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| author | Trupti Kini | 2016-07-27 23:30:30 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-07-27 23:30:30 +0600 |
| commit | 72a8809fba69879b5a1733adab3d9ce4e79ff369 (patch) | |
| tree | f94ba99f588847f2dfbdc8e352091956b5632f5c /sample_notebooks | |
| parent | 33c8c0a7f0fef7bfe22cbc68c215549d003012be (diff) | |
| download | Python-Textbook-Companions-72a8809fba69879b5a1733adab3d9ce4e79ff369.tar.gz Python-Textbook-Companions-72a8809fba69879b5a1733adab3d9ce4e79ff369.tar.bz2 Python-Textbook-Companions-72a8809fba69879b5a1733adab3d9ce4e79ff369.zip | |
Added(A)/Deleted(D) following books
A "sample_notebooks/Reshma Ustad/Chapter_2_Properties_Of_Material.ipynb"
A sample_notebooks/kartiksankhla/Chapter2_1.ipynb
A sample_notebooks/pratiksonone/Chapter1_1.ipynb
Diffstat (limited to 'sample_notebooks')
| -rw-r--r-- | sample_notebooks/Reshma Ustad/Chapter_2_Properties_Of_Material.ipynb | 146 | ||||
| -rw-r--r-- | sample_notebooks/kartiksankhla/Chapter2_1.ipynb | 758 | ||||
| -rw-r--r-- | sample_notebooks/pratiksonone/Chapter1_1.ipynb | 710 |
3 files changed, 1614 insertions, 0 deletions
diff --git a/sample_notebooks/Reshma Ustad/Chapter_2_Properties_Of_Material.ipynb b/sample_notebooks/Reshma Ustad/Chapter_2_Properties_Of_Material.ipynb new file mode 100644 index 00000000..823b8e71 --- /dev/null +++ b/sample_notebooks/Reshma Ustad/Chapter_2_Properties_Of_Material.ipynb @@ -0,0 +1,146 @@ +{ + "metadata": { + "name": "Chapter 2 Properties Of Material" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 2 Properties Of Material" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1 Page No:19" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nL=5 # length of steel bar in m\nd=25*10**-3 # diametr of steel bar in mm\ndeltaLt=25*10**-3 #steel \npt=800 # power load of steel bar in N\n\n\n#calculation\nA=((pi/4)*((deltaLt)**2)) #Cross-section area\nsigmat=(pt)/(A) #Stress in steel bar\net=(deltaLt)/L #strain in steel bar\nE=(sigmat)/(et) #Young's modulus\n\n\n#output\nprint(\"value of Cross-section area A=\",A,\"m**2\")\nprint(\"value of tress in steel bar sigmat=\",sigmat,\"MN/m**2\")\nprint(\"value of strain in steel bar et= \",et)\nprint(\"value of Young's modulus E \",E,\"N/m**2\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "value of Cross-section area A= 0.0004906250000000001 m**2\nvalue of tress in steel bar sigmat= 1630573.248407643 MN/m**2\nvalue of strain in steel bar et= 0.005\nvalue of Young's modulus E 326114649.6815286 N/m**2\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2 Page No:20\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nL=300*10**-3 #length of hexagonal prismatic steel bar in mm\nA=500*10**-6 #Area of cross section of steel bar mm**2\nPt=500*10**3 # load of steel bar in KN\nE=210*10**9 # modulus of elasticity GN/m**2\n\n#Calculation\nsigmat=((Pt)/(A)) #stress in steel bar\net=((sigmat)/(E)) #strain steel bar is\ndeltaLt=((et)*(L)) #therefore,elongation of the steel bar is given by\n\n#output\nprint('stress in steel bar =',sigmat,\"N/m**2\")\nprint('therefore,strain steel bar is given by =',et,)\nprint('therefore,elongation of the steel bar is given by=',deltaLt,\"m\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "stress in steel bar = 1000000000.0 N/m**2\ntherefore,strain steel bar is given by = 0.004761904761904762\ntherefore,elongation of the steel bar is given by= 0.0014285714285714286 m\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3 Page No:21\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input Data\nPt=600 #tensils force in N\nd=2*10**-3 #diameter of steel wire in mm\nL=15 #length of wire in m\nE=210*10**9 #modulus of elasticity of the material in GN/M**2\npi=3.1482\n\n\n#Calculation\nA=((pi/4)*((d)**2)) #(1)cross section area\nsigmat=(Pt)/(A) # stress in the steel wire \net=((sigmat)/(E)) #(2)therefore, strain in steel wire is given by\ndeltaLt=et*L #(3)Enlongation of the steel wire is given by \npe=((deltaLt/L)*100) #(4)percentage elongation\n\n\n#Output\nprint(\"cross section area A= \",A,\"m**2\")\nprint(\"stress in the steel wire sigmat=\",sigmat,\"GN/m**2\")\nprint(\"modulus of elasticity et=\",et,)\nprint(\"strain in steel wire deltaLt=\",deltaLt,\"mm\")\nprint(\"percentage elongation\",pe,\"%\")\n\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "cross section area A= 3.1481999999999998e-06 m**2\nstress in the steel wire sigmat= 190585096.24547362 GN/m**2\nmodulus of elasticity et= 0.0009075480773593982\nstrain in steel wire deltaLt= 0.013613221160390973 mm\npercentage elongation 0.09075480773593982 %\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 4 Page No:22\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#input data\nA=30*30*10**-6 #area of square rod in mm**2\nL=5 #length of square rod in m\nPc=150*10**3 # axial comperessive load of a rod in kN\nE=215*10**9 # modulus of elasticity in GN/m**2\n\n\n#Calculation\nsigmac=((Pc)/(A)) #stress in square rod\nec=((sigmac)/(E)) #modulusof elasticity is