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Diffstat (limited to 'sample_notebooks/ManikandanD')
| -rwxr-xr-x | sample_notebooks/ManikandanD/Chapter1.ipynb | 248 | ||||
| -rwxr-xr-x | sample_notebooks/ManikandanD/chapter_1_.ipynb | 256 |
2 files changed, 504 insertions, 0 deletions
diff --git a/sample_notebooks/ManikandanD/Chapter1.ipynb b/sample_notebooks/ManikandanD/Chapter1.ipynb new file mode 100755 index 00000000..56736e6b --- /dev/null +++ b/sample_notebooks/ManikandanD/Chapter1.ipynb @@ -0,0 +1,248 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c711f301926e36cf02f99393ea24acbb9b052c829159da80a195eb0ad85a92da"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1:Bonding in Solids"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3 , Page no:15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=2; #in angstrom(distance)\n",
+ "e=1.6E-19; # in C (charge of electron)\n",
+ "E_o= 8.85E-12;# absolute premittivity\n",
+ "\n",
+ "#calculate\n",
+ "r=2*1*10**-10; # since r is in angstrom\n",
+ "V=-e**2/(4*3.14*E_o*r); # calculate potential\n",
+ "V1=V/e; # changing to eV\n",
+ "\n",
+ "#result\n",
+ "print \"\\nThe potential energy is V = \",V,\"J\";\n",
+ "print \"In electron-Volt V = \",round(V,3),\"eV\"; \n",
+ "print \"Note: the answer in the book is wrong due to calculation mistake\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The potential energy is V = -1.15153477995e-18 J\n",
+ "In electron-Volt V = -0.0 eV\n",
+ "Note: the answer in the book is wrong due to calculation mistake\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4 , Page no:15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "#given\n",
+ "r0=0.236; #in nanometer(interionic distance)\n",
+ "e=1.6E-19; # in C (charge of electron)\n",
+ "E_o= 8.85E-12;# absolute premittivity\n",
+ "N=8; # Born constant\n",
+ "IE=5.14;# in eV (ionisation energy of sodium)\n",
+ "EA=3.65;# in eV (electron affinity of Chlorine)\n",
+ "pi=3.14; # value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r0=r0*1E-9; # since r is in nanometer\n",
+ "PE=(e**2/(4*pi*E_o*r0))*(1-1/N); # calculate potential energy\n",
+ "PE=PE/e; #changing unit from J to eV\n",
+ "NE=IE-EA;# calculation of Net energy\n",
+ "BE=PE-NE;# calculation of Bond Energy\n",
+ "\n",
+ "#result\n",
+ "print\"The potential energy is PE= \",round(PE,2),\"eV\";\n",
+ "print\"The net energy is NE= \",round(NE,2),\"eV\";\n",
+ "print\"The bond energy is BE= \",round(BE,2),\"eV\";\n",
+ "# Note: (1)-In order to make the answer prcatically feasible and avoid the unusual answer, I have used r_0=0.236 nm instead of 236 nm. because using this value will give very much irrelevant answer.\n",
+ "# Note: (2) There is slight variation in the answer due to round off."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential energy is PE= 5.34 eV\n",
+ "The net energy is NE= 1.49 eV\n",
+ "The bond energy is BE= 3.85 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5 , Page no:16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=.41; #in mm(lattice constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "n=0.5; #repulsive exponent value\n",
+ "alpha=1.76; #Madelung constant\n",
+ "pi=3.14; # value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r=.41*1E-3; #since r is in mm\n",
+ "Beta=72*pi*E_o*r**4/(alpha*e**2*(n-1)); #calculation compressibility\n",
+ "\n",
+ "#result\n",
+ "print\"The compressibility is Beta=\",round(Beta);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The compressibility is Beta= -2.50967916144e+15\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6 , Page no:16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=3.56; #in Angstrom\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "IE=3.89; #in eV (ionisation energy of Cs)\n",
+ "EA=-3.61; #in eV (electron affinity of Cl)\n",
+ "n=10.5; #Born constant\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "alpha=1.763; #Madelung constant\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r_0=r_0*1E-10; #since r is in nanometer\n",
+ "U=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
+ "U=U/e; #changing unit from J to eV\n",
+ "ACE=U+EA+IE; #calculation of atomic cohesive energy\n",
+ "\n",
+ "#result\n",
+ "print\"The ionic cohesive energy is \",round(U),\"eV\";\n",
+ "print\"The atomic cohesive energy is\",round(ACE),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ionic cohesive energy is -6.0 eV\n",
+ "The atomic cohesive energy is -6.0 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.7 , Page no:17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=2.81; #in Angstrom\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "n=9; #Born constant\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "alpha=1.748; #Madelung constant\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r_0=r_0*1E-10; #since r is in nanometer\n",
+ "V=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
+ "V=V/e; #changing unit from J to eV\n",
+ "V_1=V/2; #Since only half of the energy contribute per ion to the cohecive energy therfore\n",
+ "\n",
+ "#result\n",
+ "print\"The potential energy is V=\",round(V,2),\"eV\";\n",
+ "print\"The energy contributing per ions to the cohesive energy is \",round(V_1,2),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential energy is V= -7.96 eV\n",
