From d36fc3b8f88cc3108ffff6151e376b619b9abb01 Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:40:35 +0530 Subject: Revised list of TBCs --- .../Raj Phani_version_backup/chapter1.ipynb | 993 +++++++++++++++++ .../Raj Phani_version_backup/chapter_1.ipynb | 1118 ++++++++++++++++++++ .../Raj Phani_version_backup/chapter_1_1.ipynb | 1118 ++++++++++++++++++++ 3 files changed, 3229 insertions(+) create mode 100755 sample_notebooks/Raj Phani/Raj Phani_version_backup/chapter1.ipynb create mode 100755 sample_notebooks/Raj Phani/Raj Phani_version_backup/chapter_1.ipynb create mode 100755 sample_notebooks/Raj Phani/Raj Phani_version_backup/chapter_1_1.ipynb (limited to 'sample_notebooks/Raj Phani/Raj Phani_version_backup') diff --git a/sample_notebooks/Raj Phani/Raj Phani_version_backup/chapter1.ipynb b/sample_notebooks/Raj Phani/Raj Phani_version_backup/chapter1.ipynb new file mode 100755 index 00000000..af42a9cd --- /dev/null +++ b/sample_notebooks/Raj Phani/Raj Phani_version_backup/chapter1.ipynb @@ -0,0 +1,993 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# chapter 1: Atomic Nucleus" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.1;pg no:2" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.1, Page:2 \n", + " \n", + "\n", + "\n", + " The electric field in V/m is = 20000.0\n", + "\n", + " The force in N/C is = 20000.0\n", + "\n", + " The force on metal sphere in N is = 7.6e-05\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.1, Page:2 \\n \\n\"\n", + "#Given:\n", + "v=1000# potential\n", + "d=0.05# distance\n", + "q=3.8*10**-9# charge\n", + "#solution:\n", + "e=v/d;#electric field\n", + "f=e;# force\n", + "f1=f*q;# force on metal sphere\n", + "print\"\\n The electric field in V/m is =\",e\n", + "print\"\\n The force in N/C is =\",f\n", + "print\"\\n The force on metal sphere in N is =\",f1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.2;pg no:2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.2, Page:2 \n", + " \n", + "\n", + "The potential in V is = 80.0\n" + ] + } + ], + "source": [ + "#cal of potential\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.2, Page:2 \\n \\n\"\n", + "#Given:\n", + "energy=2*10**-6\n", + "c=2.5*10**-8# velocity of light\n", + "#solution:\n", + "v=energy/c# potential\n", + "print\"The potential in V is =\",v" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.3;pg no:3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.3, Page:3 \n", + "\n", + "The wavelength in Angstroms is = 3.88\n", + "The photon wavelength in Angstroms is = 1242.38\n" + ] + } + ], + "source": [ + "#cal of elecrtron and photon wavelength\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.3, Page:3 \\n\"\n", + "#Given:\n", + "energy=10 #in electron volts\n", + "m=9.1*10**-31# mass of electron in kg\n", + "h=6.626*10**-34# planck's constant J.s\n", + "c=3*10**8# speed of light in m/s\n", + "#solution (a):\n", + "energy1=energy*1.6*10**-19# energy in J\n", + "p=(2*m*energy1)**0.5# momentum\n", + "wavelength=h/p*(10)**10\n", + "print\"The wavelength in Angstroms is =\",round(wavelength,2)\n", + "#solution (b):\n", + "wavelength1=h*c/energy1*(10)**10;#photon wavelength\n", + "print\"The photon wavelength in Angstroms is =\",round(wavelength1,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.4;pg no:3" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.4, Page:3 \n", + " \n", + "\n", + "The energy in eV is = 150.77\n" + ] + } + ], + "source": [ + "#cal of kinetic energy of an electron\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.4, Page:3 \\n \\n\"\n", + "#Given:\n", + "wavelength=10**-10\n", + "m=9.1*10**-31\n", + "h=6.626*10**-34\n", + "#solution:\n", + "p=h/wavelength\n", + "e=p*p/(2*m) # energy in J\n", + "e1=e/(1.6*10**-19)# energy in eV\n", + "print\"The energy in eV is =\",round(e1,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.5;pg no:3" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.5, Page:3 \n", + " \n", + "\n", + "The wavelength in 10^-5 Angstroms is = 0.66\n", + "The wavelength in 10^-5 Angstroms is = 0.65\n" + ] + } + ], + "source": [ + "#cal of wavelength of oxygen and nitrogen nucleus\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.5, Page:3 \\n \\n\"\n", + "#Given:\n", + "m=1.66*10**-27# 1u=1.66*10^-27 kg\n", + "h=6.6262*10**-34#planck's constant in J.s\n", + "energy1=120# in Mev for oxygen\n", + "energy2=140# in MeV for nitrogen\n", + "#solution(a):\n", + "p=(2*m*16*energy1*(1.6022*10**-13))**0.5\n", + "wavelength1=h/p*(10)**15#wavelength in 10^-5 Angstroms\n", + "print\"The wavelength in 10^-5 Angstroms is =\",round(wavelength1,2)\n", + "#solution (b):\n", + "p=(2*m*14*energy2*(1.6022*10**-13))**0.5\n", + "wavelength2=h/p*(10)**15#wavelength in 10^-5 Angstroms\n", + "print\"The wavelength in 10^-5 Angstroms is =\",round(wavelength2,2)\n", + "# 1 Angstrom = 10^-10 m" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.6;pg no:3" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.6, Page:3 \n", + " \n", + "\n", + "The energy in eV is = 8275.0\n" + ] + } + ], + "source": [ + "#cal of energy of a gamma photon\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.6, Page:3 \\n \\n\"\n", + "#Given:\n", + "wavelength=1.5*10**-10\n", + "h=6.62*10**-34\n", + "c=3*10**8\n", + "#solution:\n", + "e=(h*c)/wavelength# energy in J\n", + "e1=e/(1.6*10**-19)# energy in eV\n", + "print\"The energy in eV is =\",e1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.7;pg no:4" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.7, Page:4 \n", + " \n", + "\n", + "\n", + " The threshold frequency in s^-1 is = 1.23634168427e+15\n", + "\n", + " The threshold wavelength in Angstroms is = 2426.51\n", + "\n", + " The energy of photoelectrone in eV is = 3.91\n" + ] + } + ], + "source": [ + "#cal of threshold frequency,wavelength,energy of photoelectrone\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.7, Page:4 \\n \\n\"\n", + "#Given:\n", + "E=5.12*1.6*10**-19# energy in J\n", + "h=6.626*10**-34\n", + "c=3*10**8\n", + "wavelength=200*10**-9\n", + "w=2.3# in eV\n", + "#solution:\n", + "tf=E/h# (part a)\n", + "print\"\\n The threshold frequency in s^-1 is =\",round(tf,2)\n", + "tl=c/tf*10**10# (part b)\n", + "print\"\\n The threshold wavelength in Angstroms is =\",round(tl,2)\n", + "e=(h*c)/(wavelength*1.6*10**-19)# photon energy in eV (part c)\n", + "pe=e-w\n", + "print\"\\n The energy of photoelectrone in eV is =\",round(pe,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.8;pg no:4" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.8, Page:4 \n", + " \n", + "\n", + "\n", + " The velocity of alpha particles for 1 MeV in mega m/s is = 6.94\n", + "\n", + " The velocity of alpha particles for 2 MeV in mega m/s is = 9.82\n", + "\n", + " The velocity of deuteron particles for 1 MeV in mega m/s is = 9.82\n", + "\n", + " The velocity of deuteron particles for 2 MeV in mega m/s is = 13.88\n", + "\n", + " The velocity of proton particles for 