From db0855dbeb41ecb8a51dde8587d43e5d7e83620f Mon Sep 17 00:00:00 2001 From: Thomas Stephen Lee Date: Fri, 28 Aug 2015 16:53:23 +0530 Subject: add books --- ...ter_2_Nuclear_Sturcture_and_Radioactivity.ipynb | 234 +++++++++++++++++++++ .../Chapter_5_Imperfection_in_Solids.ipynb | 227 ++++++++++++++++++++ .../Chapter_5_Imperfection_in_Solids_1.ipynb | 227 ++++++++++++++++++++ 3 files changed, 688 insertions(+) create mode 100644 sample_notebooks/makarala shamukha venkatasahithi/Chapter_2_Nuclear_Sturcture_and_Radioactivity.ipynb create mode 100755 sample_notebooks/makarala shamukha venkatasahithi/Chapter_5_Imperfection_in_Solids.ipynb create mode 100755 sample_notebooks/makarala shamukha venkatasahithi/Chapter_5_Imperfection_in_Solids_1.ipynb (limited to 'sample_notebooks/makarala shamukha venkatasahithi') diff --git a/sample_notebooks/makarala shamukha venkatasahithi/Chapter_2_Nuclear_Sturcture_and_Radioactivity.ipynb b/sample_notebooks/makarala shamukha venkatasahithi/Chapter_2_Nuclear_Sturcture_and_Radioactivity.ipynb new file mode 100644 index 00000000..505cf999 --- /dev/null +++ b/sample_notebooks/makarala shamukha venkatasahithi/Chapter_2_Nuclear_Sturcture_and_Radioactivity.ipynb @@ -0,0 +1,234 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 Nuclear Sturcture and Radioactivity" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_1 pgno:25" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Half life of radioactive nuclide=t1/2=minutes 14.7674928978\n", + "\n", + "Time required for the activity to decrease to 25percent of the initial activity=t1=minutes 68.0335182976\n", + "\n", + "Time required for the activity to decrease to 10percent of the initial activity=t2=minutes 113.001227913\n" + ] + } + ], + "source": [ + "from math import log\n", + "N0=3396.;#no. of counts per minute given by radioactive nuclide at a given time#\n", + "N=1000.;#no. of counts per minute given by radioactive nuclide one hour later#\n", + "thalf=0.693*60/(2.303*log(N0/N));#half life of nuclide in minutes#\n", + "print'Half life of radioactive nuclide=t1/2=minutes',thalf\n", + "t1=2.303*log(100/25)*thalf/0.693;#time required for the activity to decrease to 25% of the initial activity in minutes#\n", + "print'\\nTime required for the activity to decrease to 25percent of the initial activity=t1=minutes',t1\n", + "t2=2.303*log(100/10)*thalf/0.693;#time required for the activity to decrease to 10% of the initial activity in minutes#\n", + "print'\\nTime required for the activity to decrease to 10percent of the initial activity=t2=minutes',t2\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_2 pgno:27" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Half life of 226Ra molecule=t1/2=years 1584.62090409\n" + ] + } + ], + "source": [ + "R=3.7*10**10;#no. of alpha particles per second emitted by 1g of 226Ra#\n", + "N=(6.023*10**23)/226;#no. of atoms of 226Ra#\n", + "yr=3.15*10**7;#no of seconds in a year#\n", + "thalf=0.693*N/(R*yr);#half life of 226Ra in years#\n", + "print'Half life of 226Ra molecule=t1/2=years',thalf#here the answer written in textbook is wrongly printed actual answer will be the one we are getting here#\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_3 pgno:29" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Weight of 1 Ci of 24Na=w=micrograms=1.13*10**-7grams 0.113352495089\n" + ] + } + ], + "source": [ + "thalf=14.8*60*60;#half life of 24Na atom in seconds#\n", + "L=6.023*10**23;#Avagadro number#\n", + "v=3.7*10**10;#1 Ci of radioactivity in disintegrations per second#\n", + "w=(24*10**6*v*thalf)/(0.693*L);#weight of 1 Ci of 24Na in grams#\n", + "print'Weight of 1 Ci of 24Na=w=micrograms=1.13*10**-7grams',w\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_4 pgno:30" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "dM value of H atom=dM=amu 0.00239\n", + "\n", + "Binding energy of H atom=BE=MeV 2.22509\n" + ] + } + ], + "source": [ + "Mp=1.00728;#mass of proton in amu#\n", + "Mn=1.00866;#mass of neutronin amu#\n", + "MH=2.01355;#isotopic mass of H atom in amu#\n", + "dM=((1*Mp)+(1*Mn)-MH);#dM value of H atom in amu#\n", + "print'dM value of H atom=dM=amu',dM\n", + "BE=dM*931;#binding energy of H atom in MeV#\n", + "print'\\nBinding energy of H atom=BE=MeV',BE\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_5 pgno:32" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Age of the specimen=t=%fyears 36120.0499843\n" + ] + } + ], + "source": [ + "from math import log\n", + "N0=15.3;#decay rate of Contemporary Carbon in disintegrations/min/gram#\n", + "N=2.25;#decay rate of 14C specimen in disintegrtions/min/gram#\n", + "thalf=5670.;#half life of nuclide in years#\n", + "t=2.303*log(N0/N)*thalf/0.693;#Age of the specimen in years#\n", + "print'Age of the specimen=t=years',t#here the answer given in textbook is actually wrong we get twice that of the answer which is shown through execution#\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_6 pgno:33" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Here N0 and N must be in terms of Uranium.N is proportional to 1gram og Uranium\n", + "\n", + "N0 can be calculated from the given data.0.0453grams of 206Pb corresponds to 238*0.0453/206=0.0523grams of 238U,i.e 0.0453 grams of 206Pb must have been formed by the decaying of 0.523grams of 238U.\n", + "Since N is proportional to 1,N0 is proportional to 1.0523.\n", + "\n", + "Age of the mineral=t=years=7.62*10**8years 762356478.526\n" + ] + } + ], + "source": [ + "from math import log\n", + "thalf=4.5*10**9;#half life of Uranium in years#\n", + "print'Here N0 and N must be in terms of Uranium.N is proportional to 1gram og Uranium'\n", + "print'\\nN0 can be calculated from the given data.0.0453grams of 206Pb corresponds to 238*0.0453/206=0.0523grams of 238U,i.e 0.0453 grams of 206Pb must have been formed by the decaying of 0.523grams of 238U.\\nSince N is proportional to 1,N0 is proportional to 1.0523.'\n", + "N0=1.0523;\n", + "N=1;\n", + "t=2.303*log(N0/N)*thalf/0.693;#Age of the mineral in years#\n", + "print'\\nAge of the mineral=t=years=7.62*10**8years',t#here also the answer given in textbook is wrong the one resulted through execution is the right one#\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/sample_notebooks/makarala shamukha venkatasahithi/Chapter_5_Imperfection_in_Solids.ipynb b/sample_notebooks/makarala shamukha venkatasahithi/Chapter_5_Imperfection_in_Solids.ipynb new file mode 100755 index 00000000..0e2a1db4 --- /dev/null +++ b/sample_notebooks/makarala shamukha venkatasahithi/Chapter_5_Imperfection_in_Solids.ipynb @@ -0,0 +1,227 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 Imperfection in Solids" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_1 pgno:56" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.1\n", + "\n", + "\n", + " Equilibrium number of vacancies/m**3 is for 1273K 2.18444488963e+25\n" + ] + } + ], + "source": [ + "# given that\n", + "Na=6.023*10**23 #Avogadro No.