+ "The energy contributing per ions to the cohesive energy is -3.98 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/sample_notebooks/ManikandanD/chapter_1_.ipynb b/sample_notebooks/ManikandanD/chapter_1_.ipynb new file mode 100755 index 00000000..e79571af --- /dev/null +++ b/sample_notebooks/ManikandanD/chapter_1_.ipynb @@ -0,0 +1,256 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:d769dfed1de81f32faa9bbbfcfead0c5e629ef3b47c5b247ff782d0972a27a01"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1:Bonding in Solids"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3 , Page no:15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=2; #in angstrom(distance)\n",
+ "e=1.6E-19; # in C (charge of electron)\n",
+ "E_o= 8.85E-12;# absolute premittivity\n",
+ "\n",
+ "#calculate\n",
+ "r=2*1*10**-10; # since r is in angstrom\n",
+ "V=-e**2/(4*3.14*E_o*r); # calculate potential\n",
+ "V1=V/e; # changing to eV\n",
+ "\n",
+ "#result\n",
+ "print \"\\nThe potential energy is V = \",V,\"J\";\n",
+ "print \"In electron-Volt V = \",round(V,3),\"eV\"; \n",
+ "print \"Note: the answer in the book is wrong due to calculation mistake\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The potential energy is V = -1.15153477995e-18 J\n",
+ "In electron-Volt V = -0.0 eV\n",
+ "Note: the answer in the book is wrong due to calculation mistake\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4 , Page no:15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "#given\n",
+ "r0=0.236; #in nanometer(interionic distance)\n",
+ "e=1.6E-19; # in C (charge of electron)\n",
+ "E_o= 8.85E-12;# absolute premittivity\n",
+ "N=8; # Born constant\n",
+ "IE=5.14;# in eV (ionisation energy of sodium)\n",
+ "EA=3.65;# in eV (electron affinity of Chlorine)\n",
+ "pi=3.14; # value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r0=r0*1E-9; # since r is in nanometer\n",
+ "PE=(e**2/(4*pi*E_o*r0))*(1-1/N); # calculate potential energy\n",
+ "PE=PE/e; #changing unit from J to eV\n",
+ "NE=IE-EA;# calculation of Net energy\n",
+ "BE=PE-NE;# calculation of Bond Energy\n",
+ "\n",
+ "#result\n",
+ "print\"The potential energy is PE= \",round(PE,2),\"eV\";\n",
+ "print\"The net energy is NE= \",round(NE,2),\"eV\";\n",
+ "print\"The bond energy is BE= \",round(BE,2),\"eV\";\n",
+ "# Note: (1)-In order to make the answer prcatically feasible and avoid the unusual answer, I have used r_0=0.236 nm instead of 236 nm. because using this value will give very much irrelevant answer.\n",
+ "# Note: (2) There is slight variation in the answer due to round off."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential energy is PE= 5.34 eV\n",
+ "The net energy is NE= 1.49 eV\n",
+ "The bond energy is BE= 3.85 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5 , Page no:16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=.41; #in mm(lattice constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "n=0.5; #repulsive exponent value\n",
+ "alpha=1.76; #Madelung constant\n",
+ "pi=3.14; # value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r=.41*1E-3; #since r is in mm\n",
+ "Beta=72*pi*E_o*r**4/(alpha*e**2*(n-1)); #calculation compressibility\n",
+ "\n",
+ "#result\n",
+ "print\"The compressibility is\tBeta=\",round(Beta);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The compressibility is\tBeta= -2.50967916144e+15\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6 , Page no:16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=3.56; #in Angstrom\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "IE=3.89; #in eV (ionisation energy of Cs)\n",
+ "EA=-3.61; #in eV (electron affinity of Cl)\n",
+ "n=10.5; #Born constant\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "alpha=1.763; #Madelung constant\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r_0=r_0*1E-10; #since r is in nanometer\n",
+ "U=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
+ "U=U/e; #changing unit from J to eV\n",
+ "ACE=U+EA+IE; #calculation of atomic cohesive energy\n",
+ "\n",
+ "#result\n",
+ "print\"The ionic cohesive energy is \",round(U),\"eV\";\n",
+ "print\"The atomic cohesive energy is\",round(ACE),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ionic cohesive energy is -6.0 eV\n",
+ "The atomic cohesive energy is -6.0 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.7 , Page no:17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=2.81; #in Angstrom\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "n=9; #Born constant\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "alpha=1.748; #Madelung constant\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r_0=r_0*1E-10; #since r is in nanometer\n",
+ "V=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
+ "V=V/e; #changing unit from J to eV\n",
+ "V_1=V/2; #Since only half of the energy contribute per ion to the cohecive energy therfore\n",
+ "\n",
+ "#result\n",
+ "print\"The potential energy is V=\",round(V,2),\"eV\";\n",
+ "print\"The energy contributing per ions to the cohesive energy is \",round(V_1,2),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential energy is V= -7.96 eV\n",
+ "The energy contributing per ions to the cohesive energy is -3.98 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |