From d36fc3b8f88cc3108ffff6151e376b619b9abb01 Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:40:35 +0530 Subject: Revised list of TBCs --- sample_notebooks/ManikandanD/chapter1.ipynb | 256 ++++++++++++++++++++++++++++ 1 file changed, 256 insertions(+) create mode 100755 sample_notebooks/ManikandanD/chapter1.ipynb (limited to 'sample_notebooks/ManikandanD/chapter1.ipynb') diff --git a/sample_notebooks/ManikandanD/chapter1.ipynb b/sample_notebooks/ManikandanD/chapter1.ipynb new file mode 100755 index 00000000..e79571af --- /dev/null +++ b/sample_notebooks/ManikandanD/chapter1.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 -- cgit