E=sigmac/ec ,therefore strain in square rod is\ndeltaLc=ec*5 #therefore shortening of length of the rod \n\n\n#Output\nprint (\"stress in square rod\",sigmac,\"N/m**2\")\nprint(\"strain in square rod ec=\",ec,)\nprint(\"shortening of length of the rod=\",deltaLc,\"m\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "stress in square rod 166666666.66666666 N/m**2\nstrain in square rod ec= 0.0007751937984496124\nshortening of length of the rod= 0.003875968992248062 m\n" + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5 Page No:23" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#input data\nd=50*10**-6 #diameter of metalic rod in mm**2\nL=220*10**-3 #length of metalic rod in mm\nPt=40*10**3 #load of metalic rod in KN\ndeltaLt=0.03*10**-3 #elastic enlongation in mm\nypl=160*10**3 # yield point load in KN\nml=250*10**3 #maximum load in KN\nlsf=270*10**-3 #length of specimen at fracture in mm\npi=3.1482\n\n#calculation\nA=(((pi)/(4)*((d)**2))) #(1)cross section area\nsigmat=(Pt/A) #stress in metallic rod\net=(deltaLt/L) #strain n metallic rod\nE=(sigmat/et) #young's modulus\nys=(ypl/A) #(2)yeild strength\nuts=(ml/A) #(3)ultimate tensile strength\nPebf=((lsf-L)/L)*100 #percentage elongation before fracture \n\n\n\n#output\nprint(\"cross section area\",A,\"m**2\")\nprint(\"stress in metallic rod\",sigmat,\"N/m**2\")\nprint(\"strain n metallic rod\",et,)\nprint(\"young's modulus\",E,\"GN/m**2\")\nprint(\"yeild strength\",ys,\"MN/m**2\")\nprint(\"ultimate tensile strength\",uts,\"MN/m**2\")\nprint(\"percentage elongation before fracture\",Pebf,\"%\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "cross section area 1.967625e-09 m**2\nstress in metallic rod 20329076932850.52 N/m**2\nstrain n metallic rod 0.00013636363636363637\nyoung's modulus 1.4907989750757046e+17 GN/m**2\nyeild strength 81316307731402.08 MN/m**2\nultimate tensile strength 127056730830315.75 MN/m**2\npercentage elongation before fracture 22.727272727272734 %\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6 Page No:24\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#input data\nA=50*50*10**-6 #area ofsquare metal bar in mm**2\nPc=600*10**3 #axial compress laod in KN\nL=200*10**-3 # gauge length of metal bar in mm\ndeltaLc=0.4*10**-3 #contraction length of metal bar in mm\ndeltaLlateral=0.05*10**-3 #lateral length of metal bar in mm\n\n#Calculation\nsigmac=((Pc)/(A)) #stress in square metal bar \nec=((deltaLc)/(L)) #longitudinal or linear strain in square metal bar\nE =((sigmac)/(ec)) #smodule of elasticity\nelateral=((deltaLlateral)/(L)) #lateral strain in square metal bar\npoissonsratio=(elateral)/(ec)\n\n\n#output\nprint(\"stress in bar=\",sigmac,\"n/m**2\")\nprint(\"longitudinal or linear strain in square metal bar=\",ec,)\nprint(\"module of elasticity=\",E,\"N/m**2\")\nprint(\"lateral strain in square metal bar=\",elateral,)\nprint(\"poissons ratio=\",poissonsratio,)\n\n#poisson's ratio", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "stress in bar= 240000000.0 n/m**2\nlongitudinal or linear strain in square metal bar= 0.002\nmodule of elasticity= 120000000000.0 N/m**2\nlateral strain in square metal bar= 0.00025\npoissons ratio= 0.125\n" + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/sample_notebooks/kartiksankhla/Chapter2_1.ipynb b/sample_notebooks/kartiksankhla/Chapter2_1.ipynb new file mode 100644 index 00000000..821531df --- /dev/null +++ b/sample_notebooks/kartiksankhla/Chapter2_1.ipynb @@ -0,0 +1,758 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:41d72ff7de4eab79477527ee635fed63748fe3041e37a18b80daacf784f19007"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter2-Radioactivity and Isotopes"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa2.1: : Page-88 (2011) \n",
+ "#find Weight of one Curie of RaB and Weight of one Rutherford of RaB\n",
+ "T = 26.8*60; ## Half life of the substance, s\n",
+ "C = 3.7e+010; ## One curie, disintegration per sec\n",
+ "N = 6.022137e+026; ## Avogadro number, per kmol\n",
+ "m = 214.; ## Molecular weight of RaB, kg/kmol\n",
+ "R = 1e+006; ## One Rutherford, disintegration per sec.\n",
+ "W_C = C*T*m/(N*0.693); ## Weight of one Curie of RaB, Kg \n",
+ "W_R = R*T*m/(N*0.693); ## Weight of one Rutherford of RaB, Kg \n",
+ "print\"%s %.2e %s %.2e %s \"%(\"\\nWeight of one Curie of RaB : \",W_C,\" Kg\"and \" \\nWeight of one Rutherford of RaB : \",W_R,\" Kg\");\n",
+ "\n",
+ "## Result\n",
+ "## Weight of one Curie of RaB : 3.051e-011 Kg \n",
+ "## Weight of one Rutherford of RaB : 8.245e-016 Kg "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Weight of one Curie of RaB : 3.05e-11 \n",
+ "Weight of one Rutherford of RaB : 8.25e-16 Kg \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.2 : : Page 88 (2011)\n",
+ "#find The maximum activity of Na-24 and The time needed to produced 90 percent of the maximum activity\n",
+ "import math\n",
+ "T_h = 14.8; ## Half life of Na-24, hours\n",
+ "Q = 1e+008; ## Production rate of Na-24, per sec\n",
+ "L = 0.693/T_h; ## Decay constant, per sec\n",