\n", + "rho=8.4e6 #Density of Copper in g/m**3\n", + "A=63.5 #Atomic weight of Copper\n", + "Qv=0.9 #Activation energy in eV\n", + "k=8.62*10**-5 #Boltzmann Constant in eV/K\n", + "T=1000+273#Temperature in K\n", + "from math import exp\n", + "print\"Example 5.1\\n\"\n", + "N=Na*rho/A #No. of atomic site per cubic meter\n", + "Nv=N*exp(-Qv/(k*T))\n", + "print\"\\n Equilibrium number of vacancies/m**3 is for 1273K\",Nv\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_3 pgno:57" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Example 5.3\n", + "\n", + "\n", + " Atomic of Al is 98.7039833218\n", + "\n", + " Atomic of Cu is 1.29601667817\n" + ] + } + ], + "source": [ + "# given that\n", + "C_Al=97. #Aluminium wt%\n", + "C_Cu=3. #Copper wt%\n", + "A_Al=26.98 #Atomic wt of Aluminium\n", + "A_Cu=63.55 #Atomic wt of Copper\n", + "\n", + "print\" Example 5.3\\n\"\n", + "CAl=C_Al*A_Cu*100/((C_Al*A_Cu)+(C_Cu*A_Al))\n", + "CCu=C_Cu*A_Al*100/((C_Cu*A_Al)+(C_Al*A_Cu))\n", + "print\"\\n Atomic of Al is\",CAl\n", + "print\"\\n Atomic of Cu is\",CCu\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_4 pgno:58" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.4\n", + "\n", + "\n", + " Number of Schottky defects are defects/m**3. 5.31422380078e+19\n" + ] + } + ], + "source": [ + "# given that\n", + "Na=6.023*10**23 #Avogadro No.\n", + "rho=1.955 #Density of KCl in g/cm**3\n", + "A_k= 39.10 #Atomic weight of potassium in g/mol\n", + "A_cl= 35.45 #Atomic weight of Chlorine in g/mol\n", + "Qs=2.6 #Activation energy in eV\n", + "k=8.62*10**-5 #Boltzmann Constant in eV/K\n", + "T=500+273 #Temperature in K\n", + "from math import exp\n", + "\n", + "print\"Example 5.4\\n\"\n", + "A = A_k+A_cl # Molar mass of KCl in gram\n", + "N=Na*rho*1e6/A #No. of atomic site per cubic meter\n", + "Ns=N*exp(-Qs/(2*k*T))\n", + "print\"\\n Number of Schottky defects are defects/m**3.\",Ns\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_6 pgno:58" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.6\n", + "\n", + "\n", + " Part A\n", + "\n", + " Grain size number is \n", + "6.49185309633\n", + "\n", + " Part B\n", + "\n", + " At magnification of 85x\n", + "\n", + " Number of grains per inch square are\n", + "62.2837370242\n" + ] + } + ], + "source": [ + "# given that \n", + "N=45. #Number of grains per square inch\n", + "M=85. # magnification\n", + "from math import log\n", + "print\"Example 5.6\\n\"\n", + "print\"\\n Part A\"\n", + "n=(log(N)/log(2))+1 #calculation for grain size no. N=2**(n-1)\n", + "print\"\\n Grain size number is \\n\",n\n", + "print\"\\n Part B\"\n", + "Nm=(100/M)**2*2**(n-1)\n", + "print\"\\n At magnification of 85x\\n\"\n", + "print\" Number of grains per inch square are\\n\",Nm\n", + "# answer in book is 62.6. It is because of rounding off at intermediate stages\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/sample_notebooks/makarala shamukha venkatasahithi/Chapter_5_Imperfection_in_Solids_1.ipynb b/sample_notebooks/makarala shamukha venkatasahithi/Chapter_5_Imperfection_in_Solids_1.ipynb new file mode 100755 index 00000000..0e2a1db4 --- /dev/null +++ b/sample_notebooks/makarala shamukha venkatasahithi/Chapter_5_Imperfection_in_Solids_1.ipynb @@ -0,0 +1,227 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 Imperfection in Solids" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_1 pgno:56" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.1\n", + "\n", + "\n", + " Equilibrium number of vacancies/m**3 is for 1273K 2.18444488963e+25\n" + ] + } + ], + "source": [ + "# given that\n", + "Na=6.023*10**23 #Avogadro No.