{
 "metadata": {
  "name": ""
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h1> Chapter No 1 : Semiconductor Materials and\n",
      "Their Properties<h1>"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.3, Page No 37 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Finding Density of silicon crystal\n",
      "\n",
      "#initialisation of variables\n",
      "a=5.3   #in \u00c5\n",
      "a=a*10**-10     #in m\n",
      "N_A=6.023*10**23  \n",
      "At_Si=28    #atomic weight of Si\n",
      "n = 4\n",
      "m=At_Si/N_A  #in gm\n",
      "m= m*10**-3  #in kg\n",
      "V=a**3       #in m^3\n",
      "\n",
      "#CALCULATIONS\n",
      "Rho=((m*n)/V)/1000  #in kg/m^3\n",
      "\n",
      "# Note: There is calculation error to find the value of density. So the answer in the book is wrong.#RESULTS\n",
      "#RESULTS\n",
      "print('Density of silicon crystal in kg/m^3 is = %.2f kg/m^3' %Rho)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Density of silicon crystal in kg/m^3 is = 1.25 kg/m^3\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.4, Page No 37 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "# Finding Density of the crystal \n",
      "\n",
      "#initialisation of variables\n",
      "n = 4\n",
      "r = 1.278  #in \u00c5\n",
      "\n",
      "#CALCULATIONS\n",
      "a = (4*r)/(math.sqrt(2)) #in\u00c5\n",
      "a = a * 10**-10   #in m\n",
      "V = (a)**3  #in m^3\n",
      "At_W = 63.5   #atomic weight\n",
      "N_A = 6.023*10**23\n",
      "m = At_W /N_A   #in gm\n",
      "m = m*10**-3  #in kg\n",
      "Rho = ((m*n)/V)/1000  #in kg/m^3\n",
      "\n",
      "#RESULTS\n",
      "print('Density of the crystal in kg/m^3 is =%.2f kg/m^3' %Rho)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Density of the crystal in kg/m^3 is =8.93 kg/m^3\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.5 Page No 43 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Finding Wavaelength of X ray\n",
      "\n",
      "import math\n",
      "from numpy import *\n",
      "#initialisation of variables\n",
      "d=2.82   #in \u00c5\n",
      "d=d*10**-10    #in m\n",
      "n=1\n",
      "theta1=10  #n degree\n",
      "\n",
      "#CALCULATIONS\n",
      "lembda=2*d*math.sin(math.radians(theta1))   #in m\n",
      "lembda=lembda*10**10       #in \u00c5\n",
      "\n",
      "#RESULTS\n",
      "print('Wavaelength of X ray  is =%.2f \u00c5 degrees' %lembda)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Wavaelength of X ray  is =0.98 \u00c5 degrees\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.6, Page No 43"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Finding The spacing of atomic layer\n",
      "\n",
      "import math\n",
      "#initialisation of variables\n",
      "lembda = 1.6  #in \u00c5\n",
      "theta = 14.2  #in degree\n",
      "n = 1\n",
      "\n",
      "#CALCULATIONS\n",
      "d = (n*lembda)/(2*math.sin(math.radians(theta)))   #in \u00c5\n",
      "\n",
      "#RESULTS\n",
      "print('The spacing of atomic layer in crystal is =%.2f in \u00c5 degrees' %d)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The spacing of atomic layer in crystal is =3.26 in \u00c5 degrees\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.7 Page No 43"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Finding The interplanner spacing\n",
      "\n",
      "import math\n",
      "#initialisation of variables\n",
      "n = 1\n",
      "theta1 = 30  #in degree\n",
      "lembda = 1.78  #in \u00c5\n",
      "\n",
      "#CALCULATIONS\n",
      "d = (n*lembda)/(2*math.sin(math.radians(theta1)))  #in \u00c5\n",
      "\n",
      "#RESULTS\n",
      "print('The interplanner spacing is =%.2f in \u00c5 degrees' %d)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The interplanner spacing is =1.78 in \u00c5 degrees\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.8 Page No 43"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Finding Interplaner spacing of the crystal\n",
      "\n",
      "import math\n",
      "\n",
      "#initialisation of variables\n",
      "lembda = 0.58  #in \u00c5\n",
      "n = 1;\n",
      "theta1 = 6.45   # in degree\n",
      "\n",
      "#CALCULATIONS\n",
      "d = (n*lembda)/(2*math.sin(math.radians(theta1))) #in \u00c5 \n",
      "print('Part (i) : At angle of 6.45\u00b0, Interplaner spacing of the crystal  is =%.2f \u00c5' %d)\n",
      "theta2 = 9.15  #in degree\n",
      "d1 = (n*lembda)/(2*math.sin(math.radians(theta2))) #in \u00c5 \n",
      "print('Part (ii) : At angle of 9.15\u00b0, Interplaner spacing of the crystal is =%.2f \u00c5' %d1)\n",
      "theta3 = 13 #in degree\n",
      "n2 = 1\n",
      "d2 = (n2*lembda)/(2*math.sin(math.radians(theta3)))  #in \u00c5\n",
      "print('Part (iii) : At angle of 13\u00b0, Interplaner spacing of the crystal is =%.2f \u00c5' %d2)\n",
