,\"eV above Fermi energy =\",\"{0:.3e}\".format(f_E);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At 273 K:\n", + "=========\n", + "The occupation probability of electron at 0.2 eV above Fermi energy = 2.047e-04\n", + "At 373 K:\n", + "=========\n", + "The occupation probability of electron at 0.2 eV above Fermi energy = 1.992e-03\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.12a,Page number 142" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.1*10**-31; # Mass of the electron, kg\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "r = 1.28*10**-10; # Atomic radius of Cu, m\n", + "a = 4*r/sqrt(2); # Lattice constant of Cu, m\n", + "tau = 2.7*10**-14; # Relaxation time for the electron in Cu, s\n", + "V = a**3; # Volume of the cell, metre cube\n", + "n = 4.0/V; # Concentration of free electrons in monovalent copper, \n", + "sigma = n*e**2*tau/m; # Electrical conductivity of monovalent copper, mho per m\n", + "print\"The electrical conductivity of monovalent copper =\",\"{0:.3e}\".format( sigma/100),\"mho per cm\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electrical conductivity of monovalent copper = 6.403e+05 mho per cm\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.13a,Page number 142" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "n = 18.1*10**22; # Number of electrons per unit volume, per cm cube\n", + "N = n/2; # Pauli's principle for number of energy levels, per cm cube\n", + "E_F = 11.58; # Fermi energy of Al, eV\n", + "E = E_F/N; # Interelectronic energy separation between bands of Al, eV\n", + "print\"The interelectronic energy separation between bands of Al =\",\"{0:.3e}\".format(E),\"eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The interelectronic energy separation between bands of Al = 1.280e-22 eV\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.14a,Page number 142" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.1*10**-31; # Mass of the electron, kg\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "h_cross = h/(2*pi); # Reduced Planck's constant, Js\n", + "E_F = 7; # Fermi energy of Cu, eV\n", + "V = 10**-6; # Volume of the cubic metal, metre cube\n", + "D_EF = V/(2*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\n", + "print\"The density of states in Cu contained in cubic metal =\",\"{0:.3e}\".format(D_EF),\"states/eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The density of states in Cu contained in cubic metal = 1.796e+22 states/eV\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.15a,Page number 143" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.1*10**-31; # Mass of the electron, kg\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "h_cross = h/(2*pi); # Reduced Planck's constant, Js\n", + "E_F = 7; # Fermi energy of Cu, eV\n", + "V = 10**-6; # Volume of the cubic metal, metre cube\n", + "D_EF = V/(2*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\n", + "d = 1.0/(D_EF); # Electronic energy level spacing between successive levels of Cu, eV\n", + "print\"The electronic energy level spacing between successive levels of Cu =\",\"{0:.3e}\".format(d),\"eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electronic energy level spacing between successive levels of Cu = 5.568e-23 eV\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.16a,Page number 143" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "A = [[1,2],[3,4],[5,6],[7,8]]; # Declare a 4X2 matrix \n", + "A[0][0] = 'Li'; \n", + "A[0][1] = -0.4039; # Energy of outermost atomic orbital of Li, Rydberg unit\n", + "A[1][0] = 'Na'; # \n", + "A[1][1] = -0.3777; # Energy of outermost atomic orbital of Na, Rydberg unit\n", + "A[2][0] = 'F'; # \n", + "A[2][1] = -1.2502; # Energy of outermost atomic orbital of F, Rydberg unit\n", + "A[3][0] = 'Cl'; # \n", + "A[3][1] = -0.9067; # Energy of outermost atomic orbital of Cl, Rydberg unit\n", + "cf = 13.6; # Conversion factor for Rydberg to eV\n", + "print\"________________________________________\";\n", + "print\" Atom Energy gap\";\n", + "print\"\",A[1][0],A[3][0],\" \",(A[1][1]-A[3][1])*cf,\"eV\";\n", + "print\"\",A[1][0],A[2][0],\" \",(A[1][1]-A[2][1])*cf,\"eV\";\n", + "print\"\",A[0][0],A[2][0],\" \",(A[0][1]-A[2][1])*cf,\"eV\";\n", + "\n", + "print\"________________________________________\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "________________________________________\n", + " Atom Energy gap\n", + " Na Cl 7.1944 eV\n", + " Na F 11.866 eV\n", + " Li F 11.50968 eV\n", + "________________________________________\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.18a,Page number 144" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "# For Cu\n", + "rs_a0_ratio = 2.67; # Ratio of solid radius to the lattice parameter\n", + "E_F = 50.1*(rs_a0_ratio)**(-2); # Fermi level energy of Cu, eV\n", + "T_F = 58.2*10**4*(rs_a0_ratio)**(-2); # Fermi level temperature of Cu, K\n", + "V_F = 4.20*10**8*(rs_a0_ratio)**(-1); # Fermi level velocity of electron in Cu, cm/sec\n", + "K_F = 3.63*10**8*(rs_a0_ratio)**(-1); \n", + "print\"For Cu :\";\n", + "print\"========\";\n", + "print\"E_F =\",round(E_F,3),\"eV\";\n", + "print\"T_F =\",\"{0:.3e}\".format(T_F),\"K\";\n", + "print\"V_F =\",\"{0:.3e}\".format(V_F),\"cm/sec\";\n", + "print\"K_F =\",\"{0:.3e}\".format(K_F),\"per cm\";\n", + "rs_a0_ratio = 3.07; # Ratio of solid radius to the lattice parameter\n", + "E_F = 50.1*(rs_a0_ratio)**(-2); # Fermi level energy of Nb, eV\n", + "T_F = 58.2*10**4*(rs_a0_ratio)**(-2); # Fermi level temperature of Nb, K\n", + "V_F = 4.20*10**8*(rs_a0_ratio)**(-1); # Fermi level velocity of electron in Nb, cm/sec\n", + "K_F = 3.63*10**8*(rs_a0_ratio)**(-1); \n", + "print\"For Nb :\";\n", + "print\"========\";\n", + "print\"E_F =\",round(E_F,3),\"eV\";\n", + "print\"T_F =\",\"{0:.3e}\".format(T_F),\"K\";\n", + "print\"V_F =\",\"{0:.3e}\".format(V_F),\"cm/sec\";\n", + "print\"K_F =\",\"{0:.3e}\".format(K_F),\"per cm\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For Cu :\n", + "========\n", + "E_F = 7.028 eV\n", + "T_F = 8.164e+04 K\n", + "V_F = 1.573e+08 cm/sec\n", + "K_F = 1.360e+08 per cm\n", + "For Nb :\n", + "========\n", + "E_F = 5.316 eV\n", + "T_F = 6.175e+04 K\n", + "V_F = 1.368e+08 cm/sec\n", + "K_F = 1.182e+08 per cm\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter5.ipynb b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter5.ipynb new file mode 100644 index 00000000..1d0fd9c5 --- /dev/null +++ b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter5.ipynb @@ -0,0 +1,302 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:5ddfc271fdce9ee0a4003a1cc58d7533d4c666ac6abbe00e85967c1976a0ff5f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5: Band Theory of Solids" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1,Page number 176" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "h_bar = h/(2*pi); # Reduced Planck's constant, Js\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "\n", + "# For Na\n", + "n_Na = 2.65*10**28; # electronic concentration of Na, per metre cube\n", + "k_F = (3*pi**2*n_Na)**(1.0/3); # Fermi wave vector, per cm\n", + "E_F = h_bar**2*k_F**2/(2*m*e); # Fermi energy of Na, eV\n", + "print\"The fermi energy of Na = \",round(E_F,4),\"eV\";\n", + "print\"The band structure value of Na = \",0.263*13.6,\"eV\";\n", + "\n", + "# For K\n", + "n_K = 1.4e+28; # electronic concentration of K, per metre cube\n", + "k_F = (3*pi**2*n_K)**(1.0/3); # Fermi wave vector, per cm\n", + "E_F = h_bar**2*k_F**2/(2*m*e); # Fermi energy of K, eV\n", + "print\"The fermi energy of K = \",round(E_F,4),\"eV\";\n", + "print\"The band structure value of K = \", 0.164*13.6,\"eV\";\n", + "print\"The agreement between the free electron and band theoretical values are fairly good both for Na and K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The fermi energy of Na = 3.2489 eV\n", + "The band structure value of Na = 3.5768 eV\n", + "The fermi energy of K = 2.1232 eV\n", + "The band structure value of K = 2.2304 eV\n", + "The agreement between the free electron and band theoretical values are fairly good both for Na and K\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3,Page number 177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "n_Na = 2.65*10**22; # electronic concentration of Na, per cm cube\n", + "k_F = (3*pi**2*n_Na)**(1.0/3); # Fermi wave vector, per cm\n", + "print\"The fermi momentum of Na =\",\"{0:.3e}\".format(k_F),\"per cm\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The fermi momentum of Na = 9.223e+07 per cm\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5,Page number 177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "h_bar = h/(2*pi); # Reduced Planck's constant, Js\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "V = 1.0*10**-6; # Volume of unit cube of material, metre cube\n", + "\n", + "# For Mg\n", + "E_F = 7.13*e; # Fermi energy of Mg, J\n", + "s = 2*pi**2/(e*V)*(h_bar**2/(2*m))**(3.0/2)*(E_F)**(-1.0/2); # Energy separation between levels for Mg, eV\n", + "print\"The energy separation between adjacent levels for Mg = \",\"{0:.3e}\".format(s),\"eV\";\n", + "\n", + "# For Cs\n", + "E_F = 1.58*e; # Fermi energy of Cs, J\n", + "s = 2*pi**2/(e*V)*(h_bar**2/(2*m))**(3.0/2)*(E_F)**(-1.0/2); # Energy separation between levels for Cs, eV\n", + "print\"The energy separation between adjacent levels for Cs =\",\"{0:.3e}\".format(s),\"eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy separation between adjacent levels for Mg = 5.517e-23 eV\n", + "The energy separation between adjacent levels for Cs = 1.172e-22 eV\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.9,Page number 180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "gamma_expt = 7.0*10**-4; # Experimental value of electronic specific heat, cal/mol/K-square\n", + "gamma_theory = 3.6*10**-4; # Theoretical value of electronic specific heat, cal/mol/K-square\n", + "L = (gamma_expt - gamma_theory)/gamma_theory;\n", + "print\"The electron-phonon coupling constant of superconductor = \",round(L,2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electron-phonon coupling constant of superconductor = 0.94\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.10,Page number 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given Data\n", + "\n", + "N_Ef = 1.235; # Density of states at fermi energy, electrons/atom-eV\n", + "N = 6.023*10**23; # Avogadro's number\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "e = 1.6*10**-19; # Charge on an electron, C\n", + "gama = pi**2*k**2/3*(N_Ef*N/e); # Electronic specific heat coefficient, J/g-atom-kelvin square\n", + "\n", + "print\"The electronic specific heat coefficient of superconductor = \",round(gama*1000,4),\"mJ/g-atom-kelvin square\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electronic specific heat coefficient of superconductor = 2.9127 mJ/g-atom-kelvin square\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.11,Page number 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "gamma_expt = 4.84; # Experimental value of electronic specific heat of metal, mJ/g-atom/K-square\n", + "gamma_theory = 2.991; # Theoretical value of electronic specific heat of metal, mJ/g-atom/K-square\n", + "L = (gamma_expt-gamma_theory)/gamma_theory;\n", + "print\"The electron-phonon coupling constant for metal = \",round(L,4);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electron-phonon coupling constant for metal = 0.6182\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.12,Page number 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given data\n", + "mu_B = 9.24*10**-27; # Bohr's magneton, J/T\n", + "N_Ef = 0.826; # Density of states at fermi energy, electrons/atom-eV\n", + "N = 6.023*10**23; # Avogadro's number\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J\n", + "chi_Pauli = mu_B**2*N_Ef*N/e;\n", + "print\"Pauli spin susceptibility of Mg = \",\"{0:.3e}\".format( chi_Pauli*1000),\"cgs units\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pauli spin susceptibility of Mg = 2.655e-07 cgs units\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter6.ipynb b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter6.ipynb new file mode 100644 index 00000000..634521ef --- /dev/null +++ b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter6.ipynb @@ -0,0 +1,714 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a712c503532bf01630566d0248f8b9fe4ae07a136310d2bc72cf3a40e429d9f4" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6: Semiconductor Physics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1,Page number 190" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "S = [[1,2],[3,4],[5,6],[7,8]]; # Declare a 4X2 matrix\n", + "# Enter material names\n", + "S[0][0] = 'Si'; S[1][0] = 'GaAs'; S[2][0] = 'GaP'; S[3][0] = 'ZnS';\n", + "# Enter energy band gap values\n", + "S[0][1] = 1.11; S[1][1] = 1.42; S[2][1] = 2.26; S[3][1] = 3.60;\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "c = 3*10**8; # Speed of light, m/s\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "print\"______________________________________________________\";\n", + "print\"Material E_g (eV) Critical Wavelength (micron)\";\n", + "print\"______________________________________________________\";\n", + "for i in range (0,4) :\n", + " lamda = h*c/(S[i][1]*e);\n", + " print\"\", S[i][0],\" \", S[i][1],\" \",round(lamda/10**-6,3);\n", + "\n", + "print\"______________________________________________________\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "______________________________________________________\n", + "Material E_g (eV) Critical Wavelength (micron)\n", + "______________________________________________________\n", + " Si 1.11 1.119\n", + " GaAs 1.42 0.875\n", + " GaP 2.26 0.55\n", + " ZnS 3.6 0.345\n", + "______________________________________________________\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2,Page number 192" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "c = 3*10**8; # Speed of light, m/s\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "omega = 2e+014; # Wave vector involved in phonon energy, rad per sec\n", + "f = omega/(2*pi); # Frequency of the wave, Hz \n", + "E = h*f/e; # Phonon energy involved in Si to lift the electron, eV\n", + "print\"The phonon energy involved in Si =\",round(E,4),\"eV which is insufficient to lift an electron.