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+{
+ "metadata": {
+  "name": "",
+  "signature": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter2 : Atomic model & bonding in solids"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example-2.1, page no-28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#given\n",
+      "#atomic no. of gold\n",
+      "Z=79\n",
+      "#kinetic energy of alpha particle\n",
+      "E=7.68*1.6*(10)**(-13)  #J because [1MeV=1.6*(10)**(-13)]\n",
+      "e=1.6*10**(-19)  #C\n",
+      "E0=8.854*10**(-12)  #F/m\n",
+      "#the distance of closest approach is given by:\n",
+      "d0=2*e*Z*e/(4*(math.pi)*E0*E)  #m\n",
+      "print \"The closest approach of alpha particle is %.2ef m\" %d0"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The closest approach of alpha particle is 2.96e-14f m\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example-2.2, page no-29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import *\n",
+      "from numpy import *\n",
+      "#given\n",
+      "#IN THE RUTHERFORD SCATTERING EXPERIMENT\n",
+      "#the no of particles scattered at\n",
+      "theta1=(pi)/2  #radians\n",
+      "#is\n",
+      "N90=44  #per minute\n",
+      "#the number of particles scattered particales N is given by\n",
+      "#N=C*(1/(sin(theta/2))**4)  where C is propotionality constant\n",
+      "#solving above equation for C\n",
+      "C=N90*(sin(theta1/2))**4  \n",
+      "# now to find the no of particles scatering at 75 and 135 degrees\n",
+      "theta2=75*(pi)/180  #radians\n",
+      "N75=C*(1/(sin(theta2/2))**4)  #per minute\n",
+      "theta3=135*(pi)/180  #radians\n",
+      "N135=C*(1/(sin(theta3/2))**4)  #per minute\n",
+      "print \"The no of particles scattered at 75 and 135 degrees are %d per minute and %d per minutes\" %(N75,N135)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The no of particles scattered at 75 and 135 degrees are 80 per minute and 15 per minutes\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example-2.3, page no-32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "#mass of electron\n",
+      "m=9.11*10**(-31)  #kg\n",
+      "#charge on an electron\n",
+      "e=1.6*10**(-19)  #C\n",
+      "#plank's constant\n",
+      "h=6.62*10**(-34)\n",
+      "E0=8.85*10**(-12) \n",
+      "#NO OF ELECTRONS SHELLS IN HYDROZEN ATOm\n",
+      "n=1\n",
+      "#atomic number of hydrogen\n",
+      "Z=1\n",
+      "#radius of first orbit of hydrogen is given by\n",
+      "r1=n**2*E0*h**2/((pi)*m*Z*e**2)  #m\n",
+      "print \"The radius of the first orbit of the electron in the hydrogen atom %.2e\"%(r1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The radius of the first orbit of the electron in the hydrogen atom 5.29e-11\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example-2.4, page no-32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "#mass of electron\n",
+      "m=9.11*10**(-31)  #kg\n",
+      "#charge on an electron\n",
+      "e=1.6*10**(-19)  #C\n",
+      "#plank's constant\n",
+      "h=6.62*10**(-34)\n",
+      "E0=8.85*10**(-12) \n",
+      "#NO OF ELECTRONS SHELLS IN HYDROZEN ATOm\n",
+      "n=1\n",
+      "#atomic number of hydrogen\n",
+      "Z=1\n",
+      "#ionization potential energy of hydrogen atom is given by\n",
+      "E=m*Z**2*e**4/(8*(E0)**2*h**2*n**2)  #J\n",
+      "#energy in eV\n",
+      "EV=E/e  #eV\n",
+      "print \"The ionization potential for hydrogen atom is %0.2f V\" %(EV)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ionization potential for hydrogen atom is 13.59 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example-2.5, page no-34"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example-2.6, page no-36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "#uncertainity in the momentum\n",
+      "deltap=10**-27  #kg ms**-1\n",
+      "#according to uncertainity principle\n",
+      "#deltap* deltax >=h/(2*(pi))\n",
+      "#we know that \n",
+      "h=6.626*10**-34  #Js\n",
+      "#here instead of inequality we are using only equality just for notation otherwise it is greater than equal to as mentioned above\n",
+      "#now deltax is given by\n",
+      "deltax=h/(2*(pi)*deltap)  #m\n",
+      "print \"The minimum uncertainity is %.2e m\"%(deltax)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum uncertainity is 1.05e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example-2.10, page no- 57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "#ionization potential of hydrogen\n",
+      "E1=13.6  #eV\n",
+      "#when \n",
+      "n=3\n",
+      "E3=-E1/n**2  #eV\n",
+      "#when \n",
+      "n=5\n",
+      "E5=-E1/n**2  #eV\n",
+      "print \"Energy of 3rd and 5th orbits are %0.2f eV and %0.2f eV\"%(E3,E5)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of 3rd and 5th orbits are -1.51 eV and -0.54 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example-2.11, page no-59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "#dipole moment og HF is\n",
+      "DM=6.375*10**(-30)  #Cm\n",
+      "#intermolecular distance\n",
+      "r=0.9178*10**(-10)  #m\n",
+      "#charge on an electron\n",
+      "e=1.67*10**(-19)  #C\n",
+      "#since the HF posses ionic characters\n",
+      "#so\n",
+      "#Hf in fully ionic state has dipole moment as\n",
+      "DM2=r*e  #Cm\n",
+      "#percentage ionic characters\n",
+      "percentage=DM/DM2*100  #%\n",
+      "print \"The percentage ionic character is %0.2f approx.\"%(percentage)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage ionic character is 41.59 approx.\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example-2.12, page no-60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "#elctronegativity of In\n",
+      "EnIn=1.5\n",
+      "#elctronegativity of As\n",
+      "EnAs=2.2\n",
+      "#elctronegativity of Ga\n",
+      "EnGa=1.8\n",
+      "#for InAs\n",
+      "ionic_charater1=(1-exp((-0.25)*(EnAs-EnIn)**2))*100  #in %\n",
+      "#for GaAs\n",
+      "ionic_charater2=(1-exp((-0.25)*(EnAs-EnGa)**2))*100  # in %\n",
+      "print \"Ionic character in InAs and GaAs are %0.1f %% and %0.1f %%\"%(ionic_charater1,ionic_charater2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ionic character in InAs and GaAs are 11.5 % and 3.9 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}