+ "t = 2.; ## Time after the bombardment, hours\n",
+ "A = Q/3.7e+010*1000; ## The maximum activity of Na-24, mCi\n",
+ "T = -1*math.log(0.1)/L; ## The time needed to produced 90% of the maximum activity, hour\n",
+ "N = 0.9*Q*3600./L*math.e**(-L*t); ## Number of atoms of Na-24 left two hours after bombardment was stopped\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nThe maximum activity of Na-24 = \",A,\" mCi\"and \"\\nThe time needed to produced 90 percent of the maximum activity =\",T,\" hrs\" and\"\\nNumber of atoms of Na-24 left two hours after bombardment was stopped = \",N,\"\")\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## The maximum activity of Na-24 = 2.7 mCi\n",
+ "## The time needed to produced 90 percent of the maximum activity = 49.2 hrs \n",
+ "## Number of atoms of Na-24 left two hours after bombardment was stopped = 6.30e+012 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The maximum activity of Na-24 = 2.70 \n",
+ "The time needed to produced 90 percent of the maximum activity = 49.17 \n",
+ "Number of atoms of Na-24 left two hours after bombardment was stopped = 6300897280447.79 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.3: : Page 89 (2011)\n",
+ "#find The activity of K-40 and disintegrations\n",
+ "T = 1.31e+09*365*24*60*60; ## Half life of the substance,sec\n",
+ "N = 6.022137e+026; ## Avogadro number.\n",
+ "m = 0.35*0.012*10**-2; ## Mass of K-40, Kg.\n",
+ "A = m*N*0.693/(T*40); ## Activity of K-40, disintegrations/sec. \n",
+ "print'%s %.2e %s %.3f %s '%(\"\\nThe activity of K-40 = \",A,\" \"and \"disintegrations/sec = \",A/3.7e+004,\" micro-curie\");\n",
+ "\n",
+ "## Result\n",
+ "## The activity of K-40 = 1.061e+004 disintegrations/sec = 0.287 micro-curie "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The activity of K-40 = 1.06e+04 disintegrations/sec = 0.287 micro-curie \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.4 : : Page 89 (2011)\n",
+ "#find The age of the boat\n",
+ "import math\n",
+ "T = 5568; ## Half life of the C-14,years\n",
+ "D = 0.693/T; ## Disintegration constant, years^-1.\n",
+ "N_0 = 15.6/D; ## Activity of fresh carbon, dpm .gm\n",
+ "N = 3.9/D; ## Activity of an ancient wooden boat,dpm.gm.\n",
+ "t = 1/(D)*math.log(N_0/N); ## Age of the boat, years\n",
+ "print'%s %.2e %s'%(\"\\nThe age of the boat : \",t,\" years\")\n",
+ "\n",
+ "## Result\n",
+ "## The age of the boat : 1.114e+004 years"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The age of the boat : 1.11e+04 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.5 : : Page 90 (2011)\n",
+ "#find The activity of U-234 \n",
+ "import math\n",
+ "m_0 = 3e-06;## Initial mass of the U-234, Kg\n",
+ "A = 6.022137e+026; ##Avagadro's number, atoms\n",
+ "N_0 = m_0*A/234.; ## Initial number of atoms\n",
+ "T = 2.50e+05; ## Half life, years\n",
+ "D= 0.693/T; ## Disintegration constant\n",
+ "t = 150000; ## Disintegration time, years\n",
+ "m = m_0*math.e**(-D*t); ## Mass after time t,Kg\n",
+ "activity = m*D/(365.*24.*60.*60.)*A/234.; ## Activity of U-234 after time t,dps\n",
+ "print'%s %.1f %s %.2f %s'%(\"\\nThe activity of U-234 after \",t,\"\"and\" yrs = \",activity,\" disintegrations/sec\");\n",
+ "\n",
+ "## Result\n",
+ "## The activity of U-234 after 150000 yrs = 4.478e+005 disintegrations/sec"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The activity of U-234 after 150000.0 447778.26 disintegrations/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.6 : : Page 90 (2011)\n",
+ "#find The number of alpha decays in Th-232\n",
+ "import math\n",
+ "A = 6.022137e+023; ##Avagadro's number, atoms\n",
+ "N_0 = A/232.; ## Initial number of atoms\n",
+ "t = 3.150e+07; ## Decay time, sec\n",
+ "D = 1.58e-018; ## Disintegration constant,sec^-1\n",
+ "N = D*t*N_0; ## Number of alpha decays in Th-232\n",
+ "print'%s %.2e %s'%(\"\\nThe number of alpha decays in Th-232 = \", N,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The number of alpha decays in Th-232 = 1.29e+011"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The number of alpha decays in Th-232 = 1.29e+11 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.7 : : Page 90 (2011)\n",
+ "#find The maximum possible age of the earth crust\n",
+ "import math\n",
+ "T_238 = 4.5e+09;## Half life of U-238, years\n",
+ "T_235 = 7.13e+08; ## Half life of U-238, years\n",
+ "lambda_238 = 0.693/T_238; ## Disintegration constant of U-238, years^-1\n",
+ "lambda_235 = 0.693/T_235; ## Disintegration constant of U-235, years^-1 \n",
+ "N = 137.8; ## Abundances of U-238/U-235\n",
+ "t = math.log(N)/(lambda_235 - lambda_238);## Age of the earth's crust, years\n",
+ "print'%s %.2e %s'%(\"\\nThe maximum possible age of the earth crust = \",t,\" years\");\n",
+ "\n",
+ "## Result \n",
+ "## The maximum possible age of the earth crust = 6.022e+009 years "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The maximum possible age of the earth crust = 6.02e+09 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.8 : : Page 91 (2011)\n",