\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The phonon energy involved in Si = 0.1318 eV which is insufficient to lift an electron.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3,Page number 192" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "N_A = 6.023*10**23; # Avogadro's number\n", + "# For Si\n", + "A = 28.1; # Atomic weight of Si, g/mol\n", + "a = 5.43*10**-8; # Lattice constant for Si, cm\n", + "n = 8.0/a**3; # Number of atoms per unit volume, atoms/cc\n", + "rho = n*A/N_A; # Density of Si, g/cc\n", + "print\"The density of Si =\",round(rho,3),\"atoms per cc\";\n", + "# For GaAs\n", + "A = 69.7+74.9; # Atomic weight of GaAs, g/mol\n", + "a = 5.65*10**-8; # Lattice constant for Si, cm\n", + "n = 4.0/a**3; # Number of atoms per unit volume, atoms/cc\n", + "rho = n*A/N_A; # Density of GaAs, g/cc\n", + "print\"The density of GaAs =\",round(rho,3),\"toms per cc\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The density of Si = 2.331 atoms per cc\n", + "The density of GaAs = 5.324 toms per cc\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4,Page number 196" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.11*10**-31; # Electron Rest Mass , kg\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "T = 300.0; # Room temperature, K\n", + "m_e = 0.068*m; # Mass of electron, kg\n", + "m_h = 0.56*m; # Mass of hole, kg\n", + "E_g = 1.42*1.6*10**-19; # Energy band gap for GaAs, J\n", + "n_i = 2*(2*pi*k*T/h**2)**(3.0/2)*(m_e*m_h)**(3.0/4)*exp(-E_g/(2*k*T));\n", + "print\"The Intrinsic carrier concentration of GaAs at 300 K =\",\"{0:.3e}\".format(n_i),\"per metre cube\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Intrinsic carrier concentration of GaAs at 300 K = 2.618e+12 per metre cube\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5,Page number 197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.11*10**-31; # Electron Rest Mass , kg\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "T = 300.0; # Room temperature, K\n", + "m_e = 1.1*m; # Mass of electron, kg\n", + "m_h = 0.56*m; # Mass of hole, kg\n", + "E_g = 1.1; # Energy band gap for GaAs, J\n", + "E_F = E_g/2+3.0/4*k*T/e*log(m_h/m_e); # Position of Fermi level of Si at room temperature, eV\n", + "print\"The position of Fermi level of Si at room temperature =\",round(E_F,3),\"eV\";\n", + "print\"The fermi level in this case is shifted downward from the midpoint (0.55 eV) in the forbiddem gap.\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The position of Fermi level of Si at room temperature = 0.537 eV\n", + "The fermi level in this case is shifted downward from the midpoint (0.55 eV) in the forbiddem gap.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6,Page number 197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Electronic charge, C\n", + "n_i = 2.15*10**13; # Carrier density of Ge at room temperature, per cc\n", + "mu_e = 3900.0; # Mobility of electron, cm-square/V-s\n", + "mu_h = 1900.0; # Mobility of hole, cm-square/V-s\n", + "sigma_i = e*(mu_e+mu_h)*n_i; # Intrinsic conductivity of Ge, mho per m\n", + "rho_i = 1.0/sigma_i; # Intrinsic resistivity of Ge at room temperature, ohm-m\n", + "print\"The intrinsic resistivity of Ge at room temperature =\",round(rho_i,2),\"ohm-cm\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The intrinsic resistivity of Ge at room temperature = 50.12 ohm-cm\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7,Page number 197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "e = 1.6*10**-19; # Electronic charge, C\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "T = 30.0; # Given temperature, K\n", + "n = 10**22; # Carrier density of CdS, per metre cube\n", + "mu = 10**-2; # Mobility of electron, metre-square/V-s\n", + "sigma = e*mu*n; # Conductivity of CdS, mho per m\n", + "print\"The conductivity of CdS sample =\",round(sigma,2),\"mho per m\";\n", + "m_eff = 0.1*m; # Effective mass of the charge carries, kg\n", + "t = m_eff*sigma/(n*e**2); # Average time between successive collisions, s\n", + "print\"The average time between successive collisions =\",\"{0:.3e}\".format(t),\"sec\";\n", + "# We have 1/2*m_eff*v**2 = 3/2*k*T, solving for v\n", + "v = sqrt(3*k*T/m_eff); # Velocity of charrge carriers, m/s\n", + "l = v*t; # Mean free distance travelled by the carrier, m\n", + "print\"The mean free distance travelled by the carrier =\",\"{0:.3e}\".format(l),\"m\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The conductivity of CdS sample = 16.0 mho per m\n", + "The average time between successive collisions = 5.688e-15 sec\n", + "The mean free distance travelled by the carrier = 6.644e-10 m\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8,Page number 199" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "T = [385.0, 455.0, 556.0, 714.0]; # Temperatures of Ge, K\n", + "rho = [0.028, 0.0061, 0.0013, 0.000274]; # Electrical resistivity, ohm-m\n", + "Tinv = [0.0, 0.0, 0.0, 0.0]; # Create an empty row matrix for 1/T\n", + "ln_sigma = [0.0, 0.0, 0.0, 0.0]; # Create the empty row matrix for log(sigma)\n", + "for i in xrange(len(T)):\n", + " Tinv[i] = 1/T[i];\n", + " ln_sigma[i] = log(1.0/rho[i]);\n", + "# Plot the graph\n", + "plot(Tinv, ln_sigma);\n", + "axis([0,0.003,0,9])\n", + "title('Plot of ln (sigma) vs 1/T');\n", + "xlabel('1/T');\n", + "ylabel('ln (sigma)');\n", + "show();\n", + "\n", + "\n", + "# Calculate slope\n", + "slope = (ln_sigma[2]-ln_sigma[1])/(Tinv[2]-Tinv[1]);\n", + "E_g = abs(2*slope*k); # Energy gap of Ge, J\n", + "print\"The energy gap of Ge =\",E_g/e,\"eV\";\n", + "\n", + "# Result \n", + "# The energy gap of Ge = 0.658 eV " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, + "output_type": "display_data", + "png": 