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| author | Thomas Stephen Lee | 2015-08-28 16:53:23 +0530 |
|---|---|---|
| committer | Thomas Stephen Lee | 2015-08-28 16:53:23 +0530 |
| commit | 4a1f703f1c1808d390ebf80e80659fe161f69fab (patch) | |
| tree | 31b43ae8895599f2d13cf19395d84164463615d9 /sample_notebooks/makarala shamukha venkatasahithi | |
| parent | 9d260e6fae7328d816a514130b691fbd0e9ef81d (diff) | |
| download | Python-Textbook-Companions-4a1f703f1c1808d390ebf80e80659fe161f69fab.tar.gz Python-Textbook-Companions-4a1f703f1c1808d390ebf80e80659fe161f69fab.tar.bz2 Python-Textbook-Companions-4a1f703f1c1808d390ebf80e80659fe161f69fab.zip | |
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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
+ },
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