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authorJovina Dsouza2014-07-25 00:00:07 +0530
committerJovina Dsouza2014-07-25 00:00:07 +0530
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tree896d9488c2766babf34c3c9d895b0f9f730b0e54 /Engineering_Physics/Chapter10_1.ipynb
parenta4206084fd8c2bd696ea4ae4012aa83534979456 (diff)
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-rwxr-xr-xEngineering_Physics/Chapter10_1.ipynb253
1 files changed, 9 insertions, 244 deletions
diff --git a/Engineering_Physics/Chapter10_1.ipynb b/Engineering_Physics/Chapter10_1.ipynb
index b2bd25cb..051ee9c1 100755
--- a/Engineering_Physics/Chapter10_1.ipynb
+++ b/Engineering_Physics/Chapter10_1.ipynb
@@ -1,7 +1,6 @@
{
"metadata": {
- "name": "",
- "signature": "sha256:4b61028c3be5c168cde4c3aa75ae23500168dbc119942b73de7c79e4a037fd53"
+ "name": "Chapter10"
},
"nbformat": 3,
"nbformat_minor": 0,
@@ -12,52 +11,25 @@
"cell_type": "heading",
"level": 1,
"metadata": {},
- "source": [
- "10: Statistical Mechanics"
- ]
+ "source": "10: Dielectric Materials"
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
- "source": [
- "Example number 10.1, Page number 222"
- ]
+ "source": "Example number 10.1, Page number 289"
},
{
"cell_type": "code",
"collapsed": false,
- "input": [
- "\n",
- "#importing modules\n",
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration\n",
- "k = 1.38*10**-23; #Boltzmann constant(J/K)\n",
- "e = 1.6*10**-19; #Energy equivalent of 1 eV(J/eV)\n",
- "g1 = 2; #The degeneracy of ground state\n",
- "g2 = 8; #The degeneracy of excited state\n",
- "delta_E = 10.2; #Energy of excited state above the ground state(eV)\n",
- "T = 6000; #Temperature of the state(K)\n",
- "\n",
- "#Calculation\n",
- "D_ratio = g2/g1; #Ratio of degeneracy of states\n",
- "x = k*T/e;\n",
- "N_ratio = D_ratio*math.exp(-delta_E/x); #Ratio of occupancy of the excited to the ground state\n",
- "\n",
- "#Result\n",
- "print \"The ratio of occupancy of the excited to the ground state is\",N_ratio"
- ],
+ "input": "#importing modules\nimport math\n\n#Variable declaration\nepsilon_r = 1.0000684; #dielectric constant\nN = 2.7*10**25; #number of atoms(per m**3)\nepsilon0 = 8.85*10**-12; #permittivity of free space\n\n#Calculation\nalpha_e = epsilon0*(epsilon_r-1)/N; #electronic polarizability(Fm**2)\n\n#Result\nprint \"electronic polarizability is\",alpha_e,\"Fm**2\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
- "text": [
- "The ratio of occupancy of the excited to the ground state is 1.10167326887e-08\n"
- ]
+ "text": "electronic polarizability is 2.242e-41 Fm**2\n"
}
],
"prompt_number": 1
@@ -66,229 +38,22 @@
"cell_type": "heading",
"level": 2,
"metadata": {},
- "source": [
- "Example number 10.2, Page number 222"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "a = 10/2;\n",
- "#enegy of 10 bosons is E = (10*pi**2*h**2)/(2*m*a**2) = (5*pi**2*h**2)/(m*a**2)\n",
- "\n",
- "#Result\n",
- "print \"enegy of 10 bosons is E = \",int(a),\"(pi**2*h**2)/(m*a**2)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "enegy of 10 bosons is E = 5 (pi**2*h**2)/(m*a**2)\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.3, Page number 223"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- " \n",
- "#importing modules\n",
- "import math\n",
- "\n",
- "#Variable declaration\n",
- "n1=1; #1st level\n",
- "n2=2; #2nd level\n",
- "n3=3; #3rd level\n",
- "n4=4; #4th level\n",
- "n5=5; #5th level\n",
- "\n",
- "#Calculation\n",
- "#an energy level can accomodate only 2 fermions. hence there will be 2 fermions in each level\n",
- "#thus total ground state energy will be E = (2*E1)+(2*E2)+(2*E3)+(2*E4)+E5\n",
- "#let X = ((pi**2)*(h**2)/(2*m*a**2)). E = X*((2*n1**2)+(2*n2**2)+(2*n3**2)+(2*n4**2)+(n5**2))\n",
- "A = (2*n1**2)+(2*n2**2)+(2*n3**2)+(2*n4**2)+(n5**2);\n",
- "#thus E = A*X\n",
- "\n",
- "#Result\n",