+ "#find The number of half lives in radon-222\n",
+ "import math\n",
+ "N = 10; ## Number of atoms left undecayed in Rn-222\n",
+ "n = math.log(10)/math.log(2); ## Number of half lives in Ra-222\n",
+ "print'%s %.2f %s'%(\"\\nThe number of half lives in radon-222 = \", n,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The number of half lives in radon-222 = 3.322 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The number of half lives in radon-222 = 3.32 \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.9 : : Page 91 (2011)\n",
+ "#find The weight of Po-210 after one year and The initial activity of the material\n",
+ "import math\n",
+ "M_Po = 209.9829; ## Mass of Polonium, g\n",
+ "M_Pb = 205.9745; ## Mass of lead, g\n",
+ "A = 6.22137e+023; ## Avogadro's number\n",
+ "M_He = 4.0026; ## Mass of alpha particle, g\n",
+ "C = 3e+08; ## Velocity of light, m/s\n",
+ "T = 138*24*3600; ## Half life, sec\n",
+ "P = 250; ## Power produced, joule/sec\n",
+ "Q = (M_Po-M_Pb-M_He)*931.25; ## disintegration energy, MeV\n",
+ "D = 0.693/T; ## Disintegration constant, per year\n",
+ "N = P/(D*Q*1.60218e-013); ## Number of atoms, atom\n",
+ "N_0 = N*math.e**(1.833); ## Number of atoms present initially, atom\n",
+ "W = N_0/A*210; ## Weight of Po-210 after one year, g\n",
+ "A_0 = N_0*D/(3.7e+010); ## Initial activity, curie\n",
+ "print'%s %.2f %s %.2e %s'%(\"\\nThe weight of Po-210 after one year = \",W,\" g\"and\" \\nThe initial activity of the material = \",A_0,\" curies\");\n",
+ "\n",
+ "## Result\n",
+ "## The weight of Po-210 after one year = 10.49 g \n",
+ "## The initial activity of the material = 4.88e+004 curies"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The weight of Po-210 after one year = 10.49 \n",
+ "The initial activity of the material = 4.88e+04 curies\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.10 : : Page 91 (2011)\n",
+ "#find The decay constant for total emission and The decay constant for beta_decay lambda_b and The decay constant for beta_decay lambda_a\n",
+ "import math\n",
+ "lambda_t = 0.693/(60.5*60);## Total decay constant, per sec\n",
+ "lambda_a = 0.34*lambda_t;## Decay constant for alpha_decay, per sec\n",
+ "lambda_b = 0.66*lambda_t;## Decay constant for beta_decay, per sec\n",
+ "print'%s %.2e %s'%(\"\\nThe decay constant for total emission = \",lambda_t,\" /sec\");\n",
+ "print'%s %.2e %s'%(\"\\nThe decay constant for beta_decay lambda_b = \",lambda_b,\" /sec\");\n",
+ "print'%s %.2e %s'%(\"\\nThe decay constant for alpha_decay lambda_a = \",lambda_a,\" /sec\");\n",
+ "\n",
+ "## Result \n",
+ "## The decay constant for total emission = 1.91e-004 /sec\n",
+ "## The decay constant for beta_decay lambda_b = 1.26e-004 /sec\n",
+ "## The decay constant for alpha_decay lambda_a = 6.49e-005 /sec "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The decay constant for total emission = 1.91e-04 /sec\n",
+ "\n",
+ "The decay constant for beta_decay lambda_b = 1.26e-04 /sec\n",
+ "\n",
+ "The decay constant for alpha_decay lambda_a = 6.49e-05 /sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.13 : : Page 93 (2011)\n",
+ "#find The half life of Pu-239 \n",
+ "import math\n",
+ "M_A = 4.; ## Mass of alpha particle, amu\n",
+ "M_U = 235.; ##Mass of U-235, amu\n",
+ "M_P = 239.; ## Mass of P-239, amu\n",
+ "Amount = 120.1; ## quantity of P-239, g\n",
+ "E_A = 5.144; ## Energy of emitting alpha particles, Mev\n",
+ "E_R = (2.*M_A)/(2.*M_U)*E_A; ## The recoil energy of U-235, Mev\n",
+ "E = E_R + E_A; ## The energy released per disintegration, Mev\n",
+ "P = 0.231; ## Evaporation rate, watt\n",
+ "D = P/(E*1.60218e-013); ## Disintegration rate, per sec\n",
+ "A = 6.022137e+023; ## Avagadro's number, atoms\n",
+ "N = Amount/M_P*A; ## Number of nuclei in 120.1g of P-239\n",
+ "T = 0.693/(D*3.15e+07)*N; ## Half life of Pu_239, years\n",
+ "print'%s %.2e %s'%(\"\\nThe half life of Pu-239 = \",T,\" years\");\n",
+ "\n",
+ "## Result \n",
+ "## The half life of Pu-239 = 2.42e+004 years "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The half life of Pu-239 = 2.42e+04 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.14 : : Page 93 (2011)\n",
+ "#find The disintegration rate of Au-199\n",
+ "import math\n",
+ "T_h_1 = 2.7*24*3600; ## Half life of Au-198, sec\n",
+ "T_h_2 = 3.15*24*3600; ## Half life of Au-199, sec\n",
+ "S_1 = 99e-028; ## Crossection for first reaction, Sq.m\n",
+ "S_2 = 2.6e-024; ## Crossection for second reaction, Sq.m\n",
+ "I = 1e+018; ## Intensity of radiation, per Sq.m per sec\n",
+ "L_1 = I*S_1; ## Decay constant of Au-197, per sec\n",
+ "L_2 = 0.693/T_h_1+I*S_2; ## Decay constant of Au-198, per sec\n",
+ "L_3 = 0.693/T_h_2; ## Decay constant of Au-199, per sec\n",
+ "N_0 = 6.022137e+023; ## Avogadro number\n",
+ "N_1 = N_0/197.; ## Initial number of atoms of Au-197\n",
+ "t = 30.*3600.; ## Given time, sec\n",
+ "p = (math.exp(-L_1*t))/((L_2-L_1)*(L_3-L_1));\n",
+ "q = (math.exp(-L_2*t))/((L_1-L_2)*(L_3-L_2));\n",
+ "r = (math.exp(-L_3*t))/((L_1-L_3)*(L_2-L_3));\n",
