From c7fe425ef3c5e8804f2f5de3d8fffedf5e2f1131 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 7 Apr 2015 15:58:05 +0530 Subject: added books --- Engineering_Physics_-_I/README.txt | 10 + Engineering_Physics_-_I/chapter1.ipynb | 188 ++++++ Engineering_Physics_-_I/chapter2.ipynb | 293 +++++++++ Engineering_Physics_-_I/chapter3.ipynb | 209 ++++++ Engineering_Physics_-_I/chapter4.ipynb | 532 ++++++++++++++++ Engineering_Physics_-_I/chapter5.ipynb | 700 +++++++++++++++++++++ Engineering_Physics_-_I/screenshots/Wavelength.PNG | Bin 0 -> 36766 bytes .../screenshots/Young's_modulus.PNG | Bin 0 -> 33987 bytes .../screenshots/numerical_aperture.PNG | Bin 0 -> 29899 bytes 9 files changed, 1932 insertions(+) create mode 100755 Engineering_Physics_-_I/README.txt create mode 100755 Engineering_Physics_-_I/chapter1.ipynb create mode 100755 Engineering_Physics_-_I/chapter2.ipynb create mode 100755 Engineering_Physics_-_I/chapter3.ipynb create mode 100755 Engineering_Physics_-_I/chapter4.ipynb create mode 100755 Engineering_Physics_-_I/chapter5.ipynb create mode 100755 Engineering_Physics_-_I/screenshots/Wavelength.PNG create mode 100755 Engineering_Physics_-_I/screenshots/Young's_modulus.PNG create mode 100755 Engineering_Physics_-_I/screenshots/numerical_aperture.PNG (limited to 'Engineering_Physics_-_I') diff --git a/Engineering_Physics_-_I/README.txt b/Engineering_Physics_-_I/README.txt new file mode 100755 index 00000000..9cb516db --- /dev/null +++ b/Engineering_Physics_-_I/README.txt @@ -0,0 +1,10 @@ +Contributed By: saikomal chanagam +Course: btech +College/Institute/Organization: SASTRA University +Department/Designation: ECE +Book Title: Engineering Physics - I +Author: G. SenthilKumar +Publisher: VRB Publishers, Chennai +Year of publication: 2011 +Isbn: 9789380241852 +Edition: 1 \ No newline at end of file diff --git a/Engineering_Physics_-_I/chapter1.ipynb b/Engineering_Physics_-_I/chapter1.ipynb new file mode 100755 index 00000000..4b792797 --- /dev/null +++ b/Engineering_Physics_-_I/chapter1.ipynb @@ -0,0 +1,188 @@ +{ + "metadata": { + "name": "Chapter1(s)" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 1: Ultrasonics" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1, Page No:1.29" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n# variable declaration\nP = 1; # for fundamental mode\nt = 0.1*10**-2; # thickness of piezo electric crystal\nE = 80*10**9 # young's modulus\np = 2654 # density in kg/m^3\n\n# Calculations\n\nf = (P/(2*t))*math.sqrt(E/p); # frequency of the oscillator circuit\n\n# Result\nprint 'The Frequency of the oscillator circuit %3.4g' %f,'Hz';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The Frequency of the oscillator circuit 2.745e+06 Hz\n" + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2, Page No:1.29" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n# variable declaration\nP = 1; # for fundamental mode\nt = 0.1*10**-2; # thickness of piezo electric crystal\nE = 7.9*10**10 # young's modulus\np = 2650 # density in kg/m^3\n\n# Calculations\n\nf = (P/(2*t))*math.sqrt(E/p); # frequency of the oscillator circuit\n\n#Result\nprint 'The Frequency of the vibrating crystal %0.2f'%(f/(10**6)),'MHz';\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The Frequency of the vibrating crystal 2.73 MHz\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3, Page No:1.30" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n# variable Declaration\nf = 1.5*10**6; # frequency of ultrasonics in Hz\nd6 = 2.75*10**-3; # distance between 6 consecutive nodes\n\n# Calculations\nd = d6/5; # distance b/w two nodes\nlamda = 2*d; # wavelength in m\nv = f*lamda; # velocity of ultrasonics\n\n# Result\nprint 'Velocity of ultrasonics ' ,v,'m/sec';\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Velocity of ultrasonics 1650.0 m/sec\n" + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Add_example 1, Page No: 1.31" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n# Variable Declaration\nP = 1; # for fundamental mode\nt = 1.5*10**-3; # thickness of quartz crystal\nE = 7.9*10**10 # young's modulus in N/m^2\np = 2650 # density in kg/m^3\n\n# Calculations\n\nf = (P/(2*t))*math.sqrt(E/p); # frequency of the oscillator circuit\n\n# Result\nprint 'The Fundamental Frequency of the Quartz crystal %3.2f'%(f/10**6), 'MHz';\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The Fundamental Frequency of the Quartz crystal 1.82 MHz\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Add_example 2, Page No: 1.31" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n# Variable Declaration\nv = 5000; # velocity of ultrasonics in m/s\ndf = 60*10**3; # difference b/w two adjacent harmonic freq. in Hz\n\n# Calculations\n\nd = (float(v)/(2*df)) ; # thickness of steel plate\n\n# Result\nprint 'The thickness of steel plate %3.4f'%(d),'m';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The thickness of steel plate 0.0417 m\n" + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Add_example 3, Page No: 1.32" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n# Variable Declaration\nv = 1440; # velocity of ultrasonics in sea water in m/s\nt = 0.33 # time taken b/w tx and rx in sec\n\n# Calculations\n\nd = v*t; # distance travelled by ultrasonics\nD = d/2; # depth of submerged submarine in m\n\n# Result\nprint 'Depth of submerged submarine',D,'m';\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Depth of submerged submarine 237.6 m\n" + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Add_example 4, Page No: 1.33" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n# Variable Declaration\nd = 0.55*10**-3; # distance b/w two antinodes\nf = 1.5*10**6; # freq of the crystal\n\n# Calculations\n\nlamda = 2*d; # wavelength\nv = f*lamda; # velocity of ultronics\n\n# Result\nprint 'Velocity of waves in sea water',v,'m/s';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Velocity of waves in sea water 1650.0 m/s\n" + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Add_example 5, Page No: 1.33\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n# Variable Declaration\nP = 1; # for fundamental mode\np = 2660 # density of quartz in kg/m^3\nf = 1300*10**3 # freq of quartz plate for sub division ii\nk = 2.87*10**3\n\n#f1 = (k)/t # freq for sub division i\n\n# Calculations\n\n#f = (P/(2*t))*sqrt(E/p); \nE = p*4*(k)**2; # Youngs modulus in N/m^2\nt = (float(P)/(2*f))*math.sqrt(E/p); \n\n\n# Result\nprint 'Youngs modulus of quartz plate %3.5g'%E,'Nm^-2'\nprint 'Thickness of the crystal %.4e'%t,'m';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Youngs modulus of quartz plate 8.7641e+10 Nm^-2\nThickness of the crystal 2.2077e-03 m\n" + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Engineering_Physics_-_I/chapter2.ipynb b/Engineering_Physics_-_I/chapter2.ipynb new file mode 100755 index 00000000..84198c1f --- /dev/null +++ b/Engineering_Physics_-_I/chapter2.ipynb @@ -0,0 +1,293 @@ +{ + "metadata": { + "name": "Chapter 2(s)" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 2: Lasers" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1, Page No: 2.40" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding Number of electron hole pairs\n\nimport math\n\n# Variable Declaration\nA = 4*10**-6; # Receiving area of photo detector\nI = 200; # Intensity in W/m^2\nh = 6.625*10**-34; # planck's constant\nc = 3*10**8; # vel. of light in m/s\nlamda = 0.4*10**-6; # wavelength of light in m\n\n#Calculations\nv = c/lamda; # frequency\nNOP = I*A/(h*v) # number of photons\n\n#since each photon generates an electron hole pair, the number of photons is equal to number of electron hole pairs\n\n# Result\n\nprint 'Number of electron hole pairs = %3.2e '%NOP;\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Number of electron hole pairs = 1.61e+15 \n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2, Page