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2lYWFBX788Ufk5+dj/PjxOHPmDPz8/PTv3z3uoNVqodVq5U6JiMik6HQ66HQ6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\nGIiISOL/AVf2bVIyj9ReAAAAAElFTkSuQmCC\n", + "text": [ + "<matplotlib.figure.Figure at 0x7fe3a9ddb6d0>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy gap of Ge = 0.667947295491 eV\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9,Page number 199" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "c = 3*10**8; # Speed of light, m/s\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "x = 0.07; # Al concentration in host GaAs\n", + "E_g = 1.424 + 1.266*x + 0.266*x**2; # Band gap of GaAs as a function of x, eV\n", + "# As E_g = h*c/lambda, solving for lambda\n", + "lamda = h*c/(E_g*e); # Emission wavelength of light, m\n", + "print\"The energy band gap of Al doped GaAs =\",round(E_g,3),\"eV\";\n", + "print\"The emission wavelength of light =\",round(lamda*10**6,3),\"micron\";\n", + "print\"The Al atoms go as substitutional impurity in the host material.\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy band gap of Al doped GaAs = 1.514 eV\n", + "The emission wavelength of light = 0.821 micron\n", + "The Al atoms go as substitutional impurity in the host material.\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10,Page number 200" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "x = 0.38; # Al concentration in host GaAs\n", + "E_g = 1.424 + 1.266*x + 0.266*x**2; # Band gap of GaAs as a function of x, eV\n", + "print\"The energy band gap of 38 percent Al doped in GaAs =\",round(E_g,3),\"eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy band gap of 38 percent Al doped in GaAs = 1.943 eV\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.11,Page number 200" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "rho_40 = 0.2; # Resistivity of Ge at 40 degree celsius, ohm-m\n", + "T1 = 40+273; # Temperature at which resistivity of Ge becomes 0.2 ohm-m, K\n", + "T2 = 20+273; # Temperature at which resistivity of Ge is to be calculated, K\n", + "E_g = 0.7; # Band gap of Ge, eV\n", + "# As rho = exp(E_g/(2*k*T)), so for rho_20\n", + "rho_20 = rho_40*exp(E_g/(2*k/e)*(1.0/T2-1.0/T1)); # Resistivity of Ge at 20 degree celsius, ohm-m\n", + "print\"The resistivity of Ge at 20 degree celsius =\",round(rho_20,1),\"ohm-m\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistivity of Ge at 20 degree celsius = 0.5 ohm-m\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.12,Page number 203" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "T = 300.0; # Room temperature of the material, K\n", + "K_Si = 11.7; # Dielectric constant of Si\n", + "K_Ge = 15.8; # Dielectric constant of Ge\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "m_eff = 0.2; # Effective masses of the electron in both Si and Ge, kg\n", + "E_ion_Si = 13.6*m_eff/K_Si**2; # Donor ionization energy of Si, eV\n", + "E_ion_Ge = 13.6*m_eff/K_Ge**2; # Donor ionization energy of Ge, eV\n", + "E = k*T/e; # Energy available for electrons at 300 K, eV\n", + "print\"The donor ionization energy of Si =\",round(E_ion_Si,4),\"eV\";\n", + "print\"The donor ionization energy of Ge =\",round(E_ion_Ge,4),\"eV\";\n", + "print\"The energy available for electrons at 300 K =\",round(E,4),\"eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The donor ionization energy of Si = 0.0199 eV\n", + "The donor ionization energy of Ge = 0.0109 eV\n", + "The energy available for electrons at 300 K = 0.0259 eV\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.13,Page number 203" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "epsilon = 15.8; # Dielectric constant of Ge \n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "m_e = 0.2*m; # Effective masses of the electron in Ge, kg\n", + "a_Ge = 5.65; # Lattice parameter of Ge, angstrom\n", + "A_d = 0.53*epsilon*(m/m_e); # Radius of donor atom, angstrom\n", + "print\"The radius of the orbits of fifth valence electron of acceptor impurity =\",ceil(A_d),\"angstrom\";\n", + "print\"This radius is\",ceil(A_d/a_Ge),\"times the lattice constant of Ge\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The radius of the orbits of fifth valence electron of acceptor impurity = 42.0 angstrom\n", + "This radius is 8.0 times the lattice constant of Ge\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.14,Page number 203" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "tau = 10**-12; # Life time of electron in Ge, s\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "m_e = 0.5*m; # Effective masses of the electron in Ge, kg\n", + "mu = e*tau/m_e; # Mobility of electron in Ge, m-square/V-s\n", + "n_i = 2.5*10**19; # Intrinsic carrier concentration of Ge at room temperature, per metre cube\n", + "n_Ge = 5*10**28; # Concentration of Ge atoms, per metre cube\n", + "n_e = n_Ge/10**6; # Concentration of impurity atoms, per metre cube\n", + "# From law of mass action, n_e*n_h = n_i**2, solving for n_h\n", + "n_h = n_i**2/n_e; # Concentration of holes, per metre cube\n", + "\n", + "print\"This mobility of electron in Ge =\",round(mu/10**-4,1),\"cm-square/V-s\";\n", + "print\"This concentration of holes in Ge =\",\"{0:.3e}\".format(n_h),\"per metre cube\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "This mobility of electron in Ge = 3516.5 cm-square/V-s\n", + "This concentration of holes in Ge = 1.250e+16 per metre cube\n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.15,Page number 204" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "n_i = 2.5*10**19; # Intrinsic carrier concentration of Ge at room temperature, per metre cube\n", + "n_Ge = 5*10**28; # Concentration of Ge atoms, per metre cube\n", + "delta_d = 10**6; # Rate at which pentavalent impurity is doped in pure Ge, ppm\n", + "n_e = n_Ge/delta_d; # Concentration of impurity atoms, per metre cube\n", + "# From law of mass action, n_e*n_h = n_i**2, solving for n_h\n", + "n_h = n_i**2/n_e; # Concentration of holes, per metre cube\n", + "\n", + "print\"This concentration of holes in Ge =\",\"{0:.3e}\".format(n_h),\"per metre cube\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "This concentration of holes in Ge = 1.250e+16 per metre cube\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.16,Page number 205" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Charge on an electron, C\n", + "mu = 1400*10**-4; # Mobility of electron, metre-square per volt per sec\n", + "l = 300-6; # Length of the n-type semiconductor, m\n", + "w = 100-6; # Width of the n-type semiconductor, m\n", + "t = 20-6; # Thickness of the n-type semiconductor, m\n", + "N_D = 4.5*10**21; # Doping concentration of donor impurities, per