1 MeV in mega m/s is = 13.88\n", + "\n", + " The velocity of proton particles for 2 MeV in mega m/s is = 19.63\n" + ] + } + ], + "source": [ + "#cal of velocity of alpha particles,deuteron,proton\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.8, Page:4 \\n \\n\"\n", + "#Given:\n", + "e1=1 # in MeV\n", + "e2=2 # in MeV\n", + "ma=4 # in u(amu)\n", + "md=2 # in u(amu)\n", + "mp=1 # in u(amu)\n", + "# 1u = 1.6*10^-27 Kg\n", + "#solution: part a)For alpha particles\n", + "v1a=((2*e1*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of alpha particles for 1 MeV in mega m/s is =\",round(v1a/10**6,2)# For 1 MeV\n", + "v2a=((2*e2*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of alpha particles for 2 MeV in mega m/s is =\",round(v2a/10**6,2)# For 2 MeV\n", + "#solution: part b)For deuteron particles\n", + "v1b=((2*e1*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of deuteron particles for 1 MeV in mega m/s is =\",round(v1b/10**6,2) # For 1 MeV\n", + "v2b=((2*e2*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of deuteron particles for 2 MeV in mega m/s is =\",round(v2b/10**6,2) # For 2 MeV\n", + "#solution: part c)For proton particles\n", + "v1p=((2*e1*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of proton particles for 1 MeV in mega m/s is =\",round(v1p/10**6,2) # For 1 MeV\n", + "v2p=((2*e2*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of proton particles for 2 MeV in mega m/s is =\",round(v2p/10**6,2) # For 2 MeV" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.9;pg no:5" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.9, Page:5 \n", + " \n", + "\n", + "The energy in MeV is = 934.0\n" + ] + } + ], + "source": [ + "#cal of The energy equivalence\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.9, Page:5 \\n \\n\"\n", + "#Given:\n", + "m=1./(6.023*10**23)#mass of 1 atom in g\n", + "m1=m*10**-3#mass of 1 atom in Kg\n", + "c=3.*10**8# velocity in m/s\n", + "#solution:\n", + "e=m1*c*c; # energy in J\n", + "e1=e/(1.6*10**-13)# energy in MeV\n", + "print\"The energy in MeV is =\",round(e1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.10;pg no:5" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.10, Page:5 \n", + " \n", + "\n", + "The energy in eV is = 13.26\n" + ] + } + ], + "source": [ + "#cal of The energy of formation\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.10, Page:5 \\n \\n\"\n", + "#Given:\n", + "enthalpy=1278 # enthalpy of combustion in kJ/mol\n", + "#solution:\n", + "energy=(enthalpy*1000)/(6.022*10**23*1.6*10**-19)\n", + "print\"The energy in eV is =\",round(energy,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.11;pg no:5" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.11, Page:5 \n", + " \n", + "\n", + "\n", + " The mean binding energy of helium atom in MeV is = 7.07\n", + "\n", + " The mean binding energy of oxygen atom in MeV is = 7.98\n" + ] + } + ], + "source": [ + "#cal of mean binding energy of helium and oxygen\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.11, Page:5 \\n \\n\"\n", + "#Given:\n", + "mh=1.0078\n", + "mn=1.0087\n", + "ma=4.0026\n", + "mo=15.9949\n", + "Ah=4.0026 # atomic mass of helium\n", + "Ao=15.9949 # atomic mass of oxygen\n", + "#solution:\n", + "# part (a)\n", + "B1=(2*mh+2*mn-ma)*931 # in MeV\n", + "Bh=B1/Ah\n", + "print\"\\n The mean binding energy of helium atom in MeV is =\",round(Bh,2)\n", + "# part (b)\n", + "B2=(8*mh+8*mn-mo)*931 # in MeV\n", + "Bo=B2/Ao\n", + "print\"\\n The mean binding energy of oxygen atom in MeV is =\",round(Bo,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.12;pg no:6" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.12, Page:6 \n", + "\n", + "\n", + " The mean binding energy of Be atom in MeV is = 7.059\n", + "From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\n" + ] + } + ], + "source": [ + "#cal of mean binding energy of Be atom\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.12, Page:6 \\n\"\n", + "#Given:\n", + "mh=1.0078;\n", + "mn=1.0087;\n", + "ABe=8.0053; # atomic mass of beryllium\n", + "#solution:\n", + "B1=(4*mh+4*mn-ABe)*931; # in MeV\n", + "Bh=B1/ABe;\n", + "print\"\\n The mean binding energy of Be atom in MeV is =\",round(Bh,3)\n", + "print\"From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.13;pg no:6" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.13, Page:6 \n", + "\n", + "The amount of coal required in Kg is = 2.5\n" + ] + } + ], + "source": [ + "#cal of amount of coal\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.13, Page:6 \\n\"\n", + "#Given:\n", + "e=200; # in Mev\n", + "m=0.235; # weight of uranium atom in Kg\n", + "enthalpy=393.5; # in KJ/mol\n", + "Na=6.02*10**23;\n", + "#solution:\n", + "e1=e*1.6*10**-19*10**6;\n", + "atoms=Na/m;\n", + "e2=atoms*e1;#energy released in J\n", + "m1=(e2*12)/(393.5*1000*1000);# in Kg\n", + "m2=m1/1000;# in tons\n", + "print\"The amount of coal required in Kg is =\", round(m2/1000,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.14;pg no:7" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.14, Page:7 \n", + " \n", + "\n", + "The energy release in part (a) in eV/molecule is = 2.51\n", + "The energy release in part (b) in eV/molecule is = 9.23\n" + ] + } + ], + "source": [ + "#cal of The energy releases\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.14, Page:7 \\n \\n\"\n", + "#Given:\n", + "H1=241.8; # in KJ/mol\n", + "H2=887.2; # in KJ/mol\n", + "# 1 KJ/mol = 0.0104 eV/atom\n", + "#solution: part (a)\n", + "e1=H1*0.0104;\n", + "print\"The energy release in part (a) in eV/molecule is =\",round(e1,2)\n", + "#solution: part (b)\n", + "e2=H2*0.0104;\n", + "print\"The energy release in part (b) in eV/molecule is =\",round(e2,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.15;pg no:7" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.15, Page:7 \n", + "\n", + "The energy release in part (a) in KJ/mol of carbondioxide is = 394.9\n", + "The energy release in part (b) in KJ/mol of alumina is = 1676.0\n", + "The energy release in part (c) in MJ/atom of U(235) is = 19.264\n" + ] + } + ], + "source": [ + "#cal of The energy releases\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.15, Page:7 \\n\"\n", + "#Given:\n", + "H1=4.1; # in eV/molecule\n", + "H2=17.4; # in eV/molecule\n", + "H3=200;# in MeV/atom of U\n", + "# 1 eV/atom = 96.32 KJ/mol\n", + "#solution: part (a)\n", + "e1=H1*96.32;\n", + "print\"The energy release in part (a) in KJ/mol of carbondioxide is =\",round(e1,1)\n", + "#solution: part (b)\n", + "e2=H2*96.32;\n", + "print\"The energy release in part (b) in KJ/mol of alumina is =\",round(e2,1)\n", + "#solution: part (c)\n", + "e3=H3*1000*96.32;# in MJ/atom of U(235)\n", + "print\"The energy release in part (c) in MJ/atom of U(235) is =\",round(e3/10**6,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.16;pg no:7" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.16, Page:7 \n", + " \n", + "\n", + "\n", + " The rate of energy release in MW is= 949.25\n" + ] + } + ], + "source": [ + "#cal of The rate of energy release\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.16, Page:7 \\n \\n\"\n", + "#Given:\n", + "e=200.; #MeV/ atom of U\n", + "# 1 eV = 1.6*10^-19 J\n", + "Na=6.023*10**23;\n", + "M=0.235; # mass in Kg\n", + "#solution:\n", + "e1=e*1.6*10**-19*10**6;\n", + "A=Na/M;\n", + "e2=A*e1; # energy released in MJ/day\n", + "e3=e2/(24.