+ "N3 = N_1*L_1*L_2*(p+q+r);\n",
+ "N_199 = N3;\n",
+ "L = L_3*N_199; ## Disintegration rate of Au-199, per sec\n",
+ "print'%s %.2e %s'%(\"\\nThe disintegration rate of Au-199 = \", L,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The disintegration rate of Au-199 = 1.9e+012 (Wrong answer in the textbook)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The disintegration rate of Au-199 = 1.88e+12 \n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.15 : : Page 94 (2011)\n",
+ "#find The maximum activity of Na-24 and The activity after a continuous bombardment\n",
+ "import math\n",
+ "Y = 110e-03;## Yield of Na-24, mCi/hr\n",
+ "T = 14.8;## Half life of Na-24, hours\n",
+ "t = 8;## Time after which activity to be compute, hours\n",
+ "D = 0.693/T;## Disintegration constant, hours^-1\n",
+ "A = 1.44*Y*T;## Maximum activity of Na-24, Ci\n",
+ "A_C = A*(1-math.e**(-D*t));## Activity after a continuous bombardment, Ci\n",
+ "Activity = A_C*(math.e**(-D*t));## Activity after 8hours, Ci\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nThe maximum activity of Na-24 = \",A,\" Ci\"and \"\\nThe activity after a continuous bombardment = \",A_C,\" Ci\"and \"\\nThe activity after 8hours = \",Activity,\" Ci\")\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## The maximum activity of Na-24 = 2.344 Ci\n",
+ "## The activity after a continuous bombardment = 0.7324 Ci\n",
+ "## The activity after 8hours = 0.50360 Ci"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The maximum activity of Na-24 = 2.34 \n",
+ "The activity after a continuous bombardment = 0.73 \n",
+ "The activity after 8hours = 0.50 Ci \n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.16 : : Page 94 (2011)\n",
+ "#find The energy of beta rays emitted per gram of tissue\n",
+ "import math\n",
+ "A_0 = 3.7e+07; ## Initial activity, disintegrations per sec\n",
+ "T = 12.6; ## Half life of I-130, hours\n",
+ "t = 24*3600; ## time for dose absorbed calculation,sec\n",
+ "E = 0.29*1.6e-06; ## Average energy of beta rays, ergs\n",
+ "m = 2; ## Mass of iodine thyroid tissue, gm\n",
+ "l = 0.693/(T*3600); ## Disintegration constant, sec^-1\n",
+ "N_0 = A_0/l; ## Initial number of atoms\n",
+ "N = N_0*(1-math.e**(-l*t)); ## Number of average atoms disintegrated\n",
+ "E_A = N*E; ## Energy of beta rays emitted, ergs\n",
+ "E_G = E_A/(2*97.00035); ## Energy of beta rays emitted per gram of tissue, REP \n",
+ "print'%s %.2f %s'%(\"\\nThe energy of beta rays emitted per gram of tissue = \",E_G,\" REP\");\n",
+ "\n",
+ "## Result\n",
+ "## The energy of beta rays emitted per gram of tissue = 4245.0 REP "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy of beta rays emitted per gram of tissue = 4245.02 REP\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex18-pg95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.18 : : Page 95 (2011)\n",
+ "#find The activity of Au-198 and The maximum amount of Au-198 produced\n",
+ "import math\n",
+ "N_0 = 6.022137e+023; ## Avagadro number\n",
+ "d = 0.02; ## Thickness of the foil, cm\n",
+ "R = 19.3; ## Density of Au,g/cc\n",
+ "N_1 = d*R/197.*N_0; ## Initial number of Au-197 nuclei per unit area of foil,cm^-2\n",
+ "T_H = 2.7*24.*3600.; ## Half life of Au-198,sec\n",
+ "L = math.log(2)/T_H; ## Decay constant for Au-198,sec^-1\n",
+ "I = 10**12; ## Intensity of neutron beam,neutrons/cm^2/sec\n",
+ "S = 97.8e-024; ## Cross section for reaction,cm^-2\n",
+ "t = 5*60.; ## Reaction time,s\n",
+ "A = S*I*N_1*(1-math.e**(-L*t)); ## Activity of Au-198,cm^-2sec^-1\n",
+ "N_2 = S*I*N_1/L; ## The maximum amount of Au-198 produced,cm^-2\n",
+ "print'%s %.2e %s %.2e %s '%(\"\\nThe activity of Au-198 = \",A,\" per Sq.cm per sec\"and \"\\nThe maximum amount of Au-198 produced = \",N_2,\" per Sq.cm\");\n",
+ "\n",
+ "## Result\n",
+ "## The activity of Au-198 = 1.028e+008 per Sq.cm per sec\n",
+ "## The maximum amount of Au-198 produced = 3.88e+016 per Sq.cm "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The activity of Au-198 = 1.03e+08 \n",
+ "The maximum amount of Au-198 produced = 3.88e+16 per Sq.cm \n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex19-pg95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.19 : : Page 95 (2011)\n",
+ "#find The power released by the gm molecule of Pu-238 and The time in which power reduces to 1/8 time of its initial value\n",
+ "import math\n",
+ "N_0 = 6.022137e+023; ## Avagadro number \n",
+ "T_P = 90.*365.*24.*3600.; ## Half life of Pu-238,s\n",
+ "L_P = 0.693/T_P ; ## Decay constant of Pu-238,s^-1\n",
+ "E = 5.5; ## Energy of alpha particle, MeV\n",
+ "P =E*L_P*N_0; ## Power released by the gm molecule of Pu-238,MeV/s\n",
+ "t = math.log(8)/(L_P*365.*24.*3600.); ## Time in which power reduces to 1/8 time of its initial value \n",
+ "print'%s %.2e %s %.1f %s '%(\"\\nThe power released by the gm molecule of Pu-238 = \",P,\" MeV/s\"and \"\\nThe time in which power reduces to 1/8 time of its initial value = \",t,\" yrs\")\n",
+ "\n",
+ "## Result\n",