metre-cube\n", + "V = 10; # Biasing voltage for semiconductor, V\n", + "B_prep = 1; # Perpendicular magnetic field to which the semiconductor is subjected, tesla\n", + "\n", + "# Part (a)\n", + "n = N_D; # Electron concentration in semiconductor, per cc\n", + "R_H = -1.0/(n*e); # Hall Co-efficient, per C per metre cube\n", + "\n", + "# Part (b)\n", + "rho = 1.0/(n*e*mu); # Resistivity of semiconductor, ohm-m\n", + "R = rho*l/(w*t); # Resistance of the semiconductor, ohm\n", + "I = V/R; # Current through the semiconductor, A\n", + "V_H = R_H*I*B_prep/t; # Hall voltage, V\n", + "\n", + "# Part (c)\n", + "theta_H = math.degrees(math.atan(-mu*B_prep)); # Hall angle, degrees\n", + "\n", + "\n", + "print\"Hall coefficient, R_H =\",\"{0:.3e}\".format(R_H),\"per C metre cube\";\n", + "print\"Hall voltage, V_H = \",math.fabs(V_H),\"V\";\n", + "print\"Hall angle, theta_H =\",round(theta_H,3),\"degree\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hall coefficient, R_H = -1.389e-03 per C metre cube\n", + "Hall voltage, V_H = 0.447619047619 V\n", + "Hall angle, theta_H = -7.97 degree\n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter8.ipynb b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter8.ipynb new file mode 100644 index 00000000..228490a2 --- /dev/null +++ b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter8.ipynb @@ -0,0 +1,621 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e91aa877d853d74b42c0663ae5d9a2e258f4239791004d61b638bb70dcb448fc" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8: Specific Heat of Solids" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.1,Page number 241" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "rho = 7.9*10**3; # Density of iron, kg per cubic meter\n", + "A = 56*10**-3; # Atomic weight of iron, g/mol\n", + "N_A = 6.02*10**23; # Avogadro's number, atoms per mole\n", + "mu_B = 9.3*10**-24; # Bohr magneton; # Ampere meter square\n", + "n = rho*N_A/A; # Total number of atoms per unit cell, per cubic meter\n", + "M = 2.2*n*mu_B; # Spontaneous magnetization of iron, Ampere per meter\n", + "print\"Spontaneous magnetization of iron =\",\"{0:.3e}\".format(M),\"Ampere per meter\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Spontaneous magnetization of iron = 1.738e+06 Ampere per meter\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.2,Page number 241" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "n = 3*10**28; # Spin density of electrons in a ferromagnetic material, per cubic meter\n", + "mu = 3*10**-23; # spin magnetic moment of a ferromagnetic material, Square Ampere \n", + "M_s = n*mu; # Saturation magnetization of a ferromagnetic material, Per Ampere\n", + "print\"Saturation magnetization of a ferromagnetic material =\",\"{0:.3e}\".format(M_s),\"ampere per meter\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Saturation magnetization of a ferromagnetic material = 9.000e+05 ampere per meter\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3,Page number 241" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "h_bar = 6.58*10**-16; # Planck's constant, eV.s\n", + "m = 0.511*10**6; # Mass of an electron, eV\n", + "e = 1.6*10**-12; # Energy equivalent of 1 eV, erg/eV\n", + "c = 3.0*10**10; # Speed of light, cm/s\n", + "N = 4.7*10**22; # Free electron gas concentration of Lithium, per cubic cm\n", + "mu_B = 9.27*10**-21; # Bohr magneton, Ampere cm-square\n", + "E_F = (h_bar*c)**2/(2*m)*(3*pi**2*N)**(2.0/3); # Fermi energy, eV\n", + "chi = 3*N*mu_B**2/(2*E_F*e); # Magnetic susceptibility of Lithium, cgs units\n", + "print\"Magnetic susceptibility of Lithium =\",\"{0:.3e}\".format(chi),\"cgs units\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetic susceptibility of Lithium = 7.967e-07 cgs units\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4,Page number 241" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "a_B = 0.53*10**-8; # Bohr radius, cm\n", + "N = 27*10**23; # Atomic density of He gas, per cubic cm\n", + "c = 3*10**10; # Speed of light, cm/sec\n", + "e = 1.6*10**-19; # Charge of an electron, Coulomb\n", + "m = 9.1*10**-28; # Mass of an electron, g\n", + "# As r_classic = e**2/(m*c**2), Classical radius of an electron\n", + "r_classic = 2.8*10**-13; # Classical radius of the electron, cm \n", + "chi = -2*N*r_classic/6*a_B**2; # Magnetic susceptibility of Helium, cgs units\n", + "\n", + "print\"Diamagnetic susceptibility of helium atom in ground state =\",\"{0:.3e}\".format(chi),\"emu\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diamagnetic susceptibility of helium atom in ground state = -7.079e-06 emu\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.5,Page number 242" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "chiA_He = 1.9*10**-6; # Atomic susceptibility of helium, cm cube per mole\n", + "chiA_Cu = 18*10**-6; # Atomic susceptibility of Copper, cm cube per mole\n", + "Q_sp = 1.77*10**7; # Specific charge of an electron, emu\n", + "Ne = 9650.0; # Charge of a gram ion, emu\n", + "Z_He = 2.0; # Atomic number of helium atom\n", + "Z_Cu = 29.0; # Atomic number of copper atom\n", + "R_He = sqrt(abs(-6*chiA_He/(Ne*Z_He*Q_sp))); # Magnetic susceptibility of helium atom, cgs units\n", + "R_Cu = sqrt(abs(-6*chiA_Cu/(Ne*Z_Cu*Q_sp))); # Magnetic susceptibility of copper atom, cgs units\n", + "print\"Atomic radius of helium =\",\"{0:.3e}\".format(R_He),\"cm\";\n", + "print\"Atomic radius of copper =\",\"{0:.3e}\".format(R_Cu),\"cm\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Atomic radius of helium = 5.777e-09 cm\n", + "Atomic radius of copper = 4.669e-09 cm\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.6,Page number 242" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "N = 6.039*10**22; # Atomic density of Neon gas, per cubic cm \n", + "# As r_classic = e**2/(m*c**2), Classical radius of an electron\n", + "r_classic = 2.8*10**-13; # Classical radius of the electron, cm\n", + "Z = 10.0; # Atomic number of helium atom\n", + "a0 = 0.53*10**-8; # Bohr's radius, cm\n", + "n1 = 2; n2 = 2; n3 = 6; # Occupation numbers for 1s, 2s and 2p states of Ne\n", + "r_sq_1s = 0.031; # Expectation value for 1s state\n", + "r_sq_2s = 0.905; # Expectation value for 2s state\n", + "r_sq_2p = 1.126; # Expectation value for 2p state \n", + "mean_r_sq = n1*r_sq_1s + n2*r_sq_2s + n3*r_sq_2p; # Mean square radius, cm-square\n", + "Chi_A = -1.0/6*N*Z*r_classic*mean_r_sq*a0**2; # Magnetic susceptibility of helium atom, cgs units\n", + "print\"Atomic susceptibility of Ne atom =\",\"{0:.3e}\".format(Chi_A),\"emu/mole\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Atomic susceptibility of Ne atom = -6.830e-06 emu/mole\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.7,Page