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| author | Thomas Stephen Lee | 2015-09-04 22:04:10 +0530 |
|---|---|---|
| committer | Thomas Stephen Lee | 2015-09-04 22:04:10 +0530 |
| commit | 41f1f72e9502f5c3de6ca16b303803dfcf1df594 (patch) | |
| tree | f4bf726a3e3ce5d7d9ee3781cbacfe3116115a2c /Advance_Semiconductor_Devices/ChapterNo1.ipynb | |
| parent | 9c9779ba21b9bedde88e1e8216f9e3b4f8650b0e (diff) | |
| download | Python-Textbook-Companions-41f1f72e9502f5c3de6ca16b303803dfcf1df594.tar.gz Python-Textbook-Companions-41f1f72e9502f5c3de6ca16b303803dfcf1df594.tar.bz2 Python-Textbook-Companions-41f1f72e9502f5c3de6ca16b303803dfcf1df594.zip | |
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diff --git a/Advance_Semiconductor_Devices/ChapterNo1.ipynb b/Advance_Semiconductor_Devices/ChapterNo1.ipynb deleted file mode 100755 index 4d1dc38d..00000000 --- a/Advance_Semiconductor_Devices/ChapterNo1.ipynb +++ /dev/null @@ -1,800 +0,0 @@ -{
- "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": {}
- }
- ]
-}
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