\n", + "rho=8.4e6 #Density of Copper in g/m**3\n", + "A=63.5 #Atomic weight of Copper\n", + "Qv=0.9 #Activation energy in eV\n", + "k=8.62*10**-5 #Boltzmann Constant in eV/K\n", + "T=1000+273#Temperature in K\n", + "from math import exp\n", + "print\"Example 5.1\\n\"\n", + "N=Na*rho/A #No. of atomic site per cubic meter\n", + "Nv=N*exp(-Qv/(k*T))\n", + "print\"\\n Equilibrium number of vacancies/m**3 is for 1273K\",Nv\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_3 pgno:57" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Example 5.3\n", + "\n", + "\n", + " Atomic of Al is 98.7039833218\n", + "\n", + " Atomic of Cu is 1.29601667817\n" + ] + } + ], + "source": [ + "# given that\n", + "C_Al=97. #Aluminium wt%\n", + "C_Cu=3. #Copper wt%\n", + "A_Al=26.98 #Atomic wt of Aluminium\n", + "A_Cu=63.55 #Atomic wt of Copper\n", + "\n", + "print\" Example 5.3\\n\"\n", + "CAl=C_Al*A_Cu*100/((C_Al*A_Cu)+(C_Cu*A_Al))\n", + "CCu=C_Cu*A_Al*100/((C_Cu*A_Al)+(C_Al*A_Cu))\n", + "print\"\\n Atomic of Al is\",CAl\n", + "print\"\\n Atomic of Cu is\",CCu\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_4 pgno:58" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.4\n", + "\n", + "\n", + " Number of Schottky defects are defects/m**3. 5.31422380078e+19\n" + ] + } + ], + "source": [ + "# given that\n", + "Na=6.023*10**23 #Avogadro No.\n", + "rho=1.955 #Density of KCl in g/cm**3\n", + "A_k= 39.10 #Atomic weight of potassium in g/mol\n", + "A_cl= 35.45 #Atomic weight of Chlorine in g/mol\n", + "Qs=2.6 #Activation energy in eV\n", + "k=8.62*10**-5 #Boltzmann Constant in eV/K\n", + "T=500+273 #Temperature in K\n", + "from math import exp\n", + "\n", + "print\"Example 5.4\\n\"\n", + "A = A_k+A_cl # Molar mass of KCl in gram\n", + "N=Na*rho*1e6/A #No. of atomic site per cubic meter\n", + "Ns=N*exp(-Qs/(2*k*T))\n", + "print\"\\n Number of Schottky defects are defects/m**3.\",Ns\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_6 pgno:58" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.6\n", + "\n", + "\n", + " Part A\n", + "\n", + " Grain size number is \n", + "6.49185309633\n", + "\n", + " Part B\n", + "\n", + " At magnification of 85x\n", + "\n", + " Number of grains per inch square are\n", + "62.2837370242\n" + ] + } + ], + "source": [ + "# given that \n", + "N=45. #Number of grains per square inch\n", + "M=85. # magnification\n", + "from math import log\n", + "print\"Example 5.6\\n\"\n", + "print\"\\n Part A\"\n", + "n=(log(N)/log(2))+1 #calculation for grain size no. N=2**(n-1)\n", + "print\"\\n Grain size number is \\n\",n\n", + "print\"\\n Part B\"\n", + "Nm=(100/M)**2*2**(n-1)\n", + "print\"\\n At magnification of 85x\\n\"\n", + "print\" Number of grains per inch square are\\n\",Nm\n", + "# answer in book is 62.6. It is because of rounding off at intermediate stages\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} -- cgit