number 249" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "h = 6.626*10**-34; # Planck's constant, Js\n", + "h_cross = h/(2*pi); # Reduced Planck's constant, Js\n", + "m = 9.1*10**-31; # Mass of an electron, kg\n", + "mu = e*h_cross/(2*m); # Bohr magneton, J/T\n", + "mu_H = mu/e; # Magnetic energy, eV\n", + "kT = 0.025; # Energy associated with two degrees of freedom, eV\n", + "E_ratio = mu_H/kT; # Exceptional terms in Langevin's function\n", + "print\"The magnitude of mu*H/(k*T) =\",\"{0:.3e}\".format(E_ratio);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The magnitude of mu*H/(k*T) = 2.318e-03\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.8,Page number 249" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "mu = 5.78*10**-5; # Bohr magneton, eV/T\n", + "NE_F = 0.826; # Density of states at fermi level, electrons/atom-J\n", + "chi_Pauli = mu**2*NE_F/10**-4; # Pauli diamagnetism, cgs units\n", + "chi_Core = -4.2*10**-6; # Core diamagnetism, cgs units\n", + "chi_Landau = -1.0/3*chi_Pauli; # Landau diamagnetism, cgs units\n", + "chi_Total = chi_Core+ chi_Pauli+chi_Landau; # Paramagnetic susceptibility of Mg, cgs units\n", + "\n", + "print\"The paramagnetic susceptibility of Mg =\",\"{0:.3e}\".format(chi_Total),\"cgs units\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The paramagnetic susceptibility of Mg = 1.420e-05 cgs units\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.9,Page number 250" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "mu = 9.29*10**-24; # Bohr magneton, J/T\n", + "mu_0 = 1.26*10**-6; # Permeability of free space, Sq. tesla cubic meter per joule\n", + "E_F= 11.63*e; # Fermi energy, J\n", + "N = 6.02*10**28; # Atomic concentration, atoms per cubic meter \n", + "chi_Total = 2.2*10**-5; # Paramagnetic susceptibility of Mg, S.I. units\n", + "chi_Pauli = 3*N*mu**2*mu_0/(2*E_F); # Pauli diamagnetism, S.I. units\n", + "chi_dia = chi_Total - chi_Pauli; # Diamagnetic contribution to magnetic susceptibility\n", + "\n", + "print\"The Pauli spin susceptibility of Al =\",\"{0:.3e}\".format(chi_Pauli),\"S.I. units\";\n", + "print\"The diamagnetic contribution to magnetic susceptibility of Al =\",\"{0:.3e}\".format(chi_dia),\"S.I. units\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Pauli spin susceptibility of Al = 5.277e-06 S.I. units\n", + "The diamagnetic contribution to magnetic susceptibility of Al = 1.672e-05 S.I. units\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.10,Page number 250" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "a0 = 5.3; # Bohr radius, nm\n", + "rs_a0_ratio = 3.93; # Ratio of solid radius to the lattice parameter \n", + "chi_Pauli = 2.59/rs_a0_ratio; # Pauli's spin susceptibility, cgs units\n", + "\n", + "print\"The Pauli spin susceptibility for Na in terms of free electron gas parameter =\",round(chi_Pauli,3);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Pauli spin susceptibility for Na in terms of free electron gas parameter = 0.659\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.11,Page number 264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "S = 2; # Spin quantum number\n", + "J = 0; # Total quantum number\n", + "L = 2; # Orbital quantum number\n", + "g = 2; # Lande splitting factor\n", + "print\"The spectroscopic term value of Mn3+ ion =\",2*S+1,\"_D_\",J;\n", + "# For J = L - S\n", + "J = L - S;\n", + "mu_N = g*sqrt(J*(J+1)); # Effective magneton number\n", + "print\"The effective magneton number for J = L - S is\",mu_N;\n", + "# For J = S, L = 0 so that\n", + "L = 0;\n", + "J = L+S;\n", + "mu_N = g*sqrt(J*(J+1)); # Effective magneton number\n", + "print\"The effective magneton number for J = S is\",round(mu_N,2),\"\\nIt is in agreement with the experimental value of 5.0.\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The spectroscopic term value of Mn3+ ion = 5 _D_ 0\n", + "The effective magneton number for J = L - S is 0.0\n", + "The effective magneton number for J = S is 4.9 \n", + "It is in agreement with the experimental value of 5.0.\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.12,Page number 264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "mu = 9.27*10**-24; # Bohr's magneton, J/T\n", + "N_up = 5; # Number of electrons with spin up as per Hunds Rule\n", + "N_down = 1; # Number of electrons with spin down as per Hunds Rule\n", + "M = mu*(N_up-N_down); # Net magnetic moment associated with six electrons in the 3d shell, J/T\n", + " \n", + "print\"The magnetic moment of 3d electrons of Fe using Hunds rule =\",M/mu,\"Bohr magnetons\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The magnetic moment of 3d electrons of Fe using Hunds rule = 4.0 Bohr magnetons\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.13,Page number 264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "C = [[1,2,3,4],[5,6,7,8],[9,10,11,12]];\n", + "# Enter compound names\n", + "C[0][0] = 'LaCrO3';\n", + "C[1][0] = 'LaMnO3';\n", + "C[2][0] = 'LaCoO3';\n", + "# Enter Magnetic moments from Hunds rule\n", + "C[0][1] = 3.0;\n", + "C[1][1] = 4.0;\n", + "C[2][1] = 5.0;\n", + "# Enter Magnetic moments from Band theory\n", + "C[0][2] = 2.82;\n", + "C[1][2] = 3.74;\n", + "C[2][2] = 4.16;\n", + "# Enter Magnetic moments from the Experiment\n", + "C[0][3] = 2.80;\n", + "C[1][3] = 3.90;\n", + "C[2][3] = 4.60;\n", + "print\"__________________________________________________\";\n", + "print\"Compound Magnetic moment per formula unit (in BM) \";\n", + "print\" ________________________________________\";\n", + "print\" Hunds Rule Band Theory Experiment\";\n", + "print\"__________________________________________________\";\n", + "for i in range (0,3) :\n", + " print\"\",C[i][0],\" \",C[i][1],\" \",C[i][2],\" \",C[i][3]\n", + "print\"__________________________________________________\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "__________________________________________________\n", + "Compound Magnetic moment per formula unit (in BM) \n", + " ________________________________________\n", + " Hunds Rule Band Theory Experiment\n", + "__________________________________________________\n", + " LaCrO3 3.0 2.82 2.8\n", + " LaMnO3 4.0 3.74 3.9\n", + " LaCoO3 5.0 4.16 4.6\n", + "__________________________________________________\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.14,Page number 268" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "C = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]];\n", + "# Enter compound names\n", + "C[0][0] = 'LaTiO3';\n", + "C[1][0] = 'LaCrO3';\n", + "C[2][0] = 'LaFeO3';\n", + "C[3][0] = 'LaCoO3';\n", + "# Enter total energy difference w.r.t. ground state for Paramagnetics, mRyd\n", + "C[0][1] = 