- "print \"the ground state energy of the system is\",A,\"(pi**2)*(h**2)/(2*m*a**2)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the ground state energy of the system is 85 (pi**2)*(h**2)/(2*m*a**2)\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.4, Page number 223"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- " \n",
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "e = 1.6*10**-19; #Energy equivalent of 1 eV(J/eV)\n",
- "N_A = 6.02*10**23; #Avogadro's number\n",
- "h = 6.626*10**-34; #Planck's constant(Js)\n",
- "me = 9.1*10**-31; #Mass of electron(kg)\n",
- "rho = 10.5; #Density of silver(g/cm)\n",
- "m = 108; #Molecular mass of silver(g/mol)\n",
- "\n",
- "#Calculation\n",
- "N_D = rho*N_A/m; #Number density of conduction electrons(per cm**3)\n",
- "N_D = N_D*10**6; #Number density of conduction electrons(per m**3)\n",
- "E_F = ((h**2)/(8*me))*(3/math.pi*N_D)**(2/3); #fermi energy(J)\n",
- "E_F = E_F/e; #fermi energy(eV)\n",
- "E_F = math.ceil(E_F*10**2)/10**2; #rounding off the value of E_F to 2 decimals\n",
- "\n",
- "#Result\n",
- "print \"The number density of conduction electrons is\",N_D, \"per metre cube\"\n",
- "print \"The Fermi energy of silver is\",E_F, \"eV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The number density of conduction electrons is 5.85277777778e+28 per metre cube\n",
- "The Fermi energy of silver is 5.51 eV\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.5, Page number 224"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- " \n",
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "N_A = 6.02*10**23; #Avogadro's number\n",
- "k = 1.38*10**-23; #Boltzmann constant(J/K)\n",
- "T = 293; #Temperature of sodium(K)\n",
- "E_F = 3.24; #Fermi energy of sodium(eV)\n",
- "e = 1.6*10**-19; #Energy equivalent of 1 eV(J/eV)\n",
- "\n",
- "#Calculation\n",
- "C_v = math.pi**2*N_A*k**2*T/(2*E_F*e); #Molar specific heat of sodium(per mole)\n",
- "C_v = math.ceil(C_v*10**2)/10**2; #rounding off the value of C_v to 2 decimals\n",
- "\n",
- "#Result\n",
- "print \"The electronic contribution to molar specific heat of sodium is\",C_v, \"per mole\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The electronic contribution to molar specific heat of sodium is 0.32 per mole\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 10.6, Page number 224"
- ]
+ "source": "Example number 10.2, Page number 290"
},
{
"cell_type": "code",
"collapsed": false,
- "input": [
- " \n",
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "e = 1.6*10**-19; #Energy equivalent of 1 eV(J/eV)\n",
- "h = 6.626*10**-34; #Planck's constant(Js)\n",
- "m = 9.1*10**-31; #Mass of the electron(kg)\n",
- "N_D = 18.1*10**28; #Number density of conduction electrons in Al(per metre cube)\n",
- "\n",
- "#Calculation\n",
- "E_F = h**2/(8*m)*(3/math.pi*N_D)**(2/3); #N_D = N/V. Fermi energy of aluminium(J)\n",
- "E_F = E_F/e; #Fermi energy of aluminium(eV)\n",
- "E_F = math.ceil(E_F*10**3)/10**3; #rounding off the value of E_F to 3 decimals\n",
- "Em_0 = 3/5*E_F; #Mean energy of the electron at 0K(eV)\n",
- "Em_0 = math.ceil(Em_0*10**3)/10**3; #rounding off the value of Em_0 to 3 decimals\n",
- "\n",
- "#Result\n",
- "print \"The Fermi energy of aluminium is\",E_F, \"eV\"\n",
- "print \"The mean energy of the electron is\",Em_0, \"eV\""
- ],
+ "input": "#importing modules\nimport math\n\n#Variable declaration\nepsilon_r = 1.0024; #relative permittivity\nN = 2.7*10**25; #number of atoms(per m**3)\nepsilon0 = 8.85*10**-12; #permittivity of free space\n\n#Calculation\nalpha_e = epsilon0*(epsilon_r-1)/N; #electronic polarizability(Fm**2)\n\n#Result\nprint \"electronic polarizability is\",alpha_e,\"Fm**2\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
- "text": [
- "The Fermi energy of aluminium is 11.696 eV\n",
- "The mean energy of the electron is 7.018 eV\n"
- ]
+ "text": "electronic polarizability is 7.86666666667e-40 Fm**2\n"
}
],
- "prompt_number": 9
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": []
+ "prompt_number": 2
}
],
"metadata": {}