      "n=2\n",
      "d2 = (n*lembda)/(2*math.sin(math.radians(theta3)))  #in \u00c5 \n",
      "\n",
      "#RESULTS\n",
      "print('Part (iv) : : The interplaner spacing is : =%.4f \u00c5' %d2)\n",
      "print('The interplaner spacing for some other set of reflecting  is : =%.2f \u00c5' %d1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Part (i) : At angle of 6.45\u00b0, Interplaner spacing of the crystal  is =2.58 \u00c5\n",
        "Part (ii) : At angle of 9.15\u00b0, Interplaner spacing of the crystal is =1.82 \u00c5\n",
        "Part (iii) : At angle of 13\u00b0, Interplaner spacing of the crystal is =1.29 \u00c5\n",
        "Part (iv) : : The interplaner spacing is : =2.5783 \u00c5\n",
        "The interplaner spacing for some other set of reflecting  is : =1.82 \u00c5\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.9, Page No 44 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Finding The glancing angle for a cubic in degree\n",
      "\n",
      "import math\n",
      "from sympy.mpmath import *\n",
      "import cmath\n",
      "\n",
      "#initialisation of variables\n",
      "a = 2.814 #in \u00c5\n",
      "l = 0\n",
      "h = l\n",
      "k = 0\n",
      "d=a  #in \u00c5\n",
      "n = 2\n",
      "\n",
      "#CALCULATIONS\n",
      "lembda=0.710     #in \u00c5\n",
      "theta = math.degrees(math.asin(n*lembda/(2*d)))\n",
      "\n",
      "#RESULTS\n",
      "print('The glancing angle for a cubic in degree is : =%.2f  \u00c5 degrees' %theta)\n",
      "print('Anwser is in points i.e = 14 36\"40\"..')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The glancing angle for a cubic in degree is : =14.61  \u00c5 degrees\n",
        "Anwser is in points i.e = 14 36\"40\"..\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.10 Page No 44"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Finding Wavelength of X ray\n",
      "\n",
      "import math \n",
      "\n",
      "#initialisation of variables\n",
      "a = 3.65  #in \u00c5\n",
      "a = 3.65*10**-10  #in m\n",
      "h = 1\n",
      "k = 0\n",
      "l = 0\n",
      "\n",
      "#CALCULATIONS\n",
      "d= a/(math.sqrt(h**2+k**2+l**2))   #in m\n",
      "n = 1\n",
      "theta=60  #in degree\n",
      "lembda = 2*d*math.sin(math.radians(theta))   #in m\n",
      "lembda = lembda * 10**10   #in \u00c5\n",
      "\n",
      "#RESULTS\n",
      "print('Wavelength of X ray is : =%.2f  \u00c5' %lembda)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Wavelength of X ray is : =6.32  \u00c5\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.11 Page No 44"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Finding The glancing angle in degree\n",
      "\n",
      "import math\n",
      "\n",
      "#initialisation of variables\n",
      "lembda = 1.54   #in \u00c5\n",
      "density = 9.024   #in gm/cc\n",
      "n = 1\n",
      "MI = 100\n",
      "At_W = 63.54   #atomic weight\n",
      "N_A = 6.023*10**23   \n",
      "m = At_W/N_A   #in gm\n",
      "a =(density*m)**(1.0/3)  #in cm\n",
      "h = 1\n",
      "k = 0\n",
      "l = 0\n",
      "\n",
      "#CALCULATIONS\n",
      "d= a/(math.sqrt(h**2+k**2+l**2))\n",
      "theta =math.degrees((lembda*10**-8)/(2*d))   #in degree\n",
      "\n",
      "\n",
      "#RESULTS\n",
      "print('The glancing angle is = %.2f degrees' %theta)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The glancing angle is = 4.48 degrees\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.12 Page No 45 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Finding To get the 2nd order spectrum the position of the detector in degree\n",
      "\n",
      "import math\n",
      "\n",
      "#initialisation of variables\n",
      "a = 3.615   #in \u00c5\n",
      "theta=22   #in degree\n",
      "n=1\n",
      "h=1\n",
      "k=h\n",
      "l=k\n",
      "\n",
      "#CALCULATIONS\n",
      "d = a/(math.sqrt( ((h)**2) + ((k)**2) + ((l)**2) ))   #in \u00c5\n",
      "lembda = 2*d*math.sin(math.radians(theta))   #in \u00c5\n",
      "print('The wavelength of X ray is = %.2f \u00c5 ' %lembda)\n",
      "theta2 =math.degrees(math.asin(lembda/d))  #in degree\n",
      "             #in degree\n",
      "\n",
      "#RESULTS\n",
      "print('To get the 2nd order spectrum the position of the detector  is = %.2f degrees ' %theta2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The wavelength of X ray is = 1.56 \u00c5 \n",
        "To get the 2nd order spectrum the position of the detector  is = 48.52 degrees \n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.13, Page No 45"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#initialisation of variables\n",
      "n = 1\n",
      "lembda = 1.54      # in \u00c5\n",
      "theta = 21.7      # in degree\n",
      "\n",
      "#CALCULATIONS\n",
      "d = lembda/(2*math.sin(math.radians(theta)))     # in \u00c5\n",
      "h = 1\n",
      "k = h\n",
      "l = k\n",
      "a = d*math.sqrt(h**2+k**2+l**2)      # in \u00c5\n",
      "\n",
      "#RESULTS\n",
      "print('Lattice constant is = %.2f  in \u00c5 ' %a)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Lattice constant is = 3.61  in \u00c5 \n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.14, Page No 47"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#initialisation of variables\n",
      "h = 2\n",
      "k = 1\n",
      "l = 1\n",
      "a = 4.8      # in \u00c5\n",
      "\n",
      "#CALCULATIONS\n",
      "d_211 = a/(math.sqrt(h**2+k**2+l**2))       #in \u00c5\n",
      "\n",
      "#RESULTS\n",
      "print('The distance between planes is =%.2f \u00c5 ' %d_211)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The distance between planes is =1.96 \u00c5 \n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.15, Page No 47"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#initialisation of variables\n",
      "r = 1.28       # in \u00c5\n",
      "n = 4\n",
      "M = 63.5\n",
      "\n",
      "#CALCULATIONS\n",
      "a = (4*r)/(math.sqrt(2))        #in \u00c5\n",
      "a = a * 10**-8       # in cm\n",
      "N_A = 6.023*10**23     \n",
      "Rho = (n*M)/( N_A*((a)**3) )       # in gm/cc\n",
      "\n",
      "#RESULTS\n",
      "print('Density is =%.2f gm/cc  ' %Rho)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Density is =8.89 gm/cc  \n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.16, Page No 47"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#initialisation of variables\n",
      "M = 55.85\n",
      "a = 2.9        # in \u00c5\n",
      "Rho = 7.87    #in gm/cc\n",
      "\n",
      "#CALCULATIONS\n",
      "a = a * 10**-8    # in cm\n",
      "N_A = 6.023*10**23\n",
      "n = (Rho*N_A*((a)**3))/M            # atom per unit\n",
      "\n",
      "#RESULTS\n",
      "print('A lattice having =%.2f atom per unit cell is a BCC structure ' %n)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "A lattice having =2.07 atom per unit cell is a BCC structure \n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.17, Page No 47"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#initialisation of variables\n",
      "M = 60.0           # in gm/mole\n",
      "Rho = 6.23      # in gm/cc\n",
      "n = 4\n",
      "N_A = 6.023*10**23 \n",
      "\n",
      "#CALCULATIONS\n",
      "a = ((n*M)/(N_A * Rho))**(1.0/3)         # in cm\n",
      "r = (a*math.sqrt(2))/n               #radius of atom in cm\n",
      "r = r * 10**8            # in \u00c5\n",
      "\n",
      "#RESULTS\n",
      "print('Radius of atom is =%.2f \u00c5  ' %r)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Radius of atom is =1.41 \u00c5  \n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.18, Page No 47"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#initialisation of variables\n",
      "Rho = 5.96          # in gm/cc\n",
      "M = 50.0\n",
      "n = 2\n",
      "N_A = 6.023*10**23 \n",
      "\n",
      "#CALCULATIONS\n",
      "a =((n*M)/(Rho*N_A))**(1.0/3)          # in cm\n",
      "r = (a*math.sqrt(3))/4                 # in cm\n",
      "P_f = (2*(4.0/3)*math.pi*((r)**3))/((a)**3)       # packing factor\n",
      "\n",
      "\n",
      "#RESULTS\n",
      "print('Packing factor is =%.2f  ' %P_f)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Packing factor is =0.68  \n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.19, Page No 48"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#initialisation of variables\n",
      "M = 120.0\n",
      "n = 2.0\n",
      "N_A = 6.023*10**23\n",
      "\n",
      "#CALCULATIONS\n",
      "m1 = M/N_A              #mass of 1 atom in gm\n",
      "m2 = n*m1               #mass of unit cell in gm\n",
      "\n",
      "#RESULTS\n",
      "print('Number of unit cell in 20 gms of element is : =%.2f X 10^22 unit cells ' %((20/m2)/(10**22)))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Number of unit cell in 20 gms of element is : =5.02 X 10^22 unit cells \n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.20, Page No 48"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#initialisation of variables\n",
      "Rho = 2.48         # in gm/c.c\n",
      "n = 4.0 \n",
      "M = 58\n",
      "N_A = 6.023*10**23\n",
      "\n",
      "#CALCULATIONS\n",
      "a = ( (n*M)/(Rho*N_A) )**(1.0/3)      # in cm\n",
      "a = a * 10**8        # in \u00c5\n",
      "r = (a*math.sqrt(2))/n      # in \u00c5\n",
      "r = 2*r       # in \u00c5\n",
      "\n",
      "#RESULTS\n",
      "print('The center to center distance between ions  is =%.2f  \u00c5' %r)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The center to center distance between ions  is =3.80  \u00c5\n"
       ]
      }
     ],
     "prompt_number": 19
    }
   ],
   "metadata": {}
  }
 ]
}