No:2.40" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding Wavelength\n\nimport math\n\n# Variable Declaration\nEg = 2.8; # bandgap energy in eV\nh = 6.625*10**-34; # plank's constant\nc = 3*10**8; # vel. of light in m/s\nq = 1.602*10**-19; # charge of electron\n\n# Calculations\nE = Eg*q # eV to joules conversion\nlamda = h*c/E; # wavelength\n\n#Result\nprint 'wavelength = %3.1f' %(lamda*10**10), '\u00c5(Blue Colour)';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "wavelength = 4430.8 \u00c5(Blue Colour)\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3, Page No:2.41" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding Energy bandgap\n\nimport math;\n\n# Variable Declaration\nh = 6.625*10**-34; # plank's constant\nc = 3*10**8; # vel. of light in m/s\nlamda = 1.55*10**-6; # wavelength of light in m\nq = 1.6*10**-19; # charge of electron\n\n#Calculations\nEg = (h*c)/lamda; # band gap energy in joules\nE = Eg/q # bang gap energy in eV\n\n# Result\nprint 'Energy bandgap Eg = %3.4f'%E, 'eV';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Energy bandgap Eg = 0.8014 eV\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 4, Page No:2.41" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding Number of photons required to do one Joule of work\n\nimport math\n\n# Variable Declaration\nh = 6.625*10**-34; # plank's constant\nc = 3*10**8; # vel. of light in m/s\nlamda = 4961*10**-10; # wavelength of light in m\n\n# Calculations\nE = (h*c)/lamda; # energy in joules\nN = 1/E\n\n# Result\nprint 'Number of photons required to do one Joule of work = %3.4e'%N,'/m^3';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Number of photons required to do one Joule of work = 2.4961e+18 /m^3\n" + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5, Page No:2.41" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding Wavelength Limit\n# import math\n\n# Variable Declaration\nE = 0.02; # ionisation energy in eV\nh = 6.625*10**-34; # plank's constant\nc = 3*10**8; # vel. of light in m/s\nq = 1.6*10**-19; # charge of electron\n\n# Calculations\n\nlamda = h*c/(E*q) # long wavelength limit in m\n\n# Result\n\nprint 'long wavelength limit = %3.3e' %lamda,' m';\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "long wavelength limit = 6.211e-05 m\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6, Page No:2.42" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding Wavelength\n\nimport math;\n\n# Variable Declaration\nE = 1.44; # Bandgap energy in eV\nh = 6.625*10**-34; # plank's constant\nc = 3*10**8; # vel. of light in m/s\nq = 1.6*10**-19; # charge of electron\n\n# Calculations\n\nlamda = h*c/(E*q) # Wavelength of GaAs laser\n\n# Result\nprint 'Wavelength of GaAs laser = %3.1f'%(lamda*10**10),' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Wavelength of GaAs laser = 8626.3 \u00c5\n" + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 1, Page No:2.42" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding Energy of the first excited state\n\nimport math;\n\n# Variable Declaration\nh = 6.625*10**-34; # planck's constant\nc = 3*10**8; # vel. of light in m/s\nlamda = 5890*10**-10; # wavelength of light in m\nq = 1.6*10**-19; # charge of electron\n\n\n# Calculations\nEg = (h*c)/lamda; # energy in joules\nE = Eg/q # energy in eV\n\n# Result\nprint 'Energy of the first excited state = %3.3f' %E,'eV';\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Energy of the first excited state = 2.109 eV\n" + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 2, Page No:2.43" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding The ratio between the stimulated emission and apontaneous emission\n\nimport math;\n\n# Variable Declaration\nh = 6.625*10**-34; # planck's constant\nc = 3*10**8; # vel. of light in m/s\nlamda = 5890*10**-10; # wavelength of light in m\nk = 1.38*10**-23; # Boltzmann constant\nTc = 280 # Temperature in centigrades\n\n# Calculations\nT = Tc+273; # temperature in kelvin\nR = 1/((math.exp((h*c)/(k*T*lamda))) - 1); # ratio of stimulated emission to spontaneous emission\n\n# Result\nprint 'The ratio between the stimulated emission and apontaneous emission = %3.3e' %R;", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The ratio between the stimulated emission and apontaneous emission = 6.264e-20\n" + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 3, Page No:2.43" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding The No. of Photons emitted per minute\n\nimport math;\n\n# Variable Declaration\nh = 6.625*10**-34; # planck's constant\nc = 3*10**8; # vel. of light in m/s\nlamda = 6328*10**-10; # wavelength of He-Ne laser source in m\nq = 1.6*10**-19; # charge of electron\nP = 3*10**-3 # output power of the He-Ne source in watts or J/sec\n\n\n# Calculations\nv = c/lamda # frequency of the photon emitted by the laser beam\nE = h*v; # energy of a photon in joules\nPo = P*60; # conversion from J/sec to J/min\nN = Po/E; # No of photons emitted per minute \n\n# Result\nprint 'The No. of Photons emitted per minute = %3.3e' %N, 'photons/minute';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The No. of Photons emitted per minute = 5.731e+17 photons/minute\n" + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 4, Page No:2.44" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding The No. of Photons emitted per hour\n\nimport math;\n\n# Variable Declaration\nh = 6.625*10**-34; # planck's constant\nc = 3*10**8; # vel. of light in m/s\nlamda = 9.6*10**-6; # wavelength of CO2 laser source in m\nq = 1.6*10**-19; # charge of electron\nP = 10*10**3 # output power of the CO2 laser source in watts or J/sec\n\n\n# Calculations\nv = c/lamda # frequency of the photon emitted by the laser beam\nE = h*v; # energy of a photon in joules\nPo = P*60*60; # conversion fro J/sec to J/hour\nN = Po/E; # No of photons emitted per hour \n\n# Result\nprint 'The No. of Photons emitted per hour = %3.3e'%N,' photons/hour';\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The No. of Photons emitted per hour = 1.739e+27 photons/hour\n" + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 5, Page No:2.45" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nh = 6.625*10**-34; # planck's constant\nc = 3*10**8; # vel. of light in m/s\nlamda = 10*10**-2; # wavelength for microwave region in m\nT = 300 # Temperature in Kelvin\nKb = 1.38*10**-23 # Boltzmann constant\n\n# Calculations\n# let R = Rsp/Rst \nR = math.exp((h*c)/(lamda*Kb*T)) - 1 ; # ratio of spontaneous to stimulated emission\nif R<1:\n print 'Since the spontaneous emission is lesser than stimulated emission \\n hence MASER action is possible at thermal equilibrium'", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Since the spontaneous emission is lesser than stimulated emission \n hence MASER action is possible at thermal equilibrium\n" + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 6, Page No:2.45" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nh = 6.625*10**-34; # planck's constant\nc = 3*10**8; # vel. of light in m/s\nlamda = 5000*10*8-10; # wavelength for optical region in m\nT = 300 # Temperature in Kelvin\nKb = 1.38*10**-23 # Boltzmann constant\n\n# Calculations\n# let R = Rsp/Rst \nR = math.exp((h*c)/(lamda*Kb*T)) - 1; # ratio of spontaneous to stimulated emission\nif R<1:\n print 'Since the spontaneous emission is lesser than stimulated emission \\n hence LASER action is possible at thermal equilibrium'\nelse:\n print 'Since the spontaneous emission is more predominant than stimulated emission\\nhence LASER action is not possible at optical frequencies under thermal equilibrium';\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Since the spontaneous emission is lesser than stimulated emission \n hence LASER action is possible at thermal equilibrium\n" + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 7, Page