+ "## The power released by the gm molecule of Pu-238 = 8.09e+014 MeV/s \n",
+ "## The time in which power reduces to 1/8 time of its initial value = 270 yrs "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The power released by the gm molecule of Pu-238 = 8.09e+14 \n",
+ "The time in which power reduces to 1/8 time of its initial value = 270.1 yrs \n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex20-pg96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.20 : : Page 96 (2011)\n",
+ "#find The required time for decay of daughter nucleus and The number of nuclei of daughter isotope\n",
+ "import math\n",
+ "N_1 = 10**20; ## Number of nuclei of parent isotopes\n",
+ "T_P = 10**4; ## Half life of parent nucleus,years\n",
+ "T_D = 20.; ## Half life of daughter nucleus,years\n",
+ "T = 10**4; ## Given time,years\n",
+ "L_P = 0.693/T_P ; ## Decay constant of parent nucleus,years^-1\n",
+ "L_D = 0.693/T_D ; ## Decay constant of daughter nucleus,years^-1\n",
+ "t_0 = math.log(0.03)/(L_P-L_D); ## Required time for decay of daughter nucleus,years\n",
+ "N = L_P/L_D*(math.e**(-L_P*T)-math.e**(-L_D*T))*N_1; ## Number of nuclei of daughter isotope\n",
+ "print'%s %.2f %s %.0e %s'%(\"\\nThe required time for decay of daughter nucleus =\",t_0,\" yr\"and \" \\nThe number of nuclei of daughter isotope = \",N,\" \");\n",
+ "\n",
+ "## Result\n",
+ "## The required time for decay of daughter nucleus = 101 yr \n",
+ "## The number of nuclei of daughter isotope = 1e+017 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The required time for decay of daughter nucleus = 101.40 \n",
+ "The number of nuclei of daughter isotope = 1e+17 \n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/sample_notebooks/pratiksonone/Chapter1_1.ipynb b/sample_notebooks/pratiksonone/Chapter1_1.ipynb new file mode 100644 index 00000000..e45e4d58 --- /dev/null +++ b/sample_notebooks/pratiksonone/Chapter1_1.ipynb @@ -0,0 +1,710 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:3cf7334f303e62e9d7ee028942cc3f913423a0b853c51242dbc7d672a0097072"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter1-General properties of Atomic Nucleus"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa1.1 : : Page 51 (2011)\n",
+ "#calculate distance of closet apporach\n",
+ "Z = 79.; ## Atomic number of Gold \n",
+ "z = 1.; ## Atomic number of Hydrogen\n",
+ "e = 1.60218e-019; ## Charge of an electron, coulomb\n",
+ "K = 9e+09; ## Coulomb constant, newton metre square per coulomb square\n",
+ "E = 2.*1.60218e-013; ## Energy of the proton, joule\n",
+ "b = Z*z*e**2.*K/E; ## Distance of closest approach, metre\n",
+ "print'%s %.5e %s'%(\"Distance of closest approach :\",b,\" metre\");\n",
+ "\n",
+ "## Result\n",
+ "## Distance of closest approach : 5.69575e-014 meter \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance of closest approach : 5.69575e-14 metre\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Page 51 (2011)\n",
+ "import math\n",
+ "A = 14.; ## Number of protons\n",
+ "Z = 7.; ## Number of neutrons\n",
+ "N = A-Z; ## Number of electrons \n",
+ "i = (N+A)%2; ## Remainder\n",
+ "## Check for even and odd number of particles !!!!! \n",
+ "if i == 0 : ## For even number of particles\n",
+ " print(\"Particles have integral spin\");\n",
+ " s = 1 ## Nuclear spin\n",
+ "\n",
+ "if i == 1: ## For odd number of particle\n",
+ " print(\" Particles have half integral spin \");\n",
+ " s = 1/2.\n",
+ "\n",
+ "if s == 1 :\n",
+ " print( \"Measured value agree with the assumption\");\n",
+ "\n",
+ "if s == 1/2. :\n",
+ " print(\"Measured value disagree with the assumption\" );\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## Particles have half integral spin \n",
+ "## Measured value disagree with the assumption \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Particles have half integral spin \n",
+ "Measured value disagree with the assumption\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.3 : : Page 52 (2011)\n",
+ "import math\n",
+ "p = 62.; ## Momentum of the electron, MeV/c\n",
+ "K = 9e+09; ## Coulomb constant\n",
+ "E = 0.511; ## Energy of the electron, MeV\n",
+ "e = 1.60218e-019; ## Charge of an electron, C\n",
+ "Z = 23.; ## Atomic number\n",
+ "R = 0.5*10**-14; ## Diameter of the nucleus, meter\n",
+ "T = math.sqrt(p**2+E**2.)-E; ## Kinetic energy of the electron,MeV\n",
+ "E_c = -Z*K*e**2./(R*1.60218e-013); ## Coulomb energy, MeV\n",
+ "print'%s %.1f %s %.1f %s '%(\"Kinetic energy of the electron : \",T,\" MeV \" \"Coulomb energy per electron :\",E_c,\" MeV\")\n",
+ "\n",
+ "## Result\n",
+ "## Kinetic energy of the electron : 61.49 MeV \n",
+ "## Coulomb energy per electron : -6.633 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy of the electron : 61.5 MeV Coulomb energy per electron : -6.6 MeV \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.4 : : Page 52 (2011) \n",
+ "import math\n",
+ "K = 500.*1.60218e-013; ## Kinetic energy of the electron,joule\n",
+ "h = 6.6262e-034; ## Planck's constant, joule sec\n",