0.014;\n", + "C[1][1] = 158.3;\n", + "C[2][1] = 20.69;\n", + "C[3][1] = 0.000;\n", + "# Enter total energy difference w.r.t. ground state for Ferromagnetics, mRyd\n", + "C[0][2] = 0.034;\n", + "C[1][2] = 13.99;\n", + "C[2][2] = 0.006;\n", + "C[3][2] = 0.010;\n", + "# Enter total energy difference w.r.t. ground state for Antiferromagnetics, mRyd\n", + "C[0][3] = 0.000;\n", + "C[1][3] = 0.000;\n", + "C[2][3] = 0.000;\n", + "C[3][3] = 0.003;\n", + "print\"______________________________________________________________\";\n", + "print\"Solid Total energy difference (mRyd) (w.r.t. ground state)\";\n", + "print\" ____________________________________________________\";\n", + "print\" Paramagnetic Ferromagnetic Antiferromagnetic \";\n", + "print\"______________________________________________________________\";\n", + "for i in range (0,4) :\n", + " print\"\",C[i][0],\" \",C[i][1],\" \",C[i][2],\" \",C[i][3]\n", + "print\"______________________________________________________________\";\n", + "print\"All the solids given above crystallize in the antiferromagnetic state except that of LaCoO3.\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "______________________________________________________________\n", + "Solid Total energy difference (mRyd) (w.r.t. ground state)\n", + " ____________________________________________________\n", + " Paramagnetic Ferromagnetic Antiferromagnetic \n", + "______________________________________________________________\n", + " LaTiO3 0.014 0.034 0.0\n", + " LaCrO3 158.3 13.99 0.0\n", + " LaFeO3 20.69 0.006 0.0\n", + " LaCoO3 0.0 0.01 0.003\n", + "______________________________________________________________\n", + "All the solids given above crystallize in the antiferromagnetic state except that of LaCoO3.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter9.ipynb b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter9.ipynb new file mode 100644 index 00000000..136ce4e5 --- /dev/null +++ b/Solid_State_Physics_Principles_And_Applications_by_R._Asokamani/chapter9.ipynb @@ -0,0 +1,585 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:b4117c20c65c9d0e0ed117019f9f9b853eec6f9bf1a68fc9915ad4bb5eef3755" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9: Superconductivity" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1,Page number 278" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "H_c0 = 0.0803; # Critical field at absolute zero, Tesla\n", + "T_c = 7.19; # Transition temperature of specimen lead, Kelvin\n", + "T = 5.0; # Temperature at which destruction of superconductivity is to be found, Kelvin\n", + "H_c = H_c0*(1.0-(T/T_c)**2); # Critical field required to destroy superconductivity, Tesla\n", + "print\"Critical field required to destroy superconductivity = \",round(H_c,4),\"T\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical field required to destroy superconductivity = 0.0415 T\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.2,Page number 278" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "H0 = 1970.0; # Critical field at absolute zero, Oe\n", + "T_c = 9.25; # Transition temperature of specimen Nb, Kelvin\n", + "T = 4.0; # Temperature at which destruction of superconductivity is to be found, Kelvin\n", + "H_c = H0*(1.0-(T/T_c)**2); # Limiting magnetic field, Oe\n", + "print\"Limiting magnetic field of Nb to serve as superconductor = \",round(H_c),\"Oe\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Limiting magnetic field of Nb to serve as superconductor = 1602.0 Oe\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.3,Page number 278" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "T_1 = 14.0; # Temperature, K\n", + "T_2 = 13.0; # Temperature, K\n", + "H_c1 = 1.4*10**5; # Critical field at T_1, K\n", + "H_c2 = 4.2*10**5; # Critical field at T_2, K#As H_c1/H_c2 = (T_c**2-T_1**2)/(T_c**2-T_2**2), solving for T_c\n", + "T_c = sqrt((H_c2/H_c1*T_1**2 - T_2**2)/2); # The superconducting transition temperature of a specimen, K\n", + "print\"Transition temperature of a specimen = \",round(T_c,4),\"K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Transition temperature of a specimen = 14.4741 K\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4,Page number 280" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, J/eV\n", + "E_g = 3.4*10**-4; # Energy gap of aluminium, eV\n", + "v_F = 2.02*10**8; # Fermi velocity of aluminium, cm/sec\n", + "h_bar = 1.05*10**-34; # Planck's constant\n", + "L = h_bar*v_F/(2*E_g*e); # Coherence Length of aluminium, cm\n", + "\n", + "print\"The coherence length of aluminium = \",\"{0:.3e}\".format(L),\"cm\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The coherence length of aluminium = 1.949e-04 cm\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.6,Page number 284" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "h = 6.6*10**-34; # Planck's constant, Js\n", + "e = 1.6*10**-19; # Energy eqivalent of 1 eV, eV/J\n", + "k = 0.86*10**-4; # Boltzmann constant, eV/K\n", + "T_c = 0.56; # Critical temperature for superconducting Zr, K\n", + "E_g = 3.52*k*T_c; # Energy gap of aluminium, J\n", + "c = 3*10**8; # Speed of light, m/s\n", + "lamda = h*c/(E_g*e); # Wavelength of photon required to break a Cooper pair, m\n", + "\n", + "print\"The wavelength of photon required to break a Cooper pair = \",\"{0:.3e}\".format(lamda),\"m\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The wavelength of photon required to break a Cooper pair = 7.300e-03 m\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.7,Page number 285" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "Lamda_0 = 390.0; # Penetration depth at absolute zero, angstorm\n", + "T_c = 7.0; # Transition temperature of Pb, K\n", + "T = 2.0; # Givn temperature, K\n", + "Lamda = Lamda_0*(1.0-(T/T_c)**2)**(-1.0/2); # London penetration depth in Pb at 2K, angstorm \n", + "print\"The London penetration depth in Pb at 2K = \",round(Lamda,4),\"angstorm\";\n", + "print\"The London penetration depth at T = T_c becomes Inf\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The London penetration depth in Pb at 2K = 406.9644 angstorm\n", + "The London penetration depth at T = T_c becomes Inf\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.8,Page number 286" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given data\n", + "\n", + "M = (199.5, 200.7, 202.0, 203.3); # Isotopic mass of Hg, amu\n", + "T_c = (4.185, 4.173, 4.159, 4.146); # Critical temperature of Hg, kelvin\n", + "alpha = 0.5; # Trial value of Isotopic exponent\n", + "# Accroding to isotopic effect, T_c = K*M**(-alpha), solving for K\n", + "K = T_c[1]/M[1]**(-alpha); # Isoptopic coefficent \n", + "Tc = zeros(3); \n", + "for i in xrange(len(Tc)):\n", + " Tc[i] = K*M[i+1]**(-alpha)\n", + " print\"Tc[\",i,\"] = \",round(Tc[i],4);\n", + "if T_c[1]-Tc[0]<0.001 and T_c[2]-Tc[1]<0.001 and T_c[3]-Tc[2]<0.001 :\n", + " print\"The isotopic exponent in Isotopic effect of Hg =\",alpha;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Tc[ 0 ] = 4.173\n", + "Tc[ 1 ] = 4.1596\n", + "Tc[ 2 ] = 4.1462\n", + "The isotopic exponent in Isotopic effect of Hg = 0.5\n" + ] + } + ], + "prompt_number": 78 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.9,Page number 286" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#GIven Data\n", + "M_1 = 202.0; # Mass of first isotope of mercury, amu\n", + "M_2 = 199.0; # Mass of second isotope of mercury, amu\n", + "T_c1 = 4.153; # Transition temperature of first isotope of mercury, K \n", + "#As T_c1/T_c2 = (M_2/M_1)**1/2, solving for T_c2\n", + "T_c2 = sqrt(M_1/M_2)*T_c1; \n", + "print\"The transition temperature of isotope of Hg whose mass number is \",M_2,\"=\",round(T_c2,4),\"K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The transition temperature of isotope of Hg whose mass number is 199.0 = 4.1842 K\n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.10,Page number 287" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "alpha = 0.5; # Isotopic exponent of Osmium\n", + "T_c = 0.655; # Transition temperature of Osmium, K \n", + "M = 190.2; # Mass of Osmium, amu\n", + "K = T_c*M**alpha; # K is the constant of proportionality\n", + "\n", + "print\"The value of constant of proportionality =\",round(K,4);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of constant of proportionality = 9.0333\n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.11,Page number 298" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "k = 1.38*10**-23; # Boltzmann constant, J/mol/K\n", + "e = 1.6*10**-19; # Energy equivalent of 1 eV, eV/J\n", + "Theta_D = 96; # Debye temperature, kelvin\n", + "N0 = 0.3678; # Density of states at Fermi energy\n", + "V = 1.0; # Volume of the material, metre cube\n", + "T_c = 1.14*Theta_D*exp(-1.0/(N0*V)); # Critical temperature of the material, K\n", + "Delta_0 = k*Theta_D/sinh(1.0/(N0*V)); # Energy gap at absolute zero, J\n", + "print\"The transition temperature of a material = \",round(T_c,3),\"K\";\n", + "print\"The energy gap of a material = \",\"{0:.3e}\".format(Delta_0/e),\"eV\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The transition temperature of a material = 7.217 K\n", + "The energy gap of a material = 1.097e-03 eV\n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.12,Page number 298" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "Theta_D = 350.0; # Debye temperature, kelvin\n", + "Lamda = 0.828; # Electron-phonon coupling constant\n", + "mu_prime = 0.1373; # Reduced mass of a superconductor, amu\n", + "T_c = Theta_D/1.45*exp(-1.04*(1+Lamda)/(Lamda-mu_prime*(1+0.62*Lamda)); #Transition temperature of superconductor using McMillan formula,K\n", + "\n", + "print\"The transition temperature of the superconductor using McMillan formula = \",round(T_c,3),\"K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The transition temperature of the superconductor using McMillan formula = 11.258 K\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.13,Page number 298" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "Theta_D = 350; # Debye temperature, kelvin\n", + "Lamda = 0.641; # Electron-phonon coupling constant\n", + "mu_prime = 0.143; # Reduced mass of a superconductor, amu\n", + "T_c = Theta_D/1.45*exp(-1.04*(1+Lamda)/(Lamda-mu_prime*(1+0.62*Lamda)));#Transition temperature of superconductor using McMillan formula,K\n", + "\n", + "print\"The superconducting transition temperature of a superconductor using McMillan formula = \",round(T_c,4),\"K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The superconducting transition temperature of a superconductor using McMillan formula = 5.0426 K\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.15,Page number 314" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "Theta_D = 490; # Debye temperature, Kelvin\n", + "Lamda = 0.8; # wavelength of a superconductor, angstorm\n", + "mu_prime = 0.13; # Reduced mass of a superconductor, amu\n", + "T_c = Theta_D/1.45*exp(-1.04*(1+Lamda)/(Lamda-mu_prime*(1+0.62*Lamda)));\n", + "print\"The superconducting transition temperature of a borocarbide superconductor =\",round(T_c,4),\"K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The superconducting transition temperature of a borocarbide superconductor = 15.3526 K\n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.16,Page number 314" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "T_c = 16.5; # Transition temperature of a superconductor, K\n", + "Lamda = [0.7, 0.8, 0.9, 1.0] # Electron-phonon coupling constants at different Tc values \n", + "Theta_D = 503.0; # Debye temperature, kelvin\n", + "mu_prime = 0.13; # Reduced mass of a superconductor, amu\n", + "Tc = [0.0, 0.0, 0.0, 0.0];\n", + "print\"_____________________\";\n", + "print\"Lamda Tc\";\n", + "print\"_____________________\";\n", + "for i in xrange(len(Lamda)):\n", + " Tc[i] = Theta_D/1.45*exp(-1.04*(1+Lamda[i])/(Lamda[i]-mu_prime*(1+0.62*Lamda[i]))); \n", + " if abs(Tc[i] - 16.5) < 1.0 :\n", + " best_Lvalue = Lamda[i];\n", + " print\"\",Lamda[i],\" \",round(Tc[i],3),\"K\";\n", + "print\"_____________________\";\n", + "print\"The best electron-phonon coupling constant should be slightly above \", best_Lvalue;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "_____________________\n", + "Lamda Tc\n", + "_____________________\n", + " 0.7 11.095 K\n", + " 0.8 15.76 K\n", + " 0.9 20.407 K\n", + " 1.0 24.881 K\n", + "_____________________\n", + "The best electron-phonon coupling constant should be slightly above 0.8\n" + ] + } + ], + "prompt_number": 81 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.17,Page number 317" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "T_c = 39.4; # Transition temperature of a superconductor, K\n", + "Lamda = 1; # Electron-phonon coupling constant for a superconductor\n", + "mu_prime= 0.15; # Reduced mass of a superconductor, amu\n", + "# As T_c = Theta_D/1.45*exp(-1.04*(1+Lamda)/(Lamda-mu_prime*(1+0.62*Lamda))), solving for Theta_D\n", + "Theta_D = T_c*1.45*exp(1.04*(1+Lamda)/(Lamda-mu_prime*(1+0.62*Lamda)));\n", + "\n", + "print\"The Debye temperature of a BCS superconductor = \",round(Theta_D,3),\"K\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Debye temperature of a BCS superconductor = 891.6 K\n" + ] + } + ], + "prompt_number": 62 + } + ], + "metadata": {} + } + ] +}
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