*3600.);\n", + "print\"\\n The rate of energy release in MW is=\",round(e3/10**6,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.17;pg no:8" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.17, Page:8 \n", + " \n", + "\n", + "\n", + " The mass loss in 10^-27 Kg/He formed is = 0.0464\n" + ] + } + ], + "source": [ + "#cal of The mass loss\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.17, Page:8 \\n \\n\"\n", + "#Given:\n", + "e=26.03; # in MeV\n", + "#solution:\n", + "loss=e/931; #in atomic mass units (u)\n", + "# 1 u = 1.66*10^-27 Kg\n", + "m=(loss*1.66*10**-27)/(1*10**-27);\n", + "print\"\\n The mass loss in 10^-27 Kg/He formed is =\",round(m,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.18;pg no:8" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.18, Page:8 \n", + "\n", + "\n", + " The energy loss in MeV is = -4.0312\n" + ] + } + ], + "source": [ + "#cal of The energy loss\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.18, Page:8 \\n\"\n", + "#Given:\n", + "mh=1.007825;\n", + "mt=3.016049;\n", + "md=2.014102;\n", + "#solution:\n", + "m1=(mh+mt-2*md);\n", + "e=(-m1)*931; # in MeV\n", + "print\"\\n The energy loss in MeV is =\",round(-e,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.19;pg no:8" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.19, Page:8 \n", + "\n", + "The mean binding energy of tritium atom in MeV is = 2.811\n", + "The mean binding energy of nickel atom in MeV is = 8.716\n" + ] + } + ], + "source": [ + "#cal of mean binding energy of tritium and nickel atom\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.19, Page:8 \\n\"\n", + "#Given:\n", + "mh=1.007825;\n", + "mn=1.008665;\n", + "mt=3.016049; # atomic mass of Tritium\n", + "mNi=59.93528; # atomic mass of Nickel\n", + "#solution:\n", + "# part (a)\n", + "B1=(1*mh+2*mn-mt)*931; # in MeV\n", + "Bh=B1/mt;\n", + "print\"The mean binding energy of tritium atom in MeV is =\",round(Bh,3)\n", + "# part (b)\n", + "B2=(28*mh+32*mn-mNi)*931; # in MeV\n", + "Bo=B2/mNi;\n", + "print\"The mean binding energy of nickel atom in MeV is =\",round(Bo,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.20;pg no:9" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.20, Page:9 \n", + "\n", + "The mean binding energy of Cl (35) atom in MeV is = 8.5281\n", + "The mean binding energy of Cl (37) atom in MeV is = 8.5784\n", + "The increase in mean binding energy of Cl atom in MeV is = 0.05\n" + ] + } + ], + "source": [ + "#cal of mean binding energy of Cl\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.20, Page:9 \\n\"\n", + "#Given:\n", + "mh=1.00783;\n", + "mn=1.00867;\n", + "m35=34.96885; # atomic mass of Cl (35)\n", + "m37=36.96590; # atomic mass of Cl (37)\n", + "#solution:\n", + "B1=(17*mh+18*mn-m35)*931; # in MeV\n", + "Bh=B1/m35;\n", + "print\"The mean binding energy of Cl (35) atom in MeV is =\",round(Bh,4)\n", + "B2=(17*mh+20*mn-m37)*931; # in MeV\n", + "Bo=B2/m37;\n", + "print\"The mean binding energy of Cl (37) atom in MeV is =\",round(Bo,4)\n", + "Bi=Bo-Bh;\n", + "print\"The increase in mean binding energy of Cl atom in MeV is =\",round(Bi,2)\n", + "# NOTE: The answer depends upon how much precise value you take for atomic masses." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.21;pg no:9" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.21, Page:9 \n", + " \n", + "\n", + "\n", + " The mean binding energy of Na(22) in MeV is = 7.9236\n", + "\n", + " The mean binding energy of Na(23)in MeV is = 8.1154\n", + "\n", + " The mean binding energy of Na(24) in MeV is = 8.0717\n" + ] + } + ], + "source": [ + "#cal of mean binding energy of Na\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.21, Page:9 \\n \\n\"\n", + "#Given:\n", + "mh=1.0078;\n", + "mn=1.0087;\n", + "m22=21.99431;# atomic mass of Na 22\n", + "m23=22.9898;# atomic mass of Na 23\n", + "m24=23.9909;# atomic mass of Na 24\n", + "#solution:\n", + "# part (a)\n", + "B1=((11*mh+11*mn)-m22)*931; # in MeV\n", + "Bh=B1/m22;\n", + "print\"\\n The mean binding energy of Na(22) in MeV is =\",round(Bh,4)\n", + "# part (b)\n", + "B2=((11*mh+12*mn)-m23)*931; # in MeV\n", + "Bo=B2/m23;\n", + "print\"\\n The mean binding energy of Na(23)in MeV is =\",round(Bo,4)\n", + "# part (c)\n", + "B3=((11*mh+13*mn)-m24)*931; # in MeV\n", + "Bs=B3/m24;\n", + "print\"\\n The mean binding energy of Na(24) in MeV is =\",round(Bs,4)" + ] + }, + { + "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/Raj Phani/Raj Phani_version_backup/chapter_1.ipynb b/sample_notebooks/Raj Phani/Raj Phani_version_backup/chapter_1.ipynb new file mode 100755 index 00000000..f71cb56f --- /dev/null +++ b/sample_notebooks/Raj Phani/Raj Phani_version_backup/chapter_1.ipynb @@ -0,0 +1,1118 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# chapter1: electric charge" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.1" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false, + "scrolled": true + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.1, Page:3 \n", + " \n", + "\n", + "\n", + " The electric field in V/m is = 20000.0\n", + "\n", + " The force in N/C is = 20000.0\n", + "\n", + " The force on metal sphere in N is = 7.6e-05\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.1, Page:3 \\n \\n\"\n", + "#Given:\n", + "v=1000# potential\n", + "d=0.05# distance\n", + "q=3.8*10**-9# charge\n", + "\n", + "#solution:\n", + "e=v/d;#electric field\n", + "f=e;# force\n", + "f1=f*q;# force on metal sphere\n", + "print\"\\n The electric field in V/m is =\",e\n", + "print\"\\n The force in N/C is =\",f\n", + "print\"\\n The force on metal sphere in N is =\",f1\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.2" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.2, Page:4 \n", + " \n", + "\n", + "The potential in V is = 80.0\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.2, Page:4 \\n \\n\"\n", + "#Given:\n", + "energy=2*10**-6\n", + "c=2.5*10**-8# velocity of light\n", + "#solution:\n", + "v=energy/c# potential\n", + "print\"The potential in V is =\",v\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.3" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.3, Page:5 \n", + " \n", + "\n", + "The wavelength in Angstroms is = 3.88289589025\n", + "\n", + " The photon wavelength in Angstroms is = 9.11075e-05\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.3, Page:5 \\n \\n\"\n", + "#Given:\n", + "\n", + "energy=10 #in electron volts\n", + "m=9.1*10**-31# mass of electron in kg\n", + "h=6.626*10**-34# planck's constant J.s\n", + "c=3*10^8# speed of light in m/s\n", + "\n", + "#solution (a):\n", + "energy1=energy*1.6*10**-19# energy in J\n", + "p=(2*m*energy1)**0.5# momentum\n", + "wavelength=h/p*(10)**10\n", + "\n", + "print\"The wavelength in Angstroms is =\",wavelength\n", + "\n", + "\n", + "#solution (b):\n", + "wavelength1=h*c/energy1*(10)**10;#photon wavelength\n", + "\n", + "print\"\\n The photon wavelength in Angstroms is =\",wavelength1\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example1.4" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.4, Page:6 \n", + " \n", + "\n", + "The energy in eV is = 150.768804945\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.4, Page:6 \\n \\n\"\n", + "\n", + "#Given:\n", + "\n", + "wavelength=10**-10\n", + "m=9.1*10**-31\n", + "h=6.626*10**-34\n", + "\n", + "#solution:\n", + "\n", + "p=h/wavelength\n", + "e=p*p/(2*m) # energy in J\n", + "e1=e/(1.6*10**-19)# energy in eV\n", + "\n", + "print\"The energy in eV is =\",e1\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.5" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.5, Page:8 \n", + " \n", + "\n", + "\n", + " The wavelength in 10^-5 Angstroms is = 0.655671822473\n", + "\n", + " The wavelength in 10^-5 Angstroms is = 0.648946805494\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.5, Page:8 \\n \\n\"\n", + "\n", + "#Given:\n", + "\n", + "m=1.66*10**-27# 1u=1.66*10^-27 kg\n", + "h=6.6262*10**-34#planck's constant in J.s\n", + "energy1=120# in Mev for oxygen\n", + "energy2=140# in MeV for nitrogen\n", + "\n", + "#solution(a):\n", + "\n", + "p=(2*m*16*energy1*(1.6022*10**-13))**0.5\n", + "wavelength1=h/p*(10)**15#wavelength in 10^-5 Angstroms\n", + "\n", + "print\"\\n The wavelength in 10^-5 Angstroms is =\",wavelength1\n", + "\n", + "#solution (b):\n", + "\n", + "p=(2*m*14*energy2*(1.6022*10**-13))**0.5\n", + "wavelength2=h/p*(10)**15#wavelength in 10^-5 Angstroms\n", + "\n", + "print\"\\n The wavelength in 10^-5 Angstroms is =\",wavelength2\n", + "\n", + "# 1 Angstrom = 10^-10 m\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example1.6" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.6, Page:9 \n", + " \n", + "\n", + "The energy in eV is = 8275.0\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.6, Page:9 \\n \\n\"\n", + "\n", + "#Given:\n", + "\n", + "wavelength=1.5*10**-10\n", + "h=6.62*10**-34\n", + "c=3*10**8\n", + "\n", + "#solution:\n", + "\n", + "e=(h*c)/wavelength# energy in J\n", + "e1=e/(1.6*10**-19)# energy in eV\n", + "\n", + "print\"The energy in eV is =\",e1\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.7" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.7, Page:10 \n", + " \n", + "\n", + "\n", + " The threshold frequency in s^-1 is = 1.23634168427e+15\n", + "\n", + " The threshold wavelength in Angstroms is = 2426.51367187\n", + "\n", + " The energy of photoelectrone in eV is = 3.911875\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.7, Page:10 \\n \\n\"\n", + "\n", + "#Given:\n", + "\n", + "E=5.12*1.6*10**-19# energy in J\n", + "h=6.626*10**-34\n", + "c=3*10**8\n", + "wavelength=200*10**-9\n", + "w=2.3# in eV\n", + "\n", + "#solution:\n", + "\n", + "tf=E/h# (part a)\n", + "print\"\\n The threshold frequency in s^-1 is =\",tf\n", + "\n", + "tl=c/tf*10**10# (part b)\n", + "print\"\\n The threshold wavelength in Angstroms is =\",tl\n", + "\n", + "e=(h*c)/(wavelength*1.6*10**-19)# photon energy in eV (part c)\n", + "\n", + "pe=e-w\n", + "\n", + "print\"\\n The energy of photoelectrone in eV is =\",pe\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.8" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.8, Page:10 \n", + " \n", + "\n", + "\n", + " The velocity of alpha particles for 1 MeV in m/s is = 6941056.08394\n", + "\n", + " The velocity of alpha particles for 2 MeV in m/s is = 9816135.6511\n", + "\n", + " The velocity of deuteron particles for 1 MeV in m/s is = 9816135.6511\n", + "\n", + " The velocity of deuteron particles for 2 MeV in m/s is = 13882112.1679\n", + "\n", + " The velocity of proton particles for 1 MeV in m/s is = 13882112.1679\n", + "\n", + " The velocity of proton particles for 2 MeV in m/s is = 19632271.3022\n" + ] + } + ], + "source": [ + "#cal of velocity of alpha particles,deuteron,proton\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.8, Page:10 \\n \\n\"\n", + "#Given:\n", + "e1=1 # in MeV\n", + "e2=2 # in MeV\n", + "ma=4 # in u(amu)\n", + "md=2 # in u(amu)\n", + "mp=1 # in u(amu)\n", + "\n", + "# 1u = 1.6*10^-27 Kg\n", + "\n", + "#solution: part a)For alpha particles\n", + "\n", + "v1a=((2*e1*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of alpha particles for 1 MeV in m/s is =\",v1a# For 1 MeV\n", + "\n", + "v2a=((2*e2*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of alpha particles for 2 MeV in m/s is =\",v2a# For 2 MeV\n", + "\n", + "#solution: part b)For deuteron particles\n", + "\n", + "v1b=((2*e1*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of deuteron particles for 1 MeV in m/s is =\",v1b # For 1 MeV\n", + "\n", + "\n", + "v2b=((2*e2*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of deuteron particles for 2 MeV in m/s is =\",v2b # For 2 MeV\n", + "\n", + "#solution: part c)For proton particles\n", + "\n", + "v1p=((2*e1*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of proton particles for 1 MeV in m/s is =\",v1p # For 1 MeV\n", + "\n", + "\n", + "v2p=((2*e2*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of proton particles for 2 MeV in m/s is =\",v2p # For 2 MeV\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.9" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.9, Page: \n", + " \n", + "\n", + "The energy in MeV is = 933.919973435\n" + ] + } + ], + "source": [ + "#cal of The energy\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.9, Page: \\n \\n\"\n", + "#Given:\n", + "\n", + "m=1/(6.023*10**23)#mass of 1 atom in g\n", + "m1=m*10**-3#mass of 1 atom in Kg\n", + "c=3*10**8# velocity in m/s\n", + "#solution:\n", + "\n", + "e=m1*c*c; # energy in J\n", + "e1=e/(1.6*10**-13)# energy in MeV\n", + "\n", + "print\"The energy in MeV is =\",e1\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.10" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.1, Page:11 \n", + " \n", + "\n", + "The energy in eV is = 13.2638658253\n" + ] + } + ], + "source": [ + "#cal of The energy\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.1, Page:11 \\n \\n\"\n", + "#Given:\n", + "\n", + "enthalpy=1278 # enthalpy of combustion in kJ/mol\n", + "\n", + "#solution:\n", + "\n", + "energy=(enthalpy*1000)/(6.022*10**23*1.6*10**-19)\n", + "\n", + "print\"The energy in eV is =\",energy\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.11" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.11, Page:11 \n", + " \n", + "\n", + "\n", + " The mean binding energy of helium atom in MeV is = 7.0710038475\n", + "\n", + " The mean binding energy of oxygen atom in MeV is = 7.9800498909\n" + ] + } + ], + "source": [ + "#cal of mean binding energy of helium and oxygen\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.11, Page:11 \\n \\n\"\n", + "#Given:\n", + "mh=1.0078\n", + "mn=1.0087\n", + "ma=4.0026\n", + "mo=15.9949\n", + "Ah=4.0026 # atomic mass of helium\n", + "Ao=15.9949 # atomic mass of oxygen\n", + "\n", + "#solution:\n", + "\n", + "# part (a)\n", + "\n", + "B1=(2*mh+2*mn-ma)*931 # in MeV\n", + "Bh=B1/Ah\n", + "print\"\\n The mean binding energy of helium atom in MeV is =\",Bh\n", + "\n", + "# part (b)\n", + "\n", + "B2=(8*mh+8*mn-mo)*931 # in MeV\n", + "Bo=B2/Ao\n", + "print\"\\n The mean binding energy of oxygen atom in MeV is =\",Bo\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.12" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.12, Page:12 \n", + " \n", + "\n", + "\n", + " The mean binding energy of Be atom in MeV is = 7.05928572321\n", + "From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\n" + ] + } + ], + "source": [ + "#cal of \n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.12, Page:12 \\n \\n\"\n", + "#Given:\n", + "mh=1.0078;\n", + "mn=1.0087;\n", + "ABe=8.0053; # atomic mass of beryllium\n", + "\n", + "#solution:\n", + "\n", + "B1=(4*mh+4*mn-ABe)*931; # in MeV\n", + "Bh=B1/ABe;\n", + "print\"\\n The mean binding energy of Be atom in MeV is =\",Bh\n", + "\n", + "print\"From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.13" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.13, Page:12 \n", + " \n", + "\n", + "\n", + " The amount of coal required in Kg is = 2499.85671416\n" + ] + } + ], + "source": [ + "#cal of amount of coal\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.13, Page:12 \\n \\n\"\n", + "#Given:\n", + "\n", + "e=200; # in Mev\n", + "m=0.235; # weight of uranium atom in Kg\n", + "enthalpy=393.5; # in KJ/mol\n", + "Na=6.02*10**23;\n", + "\n", + "\n", + "#solution:\n", + "e1=e*1.6*10**-19*10**6;\n", + "atoms=Na/m;\n", + "e2=atoms*e1;#energy released in J\n", + "m1=(e2*12)/(393.5*1000*1000);# in Kg\n", + "m2=m1/1000;# in tons\n", + "print\"\\n The amount of coal required in Kg is =\", m2\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.14" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.14, Page:13 \n", + " \n", + "\n", + "\n", + " The energy release in part (a) in eV/molecule is = 2.51472\n", + "\n", + " The energy release in part (b) in eV/molecule is = 9.22688\n" + ] + } + ], + "source": [ + "#cal of The energy releases\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.14, Page:13 \\n \\n\"\n", + "#Given:\n", + "H1=241.8; # in KJ/mol\n", + "H2=887.2; # in KJ/mol\n", + "# 1 KJ/mol = 0.0104 eV/atom\n", + "\n", + "#solution: part (a)\n", + "e1=H1*0.0104;\n", + "print\"\\n The energy release in part (a) in eV/molecule is =\",e1\n", + "\n", + "#solution: part (b)\n", + "e2=H2*0.0104;\n", + "print\"\\n The energy release in part (b) in eV/molecule is =\",e2\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.15" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.15, Page:14 \n", + " \n", + "\n", + "\n", + " The energy release in part (a) in KJ/mol of carbondioxide is = 394.912\n", + "\n", + " The energy release in part (b) in KJ/mol of alumina is = 1675.968\n", + "\n", + " The energy release in part (c) in MJ/atom of U(235) is = 19264000.0\n" + ] + } + ], + "source": [ + "#cal of The energy releases\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.15, Page:14 \\n \\n\"\n", + "#Given:\n", + "H1=4.1; # in eV/molecule\n", + "H2=17.4; # in eV/molecule\n", + "H3=200;# in MeV/atom of U\n", + "\n", + "# 1 eV/atom = 96.32 KJ/mol\n", + "\n", + "#solution: part (a)\n", + "e1=H1*96.32;\n", + "print\"\\n The energy release in part (a) in KJ/mol of carbondioxide is =\",e1\n", + "\n", + "#solution: part (b)\n", + "e2=H2*96.32;\n", + "print\"\\n The energy release in part (b) in KJ/mol of alumina is =\",e2\n", + "\n", + "#solution: part (c)\n", + "e3=H3*1000*96.32;# in MJ/atom of U(235)\n", + "print\"\\n The energy release in part (c) in MJ/atom of U(235) is =\",e3\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.16" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.16, Page:15 \n", + " \n", + "\n", + "\n", + " The rate of energy release in W is 949251379.039\n" + ] + } + ], + "source": [ + "#cal of The rate of energy release\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.16, Page:15 \\n \\n\"\n", + "#Given:\n", + "e=200; #MeV/ atom of U\n", + "# 1 eV = 1.6*10^-19 J\n", + "Na=6.023*10**23;\n", + "M=0.235; # mass in Kg\n", + "\n", + "#solution:\n", + "\n", + "e1=e*1.6*10**-19*10**6;\n", + "A=Na/M;\n", + "e2=A*e1; # energy released in MJ/day\n", + "e3=e2/(24*3600);\n", + "print\"\\n The rate of energy release in W is \",e3\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.17" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.17, Page:16 \n", + " \n", + "\n", + "\n", + " The mass loss in 10^-27 Kg/He formed is = 0.046412244898\n" + ] + } + ], + "source": [ + "#cal of The mass loss\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.17, Page:16 \\n \\n\"\n", + "#Given:\n", + "e=26.03; # in MeV\n", + "\n", + "#solution:\n", + "\n", + "loss=e/931; #in atomic mass units (u)\n", + "# 1 u = 1.66*10^-27 Kg\n", + "m=(loss*1.66*10**-27)/(1*10**-27);\n", + "print\"\\n The mass loss in 10^-27 Kg/He formed is =\",m\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.18" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.18, Page:17 \n", + " \n", + "\n", + "\n", + " The energy loss in MeV is = 4.03123\n" + ] + } + ], + "source": [ + "#cal of The energy loss\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.18, Page:17 \\n \\n\"\n", + "#Given:\n", + "mh=1.007825;\n", + "mt=3.016049;\n", + "md=2.014102;\n", + "\n", + "#solution:\n", + "\n", + "m1=(mh+mt-2*md);\n", + "e=(-m1)*931; # in MeV\n", + "print\"\\n The energy loss in MeV is =\",e\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.19" + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.19, Page:18 \n", + " \n", + "\n", + "\n", + " The mean binding energy of tritium atom in MeV is = 2.81085817903\n", + "\n", + " The mean binding energy of nickel atom in MeV is = 8.71580311296\n" + ] + } + ], + "source": [ + "#cal of mean binding energy of tritium and nickel atom\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.19, Page:18 \\n \\n\"\n", + "#Given:\n", + "mh=1.007825;\n", + "mn=1.008665;\n", + "mt=3.016049; # atomic mass of Tritium\n", + "mNi=59.93528; # atomic mass of Nickel\n", + "\n", + "#solution:\n", + "\n", + "# part (a)\n", + "\n", + "B1=(1*mh+2*mn-mt)*931; # in MeV\n", + "Bh=B1/mt;\n", + "print\"\\n The mean binding energy of tritium atom in MeV is =\",Bh\n", + "\n", + "# part (b)\n", + "\n", + "B2=(28*mh+32*mn-mNi)*931; # in MeV\n", + "Bo=B2/mNi;\n", + "print\"\\n The mean binding energy of nickel atom in MeV is =\",Bo\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.20" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.20, Page:19 \n", + " \n", + "\n", + "\n", + " The mean binding energy of Cl (35) atom in MeV is = 8.52810201079\n", + "\n", + " The mean binding energy of Cl (37) atom in MeV is = 8.57839008383\n", + "\n", + " The increase in mean binding energy of Cl atom in MeV is = 0.0502880730447\n" + ] + } + ], + "source": [ + "#cal of mean binding energy of Cl\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.20, Page:19 \\n \\n\"\n", + "#Given:\n", + "mh=1.00783;\n", + "mn=1.00867;\n", + "m35=34.96885; # atomic mass of Cl (35)\n", + "m37=36.96590; # atomic mass of Cl (37)\n", + "\n", + "#solution:\n", + "\n", + "B1=(17*mh+18*mn-m35)*931; # in MeV\n", + "Bh=B1/m35;\n", + "print\"\\n The mean binding energy of Cl (35) atom in MeV is =\",Bh\n", + "\n", + "B2=(17*mh+20*mn-m37)*931; # in MeV\n", + "Bo=B2/m37;\n", + "print\"\\n The mean binding energy of Cl (37) atom in MeV is =\",Bo\n", + "\n", + "Bi=Bo-Bh;\n", + "print\"\\n The increase in mean binding energy of Cl atom in MeV is =\",Bi\n", + "\n", + "# NOTE: The answer depends upon how much precise value you take for atomic masses.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.21" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.21, Page:20 \n", + " \n", + "\n", + "\n", + " The mean binding energy of Na(22) in MeV is = 7.92358978299\n", + "\n", + " The mean binding energy of Na(23)in MeV is = 8.11544250059\n", + "\n", + " The mean binding energy of Na(24) in MeV is = 8.07172719656\n" + ] + } + ], + "source": [ + "#cal of mean binding energy of Na\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.21, Page:20 \\n \\n\"\n", + "#Given:\n", + "mh=1.0078;\n", + "mn=1.0087;\n", + "m22=21.99431;# atomic mass of Na 22\n", + "m23=22.9898;# atomic mass of Na 23\n", + "m24=23.9909;# atomic mass of Na 24\n", + "\n", + "#solution:\n", + "\n", + "# part (a)\n", + "\n", + "B1=((11*mh+11*mn)-m22)*931; # in MeV\n", + "Bh=B1/m22;\n", + "print\"\\n The mean binding energy of Na(22) in MeV is =\",Bh\n", + "\n", + "# part (b)\n", + "\n", + "B2=((11*mh+12*mn)-m23)*931; # in MeV\n", + "Bo=B2/m23;\n", + "print\"\\n The mean binding energy of Na(23)in MeV is =\",Bo\n", + "\n", + "# part (c)\n", + "\n", + "B3=((11*mh+13*mn)-m24)*931; # in MeV\n", + "Bs=B3/m24;\n", + "print\"\\n The mean binding energy of Na(24) in MeV is =\",Bs\n" + ] + }, + { + "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/Raj Phani/Raj Phani_version_backup/chapter_1_1.ipynb b/sample_notebooks/Raj Phani/Raj Phani_version_backup/chapter_1_1.ipynb new file mode 100755 index 00000000..6bb4fa49 --- /dev/null +++ b/sample_notebooks/Raj Phani/Raj Phani_version_backup/chapter_1_1.ipynb @@ -0,0 +1,1118 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# chapter1: electric charge" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.1" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false, + "scrolled": true + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.1, Page:3 \n", + " \n", + "\n", + "\n", + " The electric field in V/m is = 20000.0\n", + "\n", + " The force in N/C is = 20000.0\n", + "\n", + " The force on metal sphere in N is = 7.6e-05\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.1, Page:3 \\n \\n\"\n", + "#Given:\n", + "v=1000# potential\n", + "d=0.05# distance\n", + "q=3.8*10**-9# charge\n", + "\n", + "#solution:\n", + "e=v/d;#electric field\n", + "f=e;# force\n", + "f1=f*q;# force on metal sphere\n", + "print\"\\n The electric field in V/m is =\",e\n", + "print\"\\n The force in N/C is =\",f\n", + "print\"\\n The force on metal sphere in N is =\",f1\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.2" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.2, Page:4 \n", + " \n", + "\n", + "The potential in V is = 80.0\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.2, Page:4 \\n \\n\"\n", + "#Given:\n", + "energy=2*10**-6\n", + "c=2.5*10**-8# velocity of light\n", + "#solution:\n", + "v=energy/c# potential\n", + "print\"The potential in V is =\",v\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.3" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.3, Page:5 \n", + " \n", + "\n", + "The wavelength in Angstroms is = 3.88289589025\n", + "\n", + " The photon wavelength in Angstroms is = 9.11075e-05\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.3, Page:5 \\n \\n\"\n", + "#Given:\n", + "\n", + "energy=10 #in electron volts\n", + "m=9.1*10**-31# mass of electron in kg\n", + "h=6.626*10**-34# planck's constant J.s\n", + "c=3*10^8# speed of light in m/s\n", + "\n", + "#solution (a):\n", + "energy1=energy*1.6*10**-19# energy in J\n", + "p=(2*m*energy1)**0.5# momentum\n", + "wavelength=h/p*(10)**10\n", + "\n", + "print\"The wavelength in Angstroms is =\",wavelength\n", + "\n", + "\n", + "#solution (b):\n", + "wavelength1=h*c/energy1*(10)**10;#photon wavelength\n", + "\n", + "print\"\\n The photon wavelength in Angstroms is =\",wavelength1\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example1.4" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.4, Page:6 \n", + " \n", + "\n", + "The energy in eV is = 150.768804945\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.4, Page:6 \\n \\n\"\n", + "\n", + "#Given:\n", + "\n", + "wavelength=10**-10\n", + "m=9.1*10**-31\n", + "h=6.626*10**-34\n", + "\n", + "#solution:\n", + "\n", + "p=h/wavelength\n", + "e=p*p/(2*m) # energy in J\n", + "e1=e/(1.6*10**-19)# energy in eV\n", + "\n", + "print\"The energy in eV is =\",e1\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.5" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.5, Page:8 \n", + " \n", + "\n", + "\n", + " The wavelength in 10^-5 Angstroms is = 0.655671822473\n", + "\n", + " The wavelength in 10^-5 Angstroms is = 0.648946805494\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.5, Page:8 \\n \\n\"\n", + "\n", + "#Given:\n", + "\n", + "m=1.66*10**-27# 1u=1.66*10^-27 kg\n", + "h=6.6262*10**-34#planck's constant in J.s\n", + "energy1=120# in Mev for oxygen\n", + "energy2=140# in MeV for nitrogen\n", + "\n", + "#solution(a):\n", + "\n", + "p=(2*m*16*energy1*(1.6022*10**-13))**0.5\n", + "wavelength1=h/p*(10)**15#wavelength in 10^-5 Angstroms\n", + "\n", + "print\"\\n The wavelength in 10^-5 Angstroms is =\",wavelength1\n", + "\n", + "#solution (b):\n", + "\n", + "p=(2*m*14*energy2*(1.6022*10**-13))**0.5\n", + "wavelength2=h/p*(10)**15#wavelength in 10^-5 Angstroms\n", + "\n", + "print\"\\n The wavelength in 10^-5 Angstroms is =\",wavelength2\n", + "\n", + "# 1 Angstrom = 10^-10 m\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example1.6" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.6, Page:9 \n", + " \n", + "\n", + "The energy in eV is = 8275.0\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.6, Page:9 \\n \\n\"\n", + "\n", + "#Given:\n", + "\n", + "wavelength=1.5*10**-10\n", + "h=6.62*10**-34\n", + "c=3*10**8\n", + "\n", + "#solution:\n", + "\n", + "e=(h*c)/wavelength# energy in J\n", + "e1=e/(1.6*10**-19)# energy in eV\n", + "\n", + "print\"The