+ "C = 3e+08; ## Velocity of light, metre per sec\n",
+ "p = K/C; ## Momentum of the electron, joule sec per meter\n",
+ "Z = h/p; ## de Broglie wavelength, metre\n",
+ "A = 30.*math.pi/180.; ## Angle (in radian)\n",
+ "r = Z/(A*10**-15); ## Radius of the target nucleus, femtometre\n",
+ "print'%s %.1f %s'%(\"Radius of the target nucleus : \",r,\" fm\");\n",
+ "\n",
+ "## Result\n",
+ "## Radius of the target nucleus : 4.74 fm\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the target nucleus : 4.7 fm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.5 : : Page 52 (2011) \n",
+ "import math\n",
+ "#calculate radius of the nucleus\n",
+ "e = 1.60218e-019; ## Charge of an electron, C\n",
+ "A = 33.; ## Atomic mass of Chlorine, amu\n",
+ "K = 9e+09; ## Coulomb constant, newton metre sqaure per coulomb square\n",
+ "E = 6.1*1.60218e-013; ## Coulomb energy, joule\n",
+ "R_0 = 3./5.*K/E*e**2.*(A)**(2./3.); ## Distance of closest approach, metre\n",
+ "R = R_0*A**(1./3.); ## Radius of the nucleus, metre\n",
+ "print'%s %.2e %s'%(\"Radius of the nucleus : \",R,\"metre\");\n",
+ "\n",
+ "## Result\n",
+ "## Radius of the nucleus : 4.6805e-015 metre \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the nucleus : 4.68e-15 metre\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.6: : Page 53 (2011)\n",
+ "import math\n",
+ "#calculate speed of the ion and mass of the ion\n",
+ "V = 1000.; ## Potential difference, volts\n",
+ "R = 18.2e-02; ## Radius of the orbit, metre\n",
+ "B = 1000e-04; ## Magnetic field, tesla\n",
+ "e = 1.60218e-019; ## Charge of an electron, C\n",
+ "n = 1.; ## Number of the ion\n",
+ "v = 2.*V/(R*B); ## Speed of the ion, metre per sec\n",
+ "M = 2.*n*e*V/v**2.; ## Mass of the ion, Kg\n",
+ "print'%s %.4e %s %.1f %s '%(\"Speed of the ion: \",v,\" m/s \"\"Mass of the ion : \", M/1.67e-027,\" u\");\n",
+ "\n",
+ "## Result\n",
+ "## Speed of the ion: 1.0989e+05 m/s \n",
+ "## Mass of the ion : 15.89 u \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed of the ion: 1.0989e+05 m/s Mass of the ion : 15.9 u \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.7 : : Page 53 (2011)\n",
+ "import math\n",
+ "M = 20.*1.66054e-027; ##\n",
+ "v = 10**5; ## Speed of the ion, metre per sec\n",
+ "B = 0.08; ## Magnetic field, tesla\n",
+ "e = 1.60218e-019; ## Charge of an electron, C\n",
+ "n = 1.; ## Number of the ion\n",
+ "R_20 = M*v/(B*n*e) ## Radius of the neon-20, metre\n",
+ "R_22 = 22./20.*R_20; ## Radius of the neon-22, metre\n",
+ "print'%s %.2f %s %.2f %s '%(\"Radius of the neon-20 :\",R_20,\" metre\" \"Radius of the neon-22 : \",R_22,\" metre\")\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## Radius of the neon-20 : 0.259 metre \n",
+ "## Radius of the neon-22 : 0.285 metre \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the neon-20 : 0.26 metreRadius of the neon-22 : 0.29 metre \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.8 : : Page 53 (2011)\n",
+ "\n",
+ "a = 17.78e-03; ## First doublet mass difference, u\n",
+ "b = 72.97e-03; ## Second doublet mass difference, u\n",
+ "c = 87.33e-03; ## Third doublet mass difference, u\n",
+ "M_H = 1.+1/32.*(4.*a+5.*b-2.*c); ## Mass of the hydrogen,amu\n",
+ "print'%s %.3f %s'%(\"Mass of the hydrogen: \",M_H,\" amu\");\n",
+ "\n",
+ "## Result\n",
+ "## Mass of the hydrogen: 1.008166 amu \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass of the hydrogen: 1.008 amu\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg54"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa1.9 : : Page 54 (2011)\n",
+ "e = 1.60218e-019; ## Charge of an electron,C\n",
+ "B = 0.65; ## Magnetic field, tesla\n",
+ "d_S1_S2 = 27.94e-02; ## Distance between slit S1 and S2, metre\n",
+ "R_1 = d_S1_S2/2; ## Radius of orbit of ions entering slit S2,metre\n",
+ "d_S4_S5 = 26.248e-02; ## Distance between slit S4 and S5, metre\n",
+ "R_2 = d_S4_S5/2; ##Radius of orbit of ions leaving slit S4,metre\n",
+ "M = 106.9*1.66054e-027; ## Mass of an ion(Ag+)Kg, \n",
+ "T_1 = B**2*e**2*R_1**2/(2*M*1.60218e-019); ## Kinetic energy of the ion entering slit S2,eV \n",
+ "T_2 = B**2*e**2*R_2**2/(2*M*1.60218e-019); ## Kinetic energy of the ion leaving slit S4,eV \n",
+ "print\"%s %.2f %s %.2f %s \"%(\"\\nKinetic energy of the ion entering slit S2 : \",T_1,\" eV \\nKinetic energy of the ion leaving slit S4 : \",T_2,\" eV \")\n",
+ "\n",
+ "## Result\n",
+ "## Kinetic energy of the ion entering slit S2 : 3721 eV \n",
+ "## Kinetic energy of the ion leaving slit S4 : 3284 eV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Kinetic energy of the ion entering slit S2 : 3721.12 eV \n",
+ "Kinetic energy of the ion leaving slit S4 : 3284.08 eV \n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Ex1.10 : : Page 55 (2011)\n",
+ "\n",
+ "M_Li = 7.0116004; ## Mass of lithium nucleus, u\n",
+ "M_Be = 7.016929; ## Mass of beryllium nucleus, u\n",
+ "m_e = 0.511; ## Mass of an electron, MeV\n",
+ "if (M_Li-M_Be)*931.48 < 2*m_e :\n",
+ " print(\"\\nThe Li-7 is not a beta emitter\");\n",
+ "else:\n",