No:2.46" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n# Variable Declaration\nh = 6.625*10**-34; # plank's constant\nc = 3*10**8; # vel. of light in m/s\nlamda = 5511.11*10**-10; # wavelength of green LED light in m\nq = 1.6*10**-19; # charge of electron\n\n# Calculations\nEg = (h*c)/lamda; # band gap energy in joules\nE = Eg/q # bang gap energy in eV\n\n# Result\nprint ' Eg = %3.2f' %E,'eV';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " Eg = 2.25 eV\n" + } + ], + "prompt_number": 38 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Engineering_Physics_-_I/chapter3.ipynb b/Engineering_Physics_-_I/chapter3.ipynb new file mode 100755 index 00000000..068dcc95 --- /dev/null +++ b/Engineering_Physics_-_I/chapter3.ipynb @@ -0,0 +1,209 @@ +{ + "metadata": { + "name": "chapter 3(s)" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 3: Fibre Optics And Applications" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1, Page No: 3.40" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding Numerical aperture\n\nimport math;\n\n# Variable Declaration\nn1 = 1.6; # Refractive index of core\nn2 = 1.5; # Refractive index of cladding\n\n# Calculations\nNA = math.sqrt(n1**2 - n2**2); # Numerical Aperture of optical fiber\n\n# Result\nprint 'Numerical Aperture of the optical fiber = %3.4f' %NA;", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Numerical Aperture of the optical fiber = 0.5568\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2, Page No: 3.40" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding Numerical aperture and acceptance Angle\n\nimport math;\n\n# Variable declaration\nn1 = 1.55; # Refractive index of core\nn2 = 1.5; # Refractive index of cladding\n\n# Calculations\nNA = math.sqrt(n1**2 - n2**2); # Numerical Aperture of optical fiber\nim = math.asin(NA); # Acceptance angle\nim_d = im*180/math.pi # radian to degree conversion\n\n# Result\nprint 'Numerical Aperture of the optical fiber = %3.4f'%NA,' Acceptance angle = %3.2f' %im_d,'degrees ';\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Numerical Aperture of the optical fiber = 0.3905 Acceptance angle = 22.99 degrees \n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": " Example 3, Page No: 3.41" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding Refractive index of cladding\n\nimport math;\n\n# Variable declaration\nNA = 0.26; # Numerical aperture \nn1 = 1.5 ; # Refractive index of core\nd = 100*10^-6; # diameter of the core in m\n\n# Calculations\nn2 = math.sqrt( n1**2 - NA**2); # Refractive index of cladding\n\n# Result\nprint 'Refractive index of cladding = %3.4f' %n2;", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Refractive index of cladding = 1.4773\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 4, Page No: 3.41" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding Numerical aperture\n\nimport math;\n\n# variable Declaration\nn1 = 1.54; # Refractive index of core\nn2 = 1.5; # Refractive index of cladding\n\n# Calculations\nNA = math.sqrt(n1**2 - n2**2); # Numerical Aperture of optical fiber\n\n# Result\nprint 'Numerical Aperture of the optical fiber = %3.4f' %NA", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Numerical Aperture of the optical fiber = 0.3487\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 1, Page No: 3.42" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding Refractive index, Acceptance angle, Maximum number of modes that fibre allows\n\nimport math;\n\n# Variable Declaration\nn1 = 1.5; # Refractive index of core\nNA = float(0.26); # Numerical aperture \nd = 100*10**-6 # core diameter\nlamda = float(10**-6); # wavelength in m\n\n# Calculations\nn2 = math.sqrt( n1**2 - NA**2); # Refractive index of cladding\nim = math.asin(NA); # Acceptance angle\nim_d = im*180/math.pi # radian to degree conversion\nN = 4.9*(d*NA/lamda)**2; # maximum no of modes\n\n# Result\nprint ' Refractive index of cladding n2 = %3.4f' %n2,'\\n Acceptance angle = %3.2f' %im_d, 'degrees','\\n Maximum number of modes that fibre allows = %d '%N;\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " Refractive index of cladding n2 = 1.4773 \n Acceptance angle = 15.07 degrees \n Maximum number of modes that fibre allows = 3312 \n" + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 2, Page No: 3.43" + }, + { + "cell_type": "code", + "collapsed": false, + "input": " # Finding Numerical aperture and Critical angle\n\nimport math;\n\n# Varible Declaration\ndelta = 0.02; # relative refractive index\nn1 = 1.48; # refractive index of core\n\n# Calculations\nNA = n1*(2*delta)**0.5; # Numerical aperture\nn2 = math.sqrt( n1**2 - NA**2); # Refractive index of cladding\ncri_ang = math.asin(n2/n1); # critical angle\ncri_ang_d = cri_ang*180/math.pi; # critical angle in degrees\n\n# Result\nprint ' Numerical Aperture = %3.3f'%NA,'\\n The Critical angle = %3.2f' %cri_ang_d,' degrees';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " Numerical Aperture = 0.296 \n The Critical angle = 78.46 degrees\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 3, Page No: 3.43" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding Refractive Index\n\nimport math\n\n# Variable declaration\ndelta = 0.015; # relative refractive index\nNA = 0.27; # Numerical aperture\n\n# Calculations\n# we know that NA = n1*sqrt(2*\u0394)\nn1 = NA/math.sqrt(2*delta) # refractive index of core\nn2 = math.sqrt( n1**2 - NA**2); # Refractive index of cladding\n\n# Result\nprint ' Refractive index of the core = %3.3f' %n1,'\\n Refractive index of the cladding = %3.3f\\n' %n2;", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " Refractive index of the core = 1.559 \n Refractive index of the cladding = 1.535\n\n" + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 4, Page No: 3.44" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# No. of total modes propagating in a multimode step index fibre\n\nimport math;\n\n# variable Declaration\nNA = 0.25; # Numerical aperture \nd = 60*10**-6 # core diameter\nlamda = 2.7*10**-6; # wavelength in m\n\n# calculations\nN = 4.9*(d*NA/lamda)**2; # no of modes for step index fibre\n\n# Result\nprint 'No. of total modes propagating in a multimode step index fibre = %d' %N;", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "No. of total modes propagating in a multimode step index fibre = 151\n" + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 5, Page No: 3.44" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Finding No. of total modes propagating in the fibre\nimport math;\n\n# Variable Declaration\nNA = 0.25; # Numerical aperture \nd = 6*10**-6 # core diameter\nlamda = 1.5*10**-6; # wavelength of laser source\nn1 = 1.47; # refractive index of core\nn2 = 1.43 # refractive index of cladding\n\n# calculations\nNA = math.sqrt( n1**2 - n2**2); # Numerical Aperture\nN = 4.9*(d*NA/lamda)**2; # no of modes for step index fibre\n\n# Result\nprint 'No. of total modes propagating in the fibre = %d' %N;", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "No. of total modes propagating in the fibre = 9\n" + } + ], + "prompt_number": 30 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Engineering_Physics_-_I/chapter4.ipynb b/Engineering_Physics_-_I/chapter4.ipynb new file mode 100755 index 00000000..995bbd50 --- /dev/null +++ b/Engineering_Physics_-_I/chapter4.ipynb @@ -0,0 +1,532 @@ +{ + "metadata": { + "name": "chapter4s" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 4: Quantum Physics" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1, Page No: 4.52" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "\nimport math;\n\n# Variable Declaration\nlamda = 3*10**-10; # wavelength of incident photons in m\ntheta = 60; # viewing angle in degrees\nh = 6.625*10**-34 # plancks constant\nmo = 9.11*10**-31 # mass in Kg\nc = 3*10**8; # vel. of light \n\n# Calculatioms\n# from Compton theory ,Compton shift is given by\n# lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\n\ntheta_r = theta*math.pi/180; # degree to radian conversion\nlamda1 = lamda+( (h/(mo*c))*(1-math.cos(theta_r))) # wavelength of scattered photons\n\n# Result\nprint 'Wavelength of Scattered photons = %3.4f'%(lamda1*10**10),'\u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Wavelength of Scattered photons = 3.0121 \u00c5\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2, Page No:4.52" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "\nimport math;\n\n# Variable declaration\ntheta = 135; # angle in degrees\nh = 6.625*10**-34 # plancks constant\nmo = 9.1*10**-31 # mass in Kg\nc = 3*10**8; # vel. of light in m/s\n\n# Calculatioms\n# from Compton theory ,Compton shift is given by\n# lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\ntheta_r = theta*math.pi/180; # degree to radian conversion\nc_lamda = ( (h/(mo*c))*(1-math.cos(theta_r))) # Change in wavelength in m\n\n# Result\nprint 'Change in Wavelength = %3.5f' %(c_lamda*10**10),' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Change in Wavelength = 0.04143 \u00c5\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3, Page No:4.53" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "\nimport math;\n\n# Variable Declaration\nlamda = 0.1*10**-9; # wavelength of X-rays in m\ntheta = 90; # angle with incident beam in degrees\nh = 6.625*10**-34 # plancks constant\nmo = 9.11*10**-31 # mass in Kg\nc = 3*10**8; # vel. of light \n\n# Calculatioms\n# from Compton theory ,Compton shift is given by\n# lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\ntheta_r = theta*math.pi/180; # degree to radian conversion\nlamda1 = lamda+( (h/(mo*c))*(1-math.cos(theta_r))) #wavelength of scattered beam\n\n# Result\nprint 'Wavelength of Scattered beam = %3.4f' %(lamda1*10**10),' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Wavelength of Scattered beam = 1.0242 \u00c5\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 4, Page No:4.53" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "\nimport math;\n\n# Variable Declaration\nh = 6.625*10**-34 # plancks constant\nm = 9.11*10**-31 # mass of electron in Kg\ne = 1.6*10**-19 # charge of electron\nV = 150; # potential difference in volts\n\n# Calculations\n\nlamda = h/(math.sqrt(2*m*e*V)) # de Broglie wavelength\n\n#Result\nprint 'The de-Broglie wavelength = %d' %(lamda*10**10), '\u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The de-Broglie wavelength = 1 \u00c5\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5, Page No:4.54" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "\nimport math;\n\n# Variable Declaration\nh = 6.625*10**-34 # plancks constant\nm = 9.11*10**-31 # mass of electron in Kg\ne = 1.6*10**-19 # charge of electron\nV = 5000; # potential in volts\n\n# Calculations\n\nlamda = h/(math.sqrt(2*m*e*V)) #de Broglie wavelength\n\n# Result\nprint 'The de-Broglie wavelength of electron = %3.5f' %(lamda*10**10),' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The de-Broglie wavelength of electron = 0.17353 \u00c5\n" + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6, Page No:4.55" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "\nimport math;\n\n# Variable Declaration\nE = 100 # Energy of electron in eV\nh = 6.625*10**-34 # plancks constant\nm = 9.11*10**-31 # mass of electron in Kg\ne = 1.6*10**-19 # Charge of electron in Columbs\n\n# Calculations\n\nE1 = E*e # Energy conversion from eV to Joule\nlamda = h/(math.sqrt(2*m*E1)) # de Broglie wavelength\n\n# Result\nprint 'The de-Broglie wavelength = %3.3f' %(lamda*10**10),' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The de-Broglie wavelength = 1.227 \u00c5\n" + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7, Page No:4.55" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "\nimport math;\n\n# Variable Declaration\nm = 1.675*10**-27; # Mass of proton in kg\nc = 3*10**8; # velocity of light in m/s\nh = 6.625*10**-34 # plancks constant\n\n# Calculations\n\nvp = c/20; # velocity of proton in m/s\nlamda = h/(m*vp) # de-Broglie wavelength in m\n\n# Result\nprint 'de-Broglie wavelength = %e'%(lamda),'lamda';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "de-Broglie wavelength = 2.636816e-14 lamda\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 8, Page No:4.56" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "\nimport math;\n\n# Variable declaration\nE = 10000 # Energy of neutron in eV\nh = 6.625*10**-34 # plancks constant\nm = 1.675*10**-27 # mass of neutron in Kg\ne = 1.6*10**-19 \n\n# Calculations\n\nE1 = E*e # Energy conversion from eV to Joule\nlamda = h/(math.sqrt(2*m*E1)) # de Broglie wavelength\n\n# Result\nprint 'The de-Broglie wavelength of neutron = %3.3e' %lamda,' m';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The de-Broglie wavelength of neutron = 2.862e-13 m\n" + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10, Page No:4.58" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable decalaration\nl = 0.1*10**-9; # side of cubical box\nh = 6.625*10**-34 # plancks constant in Jsec\nm = 9.11*10**-31 # mass of electron in Kg\nKb = 1.38*10**-23 # Boltzmann constant \n\n# Calculations\n# for cubical box the energy eigen value is Enx ny nz = (h^2/(8*m*l^2))*(nx^2 + ny^2 +nz^2)\n# For the next energy level to the lowest energy level nx = 1 , ny = 1 and nz = 2\nnx = 1\nny = 1\nnz = 2\nE112 = (h**2/(8*m*l**2))*( nx**2 + ny**2 + nz**2);\n\n# We know the average energy of molecules of aperfect gas = (3/2)*(Kb*T)\nT = (2*E112)/(3*Kb); # Temperature in kelvin\n\n# Result\nprint 'E112 = %3.4e' %E112,'Joules'',\\n','Temperature of the molecules T = %3.4e' %T, 'K';\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "E112 = 3.6134e-17 Joules,\nTemperature of the molecules T = 1.7456e+06 K\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 11, Page No:4.59" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable declaration\nl = 4*10**-9; # width of infinitely deep potential\nh = 6.625*10**-34 # plancks constant in Jsec\nm = 9.11*10**-31 # mass of electron in Kg\nn = 1; # minimum energy\ne = 1.6*10**-19 # charge of electron in columbs\n\n# Calculations\nE = (h**2 * n**2)/(8*m*l**2) # Energy of electron in an infinitely deep potential well \nE1 = E/e #energy conversion from joules to eV\n\n# Result\nprint 'Minimum energy of an electron = %3.4f' %E1,' eV';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Minimum energy of an electron = 0.0235 eV\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12, Page No:4.61" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nl = 0.1*10**-9; # length of one dimensional box \nh = 6.625*10**-34 # plancks constant in Jsec\nm = 9.11*10**-31 # mass of electron in Kg\nn = 1; # for ground state\nn5 = 6; # n value for fifth excited state\ne = 1.6*10**-19 # charge of electron in columbs\n\n# Calculations\nEg = (h**2 * n**2)/(8*m*l**2 * e ) # Energy in ground state in eV \nEe = (h**2 * n5**2)/(8*m*l**2 * e) # Energy in excited state in eV\nE = Ee - Eg; # energy req to excite electrons from ground state to fifth excited state\n\n# Result\nprint 'Energy required to excite an electron from ground state to fifth excited state = %3.2f' %E, 'eV';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Energy required to excite an electron from ground state to fifth excited state = 1317.38 eV\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13, Page No:4.62" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable decalration\nl = 0.1*10**-9; # length of one dimensional box \nh = 6.625*10**-34 # plancks constant in Jsec\nm = 9.11*10**-31 # mass of electron in Kg\nn = 1; # for ground state\ne = 1.6*10**-19 # charge of electron in columbs\n\n# Calculations\nE = (h**2 * n**2)/(8*m*l**2 *e ) # Energy of electron in eV \n\n# Result\nprint 'Energy of an electron = %3.3f' %E,' eV';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Energy of an electron = 37.639 eV\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 14, Page No:4.63" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable declaration\nl = 0.5*10**-9; # width of one dimensional box in m \nh = 6.625*10**-34 # plancks constant in Jsec\nm = 9.11*10**-31 # mass of electron in Kg\nn = 1; # for ground state\ne = 1.6*10**-19 # charge of electron in columbs\n\n# Calculations\nE = (h**2 * n**2)/(8*m*l**2 *e ) # Energy of electron in eV \n\n# Result\nprint 'Least Energy of an electron = %3.4f' %E,' eV';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Least Energy of an electron = 1.5056 eV\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 1, Page No:4.64" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# variable declaration\nh = 6.625*10**-34 # plancks constant\nc = 3*10**8; # vel. of light\nlamda = 5893*10**-10; # wavelength in m\nP = 100 # power of sodium vapour lamp\n\n# Calculations\nE = (h*c)/lamda; # Energy in joules\nN = P/E # Number of photons emitted\n\n# Result\nprint 'Number of Photons emitted = %3.4e' %N,' per second';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Number of Photons emitted = 2.9650e+20 per second\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 2, Page No:4.64" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable declaration\nlamda1 = 0.022*10**-10; # wavelength of scatterd X-rays in m\ntheta = 45; # scatterring angle in degrees\nh = 6.625*10**-34 # plancks constant\nmo = 9.11*10**-31 # mass in Kg\nc = 3*10**8; # vel. of light \n\n# Calculatioms\n# from Compton theory ,Compton shift is given by\n# lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\n\ntheta_r = theta*math.pi/180; # degree to radian conversion\nlamda = lamda1-( (h/(mo*c))*(1-math.cos(theta_r))) # incident Wavelength\n\n# Result\nprint 'Wavelength of incident beam = %3.4f' %(lamda*10**10),' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Wavelength of incident beam = 0.0149 \u00c5\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 3 , Page No:4.65" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nEi = 1.02*10**6 # photon energy in eV\ntheta = 90; # scattered angle in degrees\nh = 6.625*10**-34 # plancks constant\nmo = 9.1*10**-31 # mass of electron in Kg\ne = 1.6*10**-19 # charge of electron\nc = 3*10**8; # vel. of light in m/s\n\n# Calculations\n# from Compton theory ,Compton shift is given by\n# lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\ntheta_r = theta*math.pi/180; # degree to radian conversion\nc_lamda = ( (h/(mo*c))*(1-math.cos(theta_r))) #Change in wavelength in m\ndv = c/c_lamda; # change in frequency of the scattered photon\ndE = (h*dv)/e # change in energy of scattered photon in eV\n# This change in energy is transferred as the KE of the recoil electron\nEr = dE; # Energy of recoil electron\nEs = Ei - Er # Energy of scattered photon\n\n\n# Result\nprint 'Energy of the recoil electron = %3.4f' %(Er*10**-6),' MeV','\\n','Energy of the Scattered photon = %3.4f' %(Es*10**-6),'MeV';\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Energy of the recoil electron = 0.5119 MeV \nEnergy of the Scattered photon = 0.5081 MeV\n" + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 4, Page No:4.65" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nlamda = 0.124*10**-10; # wavelength of X-rays in m\ntheta = 180; # Scattering angle in degrees\nh = 6.625*10**-34 # plancks constant\nmo = 9.11*10**-31 # mass in Kg\nc = 3*10**8; # vel. of light \n\n# Calculatioms\n# from Compton theory ,Compton shift is given by\n# lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\n\ntheta_r = theta*math.pi/180; # degree to radian conversion\nlamda1 = lamda+( (h/(mo*c))*(1-math.cos(theta_r))) # wavelength of scattered X-rays\n\n# Result\nprint 'Wavelength of Scattered X-rays = %3.4f' %(lamda1*10**10),' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Wavelength of Scattered X-rays = 0.1725 \u00c5\n" + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 5, Page No:4.65" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nh = 6.625*10**-34 # plancks constant\nm = 9.11*10**-31 # mass of electron in Kg\ne = 1.6*10**-19 # charge of electron\nV = 2000; # potential in volts\n\n# Calculations\n\nlamda = h/(math.sqrt(2*m*e*V)) # de Broglie wavelength\n\n# Result\nprint 'The de-Broglie wavelength of electron = %3.4f' %(lamda*10**10),' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The de-Broglie wavelength of electron = 0.2744 \u00c5\n" + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 6, Page No:4.66" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# variable Declaration \nh = 6.625*10**-34 # plancks constant\nm = 1.678*10**-27 # mass of proton in Kg\ne = 1.6*10**-19 # charge of electron\nKb = 1.38*10**-23; # boltzmann constant\nT = 300 # Temperature in kelvin\n\n#Calculations\nlamda = h/(math.sqrt(3*m*Kb*T)) #de Broglie wavelength\n\n#Result\nprint 'The de-Broglie wavelength = %3.4f' %(lamda*10**10),' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The de-Broglie wavelength = 1.4512 \u00c5\n" + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 7, Page No:4.66" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nh = 6.625*10**-34 # plancks constant\nm = 9.11*10**-31 # mass of electron in Kg\nlamda = 3*10**-2; # wavelength of electron wave\ne = 1.6*10**-19; # charge of electron\n\n# Calculations\nE = (h**2)/(2*m*lamda**2); # Energy in Joules\nE1 = E/e;\n\n# Result\nprint 'Energy of the electron E = %3.4e' %E1,'eV';\nprint 'Note: Calculation mistake in textbook'", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Energy of the electron E = 1.6729e-15 eV\nNote: Calculation mistake in textbook\n" + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 8, Page No:4.67" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable declaration\nh = 6.625*10**-34 # plancks constant\nm = 9.11*10**-31 # mass of electron in Kg\nc = 3*10**8; # velocity of light in m/s\n\n# Calculations\nve = 0.7071*c # velocity of electron\nlamda = h/(m*ve*math.sqrt(1-(ve/c)**2)) #de Broglie wavelength\n\n# we know Compton wavelength ,lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\n# maximum shift \u03b8 = 180\ntheta = 180\ntheta1 = theta*math.pi/180;\nd_lamda = (h/(m*c))*(1-math.cos(theta1))\nprint 'de Broglie wavelength = %e' %lamda,' m';\nprint 'compton wavelength = %e' %d_lamda,'m';\nprint 'The de-Broglie wacelength is equal to the compton wavelength';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "de Broglie wavelength = 4.848152e-12 m\ncompton wavelength = 4.848152e-12 m\nThe de-Broglie wacelength is equal to the compton wavelength\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 9, Page no:4.68" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nl = 10**-10; # side of one dimensional box \nh = 6.625*10**-34 # plancks constant in Jsec\nm = 9.11*10**-31 # mass of electron in Kg\nn1 = 1; # for 1st eigen value\nn2 = 2; # for 2nd eigen value\nn3 = 3; # for 3rd eigen value\nn4 = 4; # for 4th eigen value\ne = 1.6*10**-19 # charge of electron in columbs\n\n# Calculations\nE1 = (h**2 * n1**2)/(8*m*l**2 *e ) #first Eigen value\nE2 = (h**2 * n2**2)/(8*m*l**2 *e ) # second Eigen value\nE3 = (h**2 * n3**2)/(8*m*l**2 *e ) # third Eigen value\nE4 = (h**2 * n4**2)/(8*m*l**2 *e ) # fourth Eigen value\n \n# Result\nprint '1st Eigen value = %3.1f' %E1,'eV';\nprint '2nd Eigen value = %3.1f' %E2,'eV';\nprint '3rd Eigen value = %3.1f' %E3,'eV';\nprint '4th Eigen value = %3.1f' %E4,'eV';\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1st Eigen value = 37.6 eV\n2nd Eigen value = 150.6 eV\n3rd Eigen value = 338.8 eV\n4th Eigen value = 602.2 eV\n" + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 10 , Page No:4.68" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nl = 10**-10 ; # length of one dimensional box in m \nh = 6.625*10**-34 # plancks constant in Jsec\nm = 9.11*10**-31 # mass of electron in Kg\nn = 1; # for ground state\ne = 1.6*10**-19 # charge of electron in columbs\n\n# Calculations\nE = 2*(h**2 * n**2)/(8*m*l**2 *e ) #Energy of system having two electrons\n\n# Result\nprint 'Energy of the system having two electrons = %3.4f' %E,' eV';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Energy of the system having two electrons = 75.2789 eV\n" + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 11 , Page No:4.69" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nb = 40; # angle subtended by final images at eye in degrees\na = 10 # angle subtended by the object at the eye kept at near point in degrees\n\n# Calculations\nb_r = b*math.pi/180; # degree to radian conversion\na_r = a*math.pi/180; # degree to radian conversion\nM = math.tan(b_r)/math.tan(a_r); # magnifying power\n\n#Result\nprint 'Magnifying power = %3.3f' %M;", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Magnifying power = 4.759\n" + } + ], + "prompt_number": 38 + }, + { + "cell_type": "code", + "collapsed": false, + "input": "", + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Engineering_Physics_-_I/chapter5.ipynb b/Engineering_Physics_-_I/chapter5.ipynb new file mode 100755 index 00000000..78a1015e --- /dev/null +++ b/Engineering_Physics_-_I/chapter5.ipynb @@ -0,0 +1,700 @@ +{ + "metadata": { + "name": "Chapter5(s)" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 5: Crystal Physics" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1, Page no:5.72" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\n\n#Copper has FCC structure\nr = 1.273; # Atomic radius in angstrom\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 63.5; # Atomic weight of copper in grams\nn = 4; # No. of atoms per unit cell for FCC\n\n# Calculations\nr1 = r*10**-10; # Radius conversion from angstrom to m\na = (4*r1)/math.sqrt(2); # lattice parameter for FCC\np = (n*A)/(N*a**3); # Density of copper\n\n# Result\nprint 'Lattice Constant a = %3.1e' %a,' m','\\n', 'Density of copper = %3.1f' %p,' kg/m^3';\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Lattice Constant a = 3.6e-10 m \nDensity of copper = 9034.4 kg/m^3\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2, Page no:5.73" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\n# given intercepts 3,4 and \u221e, the recipocals of intercepts is\n# (1/3):(1/4):(1/\u221e)\n# LCM = 12\n# multiplying by LCM we get miller indices\n# miller indices of a plane are the smallest integers of the reciprocals of its intercerpts\n# therefore miller indices(h k l) is (4 3 0);\n\nh = 4; # miller indice\nk = 3; # miller indice\nl = 0; # miller indice\na = 2; # primitive vector of lattice in angstrom\n\n#Calculations\n\ndhkl = a/math.sqrt((h**2)+(k**2)+(l**2)); #interplanar distance\n\n# Result\nprint 'Miller indices = (4 3 0)';\nprint 'The interplanar distance d = %3.1f' %dhkl,' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Miller indices = (4 3 0)\nThe interplanar distance d = 0.4 \u00c5\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3, Page no:5.74" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\n# \u03b1-Iron solidifies to BCC structure\n\nr = 1.273; # Atomic radius in angstrom\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 55.85; # Atomic weight of \u03b1-Iron in kilograms\nn = 2; # No. of atoms per unit cell for BCC\np = 7860; # density in kg/m^-3\n\n#Calculations\n\n# p = (n*A)/(N*a^3); density\n\na = ((n*A)/(N*p))**(0.333); # lattice constant\na1 = a*10**10; # m to angstrom conversion\nr = (a1*math.sqrt(3))/4 # atomic radius for BCC\n\n#Output\nprint 'The Radius of the atom = %3.5f' %r,' \u00c5'\nprint 'Note : atomic wt taken as 55.58*10^-3 instead of 55.85 in calculation'", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The Radius of the atom = 1.26955 \u00c5\nNote : atomic wt taken as 55.58*10^-3 instead of 55.85 in calculation\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 4, Page no:5.75" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nlamda = 1.5418; # wavelength in \u00c5\nh = 1; # miller indice\nk = 1; # miller indice\nl = 1; # miller indice\nn = 1; # given first order\ntheta = 30; # diffraction angle in degrees\n\n# Calculations\ntheta1 = theta*math.pi/180; # degree to radian conversion\n# d = (n*lamda)/(2*sin\u03b8); by Braggs law ------------- 1\n# d = a/sqrt((h^2)+(k^2)+(l^2)); interplanar distance ------------ 2\n# equating 1 and 2\n\na = (n*lamda*math.sqrt((h**2)+(k**2)+(l**2))/(2*math.sin(theta1)))\n\n# Result\nprint 'Interatomic spacing a = %f \u00c5' %a;", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Interatomic spacing a = 2.670476 \u00c5\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5, page no:5.76" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# variable Declaration\nh1 = 1; # miller indice\nk1 = 1; # miller indice\nl1 = 1; # miller indice\nh0 = 0; # miller indice\nk0 = 0; # miller indice\nl0 = 0; # miller indice\n\n# Calculations\n# dhkl = a/sqrt((h^2)+(k^2)+(l^2)); // interplanar distance\n# assume a = 1(constant) for easier calculation in scilab\n\na = 1;\nd100 = a/math.sqrt((h1**2)+(k0**2)+(l0**2)); # interplanar distance\nd110 = a/math.sqrt((h1**2)+(k1**2)+(l0**2)); # interplanar distance\nd111 = a/math.sqrt((h1**2)+(k1**2)+(l1**2)); # interplanar distance\n\n# Result\nprint 'd100 : d110 : d111 = ','%d ' %d100,':','%3.2f' %d110,':', '%3.2f' %d111;", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "d100 : d110 : d111 = 1 : 0.71 : 0.58\n" + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6, page no:5.76" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\n# Aluminium is FCC\n\na = 0.405*10**-9; # lattice constant of aluminium\nt = 0.005*10**-2; # thickness of aluminium foil in m\ns = 25*10**-2; # side of square in m\n\n# Calculations\nVUC = a**3; # volume of unit cell\nVal = (s**2)*t # volume of aluminium foil (area*thickness)\nN = Val/VUC # Number if unit cells\n\n# Result\nprint 'Number of unit cells = %3.3e' %N", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Number of unit cells = 4.704e+22\n" + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7, page no:5.77" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# variable declaration\n# metallic iron changes from BCC to FCC form at 910 degress\nrb = 0.1258*10**-9; # atomic radius of BCC iron atom\nrf = 0.1292*10**-9; # atomic radius of FCC iron atom\n\n# Calculations\n\nab = (4*rb)/(math.sqrt(3)); # lattice constant for BCC\nVbcc = (ab**3)/2; # volume occupied by one BCC atom\naf = (4*rf)/(math.sqrt(2)) # lattice constant for FCC\nVfcc = (af**3)/4; # volume occupied by one FCC atom\ndv = ((Vbcc-Vfcc)/Vbcc)*100 # percentage change in volume\n\n# Result\nprint 'During the structural change the percentage change in volume = %3.4f' %dv;", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "During the structural change the percentage change in volume = 0.4933\n" + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 8, page no:5.78" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n#variable declaration\n#Copper Crystallines in FCC structure\n\np = 8960; # Density of copper in kg/m^3\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 63.5; # Atomic weight of copper in kg/mol\nn = 4; # No. of atoms per unit cell for FCC\n\n# Calculations\n\na = ((n*A)/(N*p))**(0.333);\n\n# Result\n\nprint 'Lattice Constant a = %3.4f' %(a*10**10),' \u00c5';\nprint 'atomic wt of copper is taken as 63.5*10^-3 instead of 63.5 in textbook';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Lattice Constant a = 3.6899 \u00c5\natomic wt of copper is taken as 63.5*10^-3 instead of 63.5 in textbook\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9, page no:5.79" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable declaration\n# (100) planes in rock crystal\nh = 1; # miller indice\nk = 0; # miller indice\nl = 0; # miller indice\na = 2.814 # lattice constant in \u00c5\n\n# Calculations\ndhkl = a/math.sqrt((h**2)+(k**2)+(l**2)); # interplanar distance\n\n#Result\nprint 'd-spacing for (100) plane in rock salt = %3.3f' %dhkl,' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "d-spacing for (100) plane in rock salt = 2.814 \u00c5\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10, page no:5.79" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\n# FCC structured crystal\n\np = 6250; # Density of crystal in kg/m^3\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 60.2; # molecular weight\nn = 4; # No. of atoms per unit cell for FCC\n\n# Calculations\n\na = ((n*A)/(N*p))**(0.333);\n\n# Result\n\nprint 'Lattice Constant a = %3.3e' %a, 'm';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Lattice Constant a = 4.087e-10 m\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 11, page no:5.80" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\n# (321) plane in simple cubic lattice\nh = 3; # miller indice\nk = 2; # miller indice\nl = 1; # miller indice\na = 4.12 # inter atomic space \u00c5\n\n# Calculations\ndhkl = a/math.sqrt((h**2)+(k**2)+(l**2)); #interplanar distance\n\n# Result\nprint 'd = %3.2f' %dhkl,' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "d = 1.10 \u00c5\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12, page no:5.81" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\n# BCC structured crystal\n\np = 7860; # Density of iron in kg/m^3\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 55.85; # Atomic weight\nn = 2; # No. of atoms per unit cell for BCC\n\n# Calculations\n\na = ((n*A)/(N*p))**(0.333); #lattice constant\n\n# Result\n\nprint 'Lattice Constant of Fe = %3.3f' %(a*10**10),' \u00c5 ';\nprint 'Note: density of iron is taken as 7.86 instead of 7860 in calculation'", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Lattice Constant of Fe = 2.932 \u00c5 \nNote: density of iron is taken as 7.86 instead of 7860 in calculation\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 14, page no:5.82" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# variable Declaration\nr = 0.123*10**-10; # Radius of the atom\n\n# Calculations\na = (4*r)/math.sqrt(3); # Lattice constant in m For a BCC structure\nV = a*a*a; # Volume of BCC\n\n# Result\nprint 'Volume of the unit cell = %3.4e' %V,' m^3';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Volume of the unit cell = 2.2920e-32 m^3\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 15, page no:5.82" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable declaration\na = 0.05; # unit cell edge of an orthorhombic crystal in nm\nb = 0.05; # unit cell edge of an orthorhombic crystal in nm\nc = 0.03; # unit cell edge of an orthorhombic crystal in nm\nIa = 0.025 # intercept on 'a' in nm\nIb = 0.02 # intercept on 'b' in nm\nIc = 0.01 # intercept on 'c' in nm\n\n# Calculations\n\nh = a/Ia; # miller indice h\nk = b/Ib; # miller indice k\nl = c/Ic # miller indice l\n\n# Result\nprint 'Miller indices (h k l) =', '%d' %h,'%d' %k, '%d' %l; ", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Miller indices (h k l) = 2 2 3\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 16, page no:5.83" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Magnesium has HCP structure\n# for HCF(Hexagonal closed packed structure) consider the relation between 'c' and 'a';\n# c/a = sqrt(8/3) = 1.6329 \n\n# Variable Declaration\nr = 0.1605*10**-9; # radius of magnesium atom in m\n\n# Calculations\n\na = 2*r # lattice constant of HCP\nc = a*math.sqrt(float(8)/3); # relation b/w c and a in HCP\nV = (3*(3**0.5))*(a*a*c)/2; #Volume of unit cell in m^3\n\n# Result\nprint 'Volume of the unit cell of magnesium = %g' %V,' m^3';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Volume of the unit cell of magnesium = 1.4033e-28 m^3\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 17, page no:5.84" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n# Variable Declaration\n# (101),(221) planes in simple cubic lattice\nh1 = 1; # miller indice\nk0 = 0; # miller indice\nl1 = 1; # miller indice\nh2 = 2; # miller indice\nk2 = 2; # miller indice\nl1 = 1; # miller indice\na = 4.2 # inter atomic space \u00c5\n\n# Calculations\n\nd101 = a/math.sqrt((h1**2)+(k0**2)+(l1**2)); # interplanar distance\nd221 = a/math.sqrt((h2**2)+(k2**2)+(l1**2)); # interplanar distance\n\n\n# Result\nprint 'd(101) = %3.4f' %d101,' \u00c5','\\n','d(221) = %3.1f' %d221,' \u00c5 ';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "d(101) = 2.9698 \u00c5 \nd(221) = 1.4 \u00c5 \n" + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 1, page no:5.85" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable declaration\n\n# Copper has FCC structure\na = 3.6; #lattice parameter of copper in \u00c5\n\n# Calculations\n\nr = a*math.sqrt(2)/4; # atomic radius of copper\n\n# Result\nprint 'Atomic Radius of copper = %3.3f' %r,'\u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Atomic Radius of copper = 1.273 \u00c5\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 2, page no:5.85" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# variable Declaration\n\n# Copper has FCC structure\n\nr = 1.278; # Atomic radius in angstrom\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 63.54; # Atomic weight of copper \nn = 4; # No. of atoms per unit cell for FCC\n\n# Calculations\nr1 = r*10**-10; # Radius conversion from angstrom to m\na = (4*r1)/math.sqrt(2); # lattice parameter for FCC\np = (n*A)/(N*a**3); # Density of copper\n\n# Result\n\nprint ' Density of copper = %3.2f' %p,' kg/m^3';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " Density of copper = 8934.43 kg/m^3\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 3, page no:5.86" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nANa = 23; # atomic wt of sodiim\nACl = 35.45 # atomic wt of chlorine\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nn = 4 # No. of atoms per unit cell for FCC\np = 2180; # density in kg/m^-3\n\n# Calculations\n\n# p = (n*A)/(N*a^3); density\nA = ANa+ACl; # atomic wt of NaCl\na = ((n*A)/(N*p))**(0.33333); # lattice constant\nr = a/2 # Distance b/w two adjacent atoms\n\n# Result\nprint 'Distance between two adjacent atoms is r = %3.2e' %r,' m';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Distance between two adjacent atoms is r = 2.81e-10 m\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 4, page no:5.87" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nr = 1.273; # Atomic radius in angstrom\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 55.85 ; # Atomic weight of Fe \nn = 2; # No. of atoms per unit cell for BCC\np = 7860; # density in kg/m^-3\n\n# Calculations\n\n# p = (n*A)/(N*a^3); density\n\na = ((n*A)/(N*p))**(0.33333); # lattice constant\na1 = a*10**10; # m to angstrom conversion\nr = (a1*math.sqrt(3))/4 # atomic radius for BCC\n\n# Result\nprint 'The Radius of the Fe = %3.3f' %r,' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The Radius of the Fe = 1.242 \u00c5\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 5, page no:5.88" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 119; # Atomic weight of pottasium bromide\nn = 4; # No. of atoms per unit cell for FCC\np = 2700; # density in kg/m^-3\n\n# Calculations\n\n# p = (n*A)/(N*a^3); density\n\na = ((n*A)/(N*p))**(0.33333); # lattice constant\na1 = a*10**10; # m to angstrom conversion\n\n# Output\nprint 'Lattice constant = %3.1f' %a1,' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Lattice constant = 6.6 \u00c5\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 6, page no:5.88" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\na = 4.3*10**-10; # Lattice constant in \u00c5\np = 960; # Density of crystal in kg/m^3\nA = 23; # Atomic wt\nN = 6.023*10**26; # avogadros no in atoms/kilomole\n\n# Calculations\n\nn = (p*N*(a**3))/A; # No. of atoms per unit cell\n\n# result\nprint 'No. of atoms per unit cell = %3.0f' %n,' (BCC)';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "No. of atoms per unit cell = 2 (BCC)\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 7, page no:5.89" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration \n\n# given crystal has BCC structure\nr = 1.2*10**-10; # atomic radius in m\n\n# Calculations\n\na = (4*r)/math.sqrt(3); # lattice constant\nV = a**3; # volume of cell\n\n# Result\nprint 'Volume of the cell = %3.3e' %V,' m^3';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Volume of the cell = 2.128e-29 m^3\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 8, page no:5.89" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\na = 4*10**-10; # lattice constant of the crystal\nh = 1 # miller indice\nk = 0 # miller indice\nl = 0 # miller indice\n\n# Calculations\n\n# in fig consider (100) plane. the no of atoms in plane ABCD\nN = 4*(float(1)/4); # Number of atoms\np = N/(a*a); # planar atomic density in atoms/m^2\np1 = p*10**-6 # planar atomic density in atoms/mm^2\n\n# Result\nprint 'planar atomic density = %3.2e' %p1,' atoms/mm^2';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "planar atomic density = 6.25e+12 atoms/mm^2\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 9, page no:5.90" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\n# in fig 5(b) the given plane is parallel to X and Z axes.Thus,its numerical intercepts on these axes is infinity\n#The numerical intercept on y axis is 1/2. Thus the numerical intercepts of plane is (\u221e 1/2 \u221e)\nprint ' Miller indices of plane shown in fig 5.(b) = (0 2 0)';\n# in fig 5(c) the given plane is parallel to Z axis.Thus its numerical intercept on z axis is infinity\n# The numerical intercept on x axis is 1 and y axis is 1/2. this numerical intercepts on plane is (1 1/2 \u221e )\nprint ' Miller indices of plane shown in fig 5.(c) = (1 2 0)'\n# in fig 5(d) the given plane is parallel to Z axis.Thus its numerical intercept on z axis is infinity\n# The numerical intercept on x axis is 1/2 and y axis is 1/2. this numerical intercepts on plane is (1/2 1/2 \u221e )\nprint ' Miller indices of plane shown in fig 5.(d) = (2 2 0)'", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " Miller indices of plane shown in fig 5.(b) = (0 2 0)\n Miller indices of plane shown in fig 5.(c) = (1 2 0)\n Miller indices of plane shown in fig 5.(d) = (2 2 0)\n" + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 11, page no:5.91" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\n\n# (311) plane in simple cubic lattice\nh = 3; # miller indice\nk = 1; # miller indice\nl = 1; # miller indice\na = 2.109*10**-10 # lattice constant in m\n\n# Calculations\ndhkl = a/math.sqrt((h**2)+(k**2)+(l**2)); # interplanar distance\n\n# Result\nprint 'd = %3.3e' %dhkl,' m';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "d = 6.359e-11 m\n" + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 12, page no:5.92" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nh = 1; # miller indice\nk = 1; # miller indice\nl = 0; # miller indice\nd = 2.86*10**-10 # interplanar distance in m\n\n# Calculations\na = d*math.sqrt((h**2)+(k**2)+(l**2)); # interplanar distance\n\n# Result\nprint 'Lattice constant a = %3.3e' %a,' m';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Lattice constant a = 4.045e-10 m\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 13, page no:5.93" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nh1 = 1;\nh0 = 0;\nk0 = 0;\nl0 = 0;\nl1 = 1;\n\n# Calculations\n\n# we know that dhkl = a/sqrt( h^2 + k^2 + l^2)\n# let sqrt( h^2 + k^2 + l^2) = p\np101 = math.sqrt( h1**2 + k0**2 + l1**2);\np100 = math.sqrt( h1**2 + k0**2 + l0**2);\np001 = math.sqrt( h0**2 + k0**2 + l1**2);\n\n# Result\nprint 'd101 : d100 : d001 :: a/%3.4f' %p101,' : ','a/%d' %p100,':',' a/%d ' %p001;", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " d101 : d100 : d001 :: a/1.4142 : a/1 : a/1 \n" + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 14, page no:5.93" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# if a plane cut intercepts of lengths l1,l2,l3 the on three crystal axes ,then\n# l1 : l2 : l3 = pa : pq :rc\n# where a,b and c are primitive vectors of the unit cell and p,q and r are numbers related to miller indices (hkl) of plane by relation\n# 1/p : 1/q : 1/r = h : k : l\n# since, the crystal is simple cubic a = b = c and given that h = 1, k = 1 and l = 1\n# p : q : r = 1/h : 1/k : 1/l = 1/1 : 1/1 : 1/1 \n# p : q : r = 1 : 1 : 1\n# similarly l1 : l2 : l3 = 1a : 1a : 1a\nprint 'ratio of intercepts on the three axes by (111) plane is l1 : l2 : l3 = 1 : 1 : 1';\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "ratio of intercepts on the three axes by (111) plane is l1 : l2 : l3 = 1 : 1 : 1\n" + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 15, page no:5.94" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\n\nr = 1.246*10**-10; # atomic radius in m\nh1 = 1 # miller indice\nh2 = 2 # miller indice\nk0 = 0 # miller indice\nk1 = 1 # miller indice\nk2 = 2 # miller indice\nl0 = 0 # miller indice\nl1 = 1 # miller indice\n\n# Calculations\na = (4*r)/math.sqrt(2); # lattice constant\nd111 = a/math.sqrt((h1**2)+(k1**2)+(l1**2)); # interplanar distance\nd200 = a/math.sqrt((h2**2)+(k0**2)+(l0**2)); # interplanar distance\nd220 = a/math.sqrt((h2**2)+(k2**2)+(l0**2)); # interplanar distance\n\n# Result\nprint 'd111 = %3.3e' %d111,' m','\\n' 'd200 = %3.4e' %d200,' m','\\n''d220 = %3.3e' %d220,' m';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "d111 = 2.035e-10 m \nd200 = 1.7621e-10 m \nd220 = 1.246e-10 m\n" + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 16, page no:5.95" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# variable Declaration\n# the intercept along X-axis be c1 = a\n# the intercept along Y-axis be c2 = b/2 and\n# the intercept along Z-axis be c3 = 3c\n# Therefore, p = c1/a = a/a = 1\n# q = c2/b = (b/2)/b = 1/2\n# r = c3/c = (3c)/c = 3\n# therefore h = 1/p = 1\n# k = 1/q = 2\n# l = 1/r = 1/3\n# lcm of 1 1 and 3 = 3\nh = 1\nk = 2\nl = float(1)/3\ns = 3 ; # lcm\nh1= s*h\nk1= s*k\nl1= s*l;\n\n# Result\nprint '(h k l) =', '%d' %h1,' %d' %k1,'%3.0f' %l1;", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(h k l) = 3 6 1\n" + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Addl_Example 17, page no:5.96" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math;\n\n# Variable Declaration\nd = 1.3*10**-10 # interplanar distance\nn = 1; # given first order\ntheta = 23; # Bragg reflection angle in degrees\n\n# Calculations\ntheta1 = theta*math.pi/180; # degree to radian conversion\n# d = (n*lamda)/(2*sin\u03b8); by Braggs law ------------- 1\nlamda = (2*d*math.sin(theta1)/n)\n\n# Result\nprint 'Wavelength of X-ray = %3.4f' %(lamda*10**10),' \u00c5';", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Wavelength of X-ray = 1.0159 \u00c5\n" + } + ], + "prompt_number": 40 + }, + { + "cell_type": "code", + "collapsed": false, + "input": "", + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Engineering_Physics_-_I/screenshots/Wavelength.PNG b/Engineering_Physics_-_I/screenshots/Wavelength.PNG new file mode 100755 index 00000000..978ed632 Binary files /dev/null and b/Engineering_Physics_-_I/screenshots/Wavelength.PNG differ diff --git a/Engineering_Physics_-_I/screenshots/Young's_modulus.PNG b/Engineering_Physics_-_I/screenshots/Young's_modulus.PNG new file mode 100755 index 00000000..8cf37fdb Binary files /dev/null and b/Engineering_Physics_-_I/screenshots/Young's_modulus.PNG differ diff --git a/Engineering_Physics_-_I/screenshots/numerical_aperture.PNG b/Engineering_Physics_-_I/screenshots/numerical_aperture.PNG new file mode 100755 index 00000000..5ebd4f18 Binary files /dev/null and b/Engineering_Physics_-_I/screenshots/numerical_aperture.PNG differ -- cgit