energy in eV is =\",e1\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.7" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.7, Page:10 \n", + " \n", + "\n", + "\n", + " The threshold frequency in s^-1 is = 1.23634168427e+15\n", + "\n", + " The threshold wavelength in Angstroms is = 2426.51367187\n", + "\n", + " The energy of photoelectrone in eV is = 3.911875\n" + ] + } + ], + "source": [ + "#cal of elelectric field and force\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.7, Page:10 \\n \\n\"\n", + "\n", + "#Given:\n", + "\n", + "E=5.12*1.6*10**-19# energy in J\n", + "h=6.626*10**-34\n", + "c=3*10**8\n", + "wavelength=200*10**-9\n", + "w=2.3# in eV\n", + "\n", + "#solution:\n", + "\n", + "tf=E/h# (part a)\n", + "print\"\\n The threshold frequency in s^-1 is =\",tf\n", + "\n", + "tl=c/tf*10**10# (part b)\n", + "print\"\\n The threshold wavelength in Angstroms is =\",tl\n", + "\n", + "e=(h*c)/(wavelength*1.6*10**-19)# photon energy in eV (part c)\n", + "\n", + "pe=e-w\n", + "\n", + "print\"\\n The energy of photoelectrone in eV is =\",pe\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.8" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.8, Page:10 \n", + " \n", + "\n", + "\n", + " The velocity of alpha particles for 1 MeV in m/s is = 6941056.08394\n", + "\n", + " The velocity of alpha particles for 2 MeV in m/s is = 9816135.6511\n", + "\n", + " The velocity of deuteron particles for 1 MeV in m/s is = 9816135.6511\n", + "\n", + " The velocity of deuteron particles for 2 MeV in m/s is = 13882112.1679\n", + "\n", + " The velocity of proton particles for 1 MeV in m/s is = 13882112.1679\n", + "\n", + " The velocity of proton particles for 2 MeV in m/s is = 19632271.3022\n" + ] + } + ], + "source": [ + "#cal of velocity of alpha particles,deuteron,proton\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.8, Page:10 \\n \\n\"\n", + "#Given:\n", + "e1=1 # in MeV\n", + "e2=2 # in MeV\n", + "ma=4 # in u(amu)\n", + "md=2 # in u(amu)\n", + "mp=1 # in u(amu)\n", + "\n", + "# 1u = 1.6*10^-27 Kg\n", + "\n", + "#solution: part a)For alpha particles\n", + "\n", + "v1a=((2*e1*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of alpha particles for 1 MeV in m/s is =\",v1a# For 1 MeV\n", + "\n", + "v2a=((2*e2*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of alpha particles for 2 MeV in m/s is =\",v2a# For 2 MeV\n", + "\n", + "#solution: part b)For deuteron particles\n", + "\n", + "v1b=((2*e1*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of deuteron particles for 1 MeV in m/s is =\",v1b # For 1 MeV\n", + "\n", + "\n", + "v2b=((2*e2*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of deuteron particles for 2 MeV in m/s is =\",v2b # For 2 MeV\n", + "\n", + "#solution: part c)For proton particles\n", + "\n", + "v1p=((2*e1*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of proton particles for 1 MeV in m/s is =\",v1p # For 1 MeV\n", + "\n", + "\n", + "v2p=((2*e2*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n", + "print\"\\n The velocity of proton particles for 2 MeV in m/s is =\",v2p # For 2 MeV\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.9" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.9, Page:10 \n", + " \n", + "\n", + "The energy in MeV is = 933.919973435\n" + ] + } + ], + "source": [ + "#cal of The energy\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.9, Page:10 \\n \\n\"\n", + "#Given:\n", + "\n", + "m=1/(6.023*10**23)#mass of 1 atom in g\n", + "m1=m*10**-3#mass of 1 atom in Kg\n", + "c=3*10**8# velocity in m/s\n", + "#solution:\n", + "\n", + "e=m1*c*c; # energy in J\n", + "e1=e/(1.6*10**-13)# energy in MeV\n", + "\n", + "print\"The energy in MeV is =\",e1\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.10" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.1, Page:11 \n", + " \n", + "\n", + "The energy in eV is = 13.2638658253\n" + ] + } + ], + "source": [ + "#cal of The energy\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.1, Page:11 \\n \\n\"\n", + "#Given:\n", + "\n", + "enthalpy=1278 # enthalpy of combustion in kJ/mol\n", + "\n", + "#solution:\n", + "\n", + "energy=(enthalpy*1000)/(6.022*10**23*1.6*10**-19)\n", + "\n", + "print\"The energy in eV is =\",energy\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.11" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.11, Page:11 \n", + " \n", + "\n", + "\n", + " The mean binding energy of helium atom in MeV is = 7.0710038475\n", + "\n", + " The mean binding energy of oxygen atom in MeV is = 7.9800498909\n" + ] + } + ], + "source": [ + "#cal of mean binding energy of helium and oxygen\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.11, Page:11 \\n \\n\"\n", + "#Given:\n", + "mh=1.0078\n", + "mn=1.0087\n", + "ma=4.0026\n", + "mo=15.9949\n", + "Ah=4.0026 # atomic mass of helium\n", + "Ao=15.9949 # atomic mass of oxygen\n", + "\n", + "#solution:\n", + "\n", + "# part (a)\n", + "\n", + "B1=(2*mh+2*mn-ma)*931 # in MeV\n", + "Bh=B1/Ah\n", + "print\"\\n The mean binding energy of helium atom in MeV is =\",Bh\n", + "\n", + "# part (b)\n", + "\n", + "B2=(8*mh+8*mn-mo)*931 # in MeV\n", + "Bo=B2/Ao\n", + "print\"\\n The mean binding energy of oxygen atom in MeV is =\",Bo\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.12" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.12, Page:12 \n", + " \n", + "\n", + "\n", + " The mean binding energy of Be atom in MeV is = 7.05928572321\n", + "From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\n" + ] + } + ], + "source": [ + "#cal of \n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.12, Page:12 \\n \\n\"\n", + "#Given:\n", + "mh=1.0078;\n", + "mn=1.0087;\n", + "ABe=8.0053; # atomic mass of beryllium\n", + "\n", + "#solution:\n", + "\n", + "B1=(4*mh+4*mn-ABe)*931; # in MeV\n", + "Bh=B1/ABe;\n", + "print\"\\n The mean binding energy of Be atom in MeV is =\",Bh\n", + "\n", + "print\"From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.13" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.13, Page:12 \n", + " \n", + "\n", + "\n", + " The amount of coal required in Kg is = 2499.85671416\n" + ] + } + ], + "source": [ + "#cal of amount of coal\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.13, Page:12 \\n \\n\"\n", + "#Given:\n", + "\n", + "e=200; # in Mev\n", + "m=0.235; # weight of uranium atom in Kg\n", + "enthalpy=393.5; # in KJ/mol\n", + "Na=6.02*10**23;\n", + "\n", + "\n", + "#solution:\n", + "e1=e*1.6*10**-19*10**6;\n", + "atoms=Na/m;\n", + "e2=atoms*e1;#energy released in J\n", + "m1=(e2*12)/(393.5*1000*1000);# in Kg\n", + "m2=m1/1000;# in tons\n", + "print\"\\n The amount of coal required in Kg is =\", m2\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.14" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.14, Page:13 \n", + " \n", + "\n", + "\n", + " The energy release in part (a) in eV/molecule is = 2.51472\n", + "\n", + " The energy release in part (b) in eV/molecule is = 9.22688\n" + ] + } + ], + "source": [ + "#cal of The energy releases\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.14, Page:13 \\n \\n\"\n", + "#Given:\n", + "H1=241.8; # in KJ/mol\n", + "H2=887.2; # in KJ/mol\n", + "# 1 KJ/mol = 0.0104 eV/atom\n", + "\n", + "#solution: part (a)\n", + "e1=H1*0.0104;\n", + "print\"\\n The energy release in part (a) in eV/molecule is =\",e1\n", + "\n", + "#solution: part (b)\n", + "e2=H2*0.0104;\n", + "print\"\\n The energy release in part (b) in eV/molecule is =\",e2\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.15" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.15, Page:14 \n", + " \n", + "\n", + "\n", + " The energy release in part (a) in KJ/mol of carbondioxide is = 394.912\n", + "\n", + " The energy release in part (b) in KJ/mol of alumina is = 1675.968\n", + "\n", + " The energy release in part (c) in MJ/atom of U(235) is = 19264000.0\n" + ] + } + ], + "source": [ + "#cal of The energy releases\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.15, Page:14 \\n \\n\"\n", + "#Given:\n", + "H1=4.1; # in eV/molecule\n", + "H2=17.4; # in eV/molecule\n", + "H3=200;# in MeV/atom of U\n", + "\n", + "# 1 eV/atom = 96.32 KJ/mol\n", + "\n", + "#solution: part (a)\n", + "e1=H1*96.32;\n", + "print\"\\n The energy release in part (a) in KJ/mol of carbondioxide is =\",e1\n", + "\n", + "#solution: part (b)\n", + "e2=H2*96.32;\n", + "print\"\\n The energy release in part (b) in KJ/mol of alumina is =\",e2\n", + "\n", + "#solution: part (c)\n", + "e3=H3*1000*96.32;# in MJ/atom of U(235)\n", + "print\"\\n The energy release in part (c) in MJ/atom of U(235) is =\",e3\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.16" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.16, Page:15 \n", + " \n", + "\n", + "\n", + " The rate of energy release in W is 949251379.039\n" + ] + } + ], + "source": [ + "#cal of The rate of energy release\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.16, Page:15 \\n \\n\"\n", + "#Given:\n", + "e=200; #MeV/ atom of U\n", + "# 1 eV = 1.6*10^-19 J\n", + "Na=6.023*10**23;\n", + "M=0.235; # mass in Kg\n", + "\n", + "#solution:\n", + "\n", + "e1=e*1.6*10**-19*10**6;\n", + "A=Na/M;\n", + "e2=A*e1; # energy released in MJ/day\n", + "e3=e2/(24*3600);\n", + "print\"\\n The rate of energy release in W is \",e3\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.17" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.17, Page:16 \n", + " \n", + "\n", + "\n", + " The mass loss in 10^-27 Kg/He formed is = 0.046412244898\n" + ] + } + ], + "source": [ + "#cal of The mass loss\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.17, Page:16 \\n \\n\"\n", + "#Given:\n", + "e=26.03; # in MeV\n", + "\n", + "#solution:\n", + "\n", + "loss=e/931; #in atomic mass units (u)\n", + "# 1 u = 1.66*10^-27 Kg\n", + "m=(loss*1.66*10**-27)/(1*10**-27);\n", + "print\"\\n The mass loss in 10^-27 Kg/He formed is =\",m\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.18" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.18, Page:17 \n", + " \n", + "\n", + "\n", + " The energy loss in MeV is = 4.03123\n" + ] + } + ], + "source": [ + "#cal of The energy loss\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.18, Page:17 \\n \\n\"\n", + "#Given:\n", + "mh=1.007825;\n", + "mt=3.016049;\n", + "md=2.014102;\n", + "\n", + "#solution:\n", + "\n", + "m1=(mh+mt-2*md);\n", + "e=(-m1)*931; # in MeV\n", + "print\"\\n The energy loss in MeV is =\",e\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.19" + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.19, Page:18 \n", + " \n", + "\n", + "\n", + " The mean binding energy of tritium atom in MeV is = 2.81085817903\n", + "\n", + " The mean binding energy of nickel atom in MeV is = 8.71580311296\n" + ] + } + ], + "source": [ + "#cal of mean binding energy of tritium and nickel atom\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.19, Page:18 \\n \\n\"\n", + "#Given:\n", + "mh=1.007825;\n", + "mn=1.008665;\n", + "mt=3.016049; # atomic mass of Tritium\n", + "mNi=59.93528; # atomic mass of Nickel\n", + "\n", + "#solution:\n", + "\n", + "# part (a)\n", + "\n", + "B1=(1*mh+2*mn-mt)*931; # in MeV\n", + "Bh=B1/mt;\n", + "print\"\\n The mean binding energy of tritium atom in MeV is =\",Bh\n", + "\n", + "# part (b)\n", + "\n", + "B2=(28*mh+32*mn-mNi)*931; # in MeV\n", + "Bo=B2/mNi;\n", + "print\"\\n The mean binding energy of nickel atom in MeV is =\",Bo\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.20" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.20, Page:19 \n", + " \n", + "\n", + "\n", + " The mean binding energy of Cl (35) atom in MeV is = 8.52810201079\n", + "\n", + " The mean binding energy of Cl (37) atom in MeV is = 8.57839008383\n", + "\n", + " The increase in mean binding energy of Cl atom in MeV is = 0.0502880730447\n" + ] + } + ], + "source": [ + "#cal of mean binding energy of Cl\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.20, Page:19 \\n \\n\"\n", + "#Given:\n", + "mh=1.00783;\n", + "mn=1.00867;\n", + "m35=34.96885; # atomic mass of Cl (35)\n", + "m37=36.96590; # atomic mass of Cl (37)\n", + "\n", + "#solution:\n", + "\n", + "B1=(17*mh+18*mn-m35)*931; # in MeV\n", + "Bh=B1/m35;\n", + "print\"\\n The mean binding energy of Cl (35) atom in MeV is =\",Bh\n", + "\n", + "B2=(17*mh+20*mn-m37)*931; # in MeV\n", + "Bo=B2/m37;\n", + "print\"\\n The mean binding energy of Cl (37) atom in MeV is =\",Bo\n", + "\n", + "Bi=Bo-Bh;\n", + "print\"\\n The increase in mean binding energy of Cl atom in MeV is =\",Bi\n", + "\n", + "# NOTE: The answer depends upon how much precise value you take for atomic masses.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 1.21" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 1.21, Page:20 \n", + " \n", + "\n", + "\n", + " The mean binding energy of Na(22) in MeV is = 7.92358978299\n", + "\n", + " The mean binding energy of Na(23)in MeV is = 8.11544250059\n", + "\n", + " The mean binding energy of Na(24) in MeV is = 8.07172719656\n" + ] + } + ], + "source": [ + "#cal of mean binding energy of Na\n", + "#intiation of all variables\n", + "# Chapter 1\n", + "print\"Example 1.21, Page:20 \\n \\n\"\n", + "#Given:\n", + "mh=1.0078;\n", + "mn=1.0087;\n", + "m22=21.99431;# atomic mass of Na 22\n", + "m23=22.9898;# atomic mass of Na 23\n", + "m24=23.9909;# atomic mass of Na 24\n", + "\n", + "#solution:\n", + "\n", + "# part (a)\n", + "\n", + "B1=((11*mh+11*mn)-m22)*931; # in MeV\n", + "Bh=B1/m22;\n", + "print\"\\n The mean binding energy of Na(22) in MeV is =\",Bh\n", + "\n", + "# part (b)\n", + "\n", + "B2=((11*mh+12*mn)-m23)*931; # in MeV\n", + "Bo=B2/m23;\n", + "print\"\\n The mean binding energy of Na(23)in MeV is =\",Bo\n", + "\n", + "# part (c)\n", + "\n", + "B3=((11*mh+13*mn)-m24)*931; # in MeV\n", + "Bs=B3/m24;\n", + "print\"\\n The mean binding energy of Na(24) in MeV is =\",Bs\n" + ] + }, + { + "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