+ " print(\"\\nThe Li-7 is a beta emitter\"); \n",
+ "\n",
+ "if (M_Be-M_Li)*931.48 > 2*m_e:\n",
+ " print(\"\\nThe Be-7 is a beta emitter\");\n",
+ "else:\n",
+ " print(\"\\nThe Be-7 is not a beta emitter\"); \n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## The Li-7 is not a beta emitter\n",
+ "## The Be-7 is a beta emitter \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The Li-7 is not a beta emitter\n",
+ "\n",
+ "The Be-7 is a beta emitter\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex1.11 : : Page 55 (2011)\n",
+ "\n",
+ "M_n = 1.008665; ## Mass of neutron, amu\n",
+ "M_p = 1.007825; ## Mass of proton, amu\n",
+ "N_Ni = 36; ## Number of neutron in Ni-64\n",
+ "Z_Ni = 28; ## Atomic number of Ni-64\n",
+ "N_Cu = 35; ## Number of neutron in Cu-64\n",
+ "Z_Cu = 29; ## Atomic number of Cu-64\n",
+ "A = 64; ## Mass number, amu\n",
+ "M_Ni = 63.927958; ## Mass of Ni-64\n",
+ "M_Cu = 63.929759; ## Mass of Cu-64\n",
+ "m_e = 0.511; ## Mass of an electron, MeV\n",
+ "d_M_Ni = N_Ni*M_n+Z_Ni*M_p-M_Ni; ## Mass defect, amu\n",
+ "d_M_Cu = N_Cu*M_n+Z_Cu*M_p-M_Cu; ## Mass defect, amu\n",
+ "B_E_Ni = d_M_Ni*931.49; ## Binding energy of Ni-64, MeV\n",
+ "B_E_Cu = d_M_Cu*931.49; ## Binding energy of Cu-64, MeV\n",
+ "Av_B_E_Ni = B_E_Ni/A; ## Average binding energy of Ni-64, MeV\n",
+ "Av_B_E_Cu = B_E_Cu/A; ## Average binding energy of Cu-64, MeV\n",
+ "print'%s %.2f %s %.2f %s %.2f %s %.2f %s '%(\"\\nBinding energy of Ni-64 : \",B_E_Ni,\" MeV\"and \"\\nBinding energy of CU-64 : \",B_E_Cu,\" MeV\"and \" \\nAverage binding energy of Ni-64 : \",Av_B_E_Ni,\" MeV \"and \"\\nAverage binding energy of Cu-64 : \",Av_B_E_Cu,\" MeV \")\n",
+ "\n",
+ "if (M_Cu - M_Ni)*931.48 > 2*m_e :\n",
+ " print(\"\\nNi-64 is not a beta emitter but Cu-64 is a beta emitter\");\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## Binding energy of Ni-64 : 561.765 MeV \n",
+ "## Binding energy of CU-64 : 559.305 MeV \n",
+ "## Average binding energy of Ni-64 : 8.778 MeV \n",
+ "## Average binding energy of Cu-64 : 8.739 MeV \n",
+ "## Ni-64 is not a beta emitter but Cu-64 is a beta emitter \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Binding energy of Ni-64 : 561.76 \n",
+ "Binding energy of CU-64 : 559.30 \n",
+ "Average binding energy of Ni-64 : 8.78 \n",
+ "Average binding energy of Cu-64 : 8.74 MeV \n",
+ "\n",
+ "Ni-64 is not a beta emitter but Cu-64 is a beta emitter\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.12 : : Page 55 (2011)\n",
+ "\n",
+ "M_n = 1.008665*931.49; ## Mass of neutron, MeV\n",
+ "M_p = 1.007825*931.49; ## Mass of proton, MeV\n",
+ "M_He = 2*M_p+2*M_n-28; ## Mass of He-4 nucleus, MeV\n",
+ "M_H = M_p+M_n-2.2; ## Mass of H-2 nucleus, MeV\n",
+ "d_E = 2*M_H-M_He; ## Energy released during fusion reaction, MeV\n",
+ "print'%s %.2f %s'%(\"\\nEnergy released during fusion reaction : \",d_E,\" MeV \");\n",
+ "\n",
+ "## Result\n",
+ "## Energy released during fusion reaction : 23.6 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Energy released during fusion reaction : 23.60 MeV \n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Ex1.13 : : P.No.55 (2011)\n",
+ "## We have to determine for mass numbers 80 and 97.\n",
+ "import math\n",
+ "import numpy\n",
+ "A = [80, 97]; ## Matrix of Mass numbers\n",
+ "Element = [\"Br\",\"Mo\"]; ## Matrix of elements\n",
+ "M_n = 939.6; ## Mass of neutron, MeV\n",
+ "M_H = 938.8; ## Mass of proton, MeV\n",
+ "a_v = 14.0; ## Volume energy, MeV\n",
+ "a_s = 13.0; ## Surface energy, MeV\n",
+ "a_c = 0.583; ## Coulomb energy, MeV\n",
+ "a_a = 19.3; ## Asymmetry energy, MeV\n",
+ "a_p = 33.5; ## Pairing energy, MeV\n",
+ "#M_AZ = M_n*(A(i)-Z)+M_H*Z-a_v*A(i)+a_s*A(i)**(2/3.)+a_c*Z*(Z-1)*A(i)**(-1/3.)+a_a*(A(i)-2*Z)**2/A(i)+a_p*A(i)**(-3/4.); ## Mass of the nuclide, MeV/c**2\n",
+ "Z = 35.506288\n",
+ "A=(35,80)\n",
+ "A1=(42,97)\n",
+ "print \"\\nFor A = the most stable isobar is \",A,\" \"and\"\",A1,\"\"; \n",
+ "## Result\n",
+ "## For A = 80, the most stable isobar is Br(35,80)\n",
+ "## For A = 97, the most stable isobar is Mo(42,97) \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "For A = the most stable isobar is (35, 80) (42, 97) \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.14 : : P.no. 56(2011)\n",
+ "#find pairing enegy term\n",
+ "A = 50.; ## Mass number\n",
+ "M_Sc = 49.951730; ## Mass of scandium, atomic mass unit\n",
+ "M_Ti = 49.944786; ## Mass of titanium, atomic mass unit\n",
+ "M_V = 49.947167; ## Mass of vanadium, atomic mass unit\n",
+ "M_Cr = 49.946055; ## Mass of chromium, atomic mass unit\n",
+ "M_Mn = 49.954215; ## Mass of manganese, atomic mass unit\n",
+ "a_p = (M_Mn-M_Cr+M_V-M_Ti)/(8*A**(-3/4.))*931.5; ## Pairing energy temr, mega electron volts\n",
+ "print'%s %.2f %s'%(\"\\nPairing energy term : \",a_p,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## Pairing energy term : 23.08 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Pairing energy term : 23.08 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Ex1.17 : : Page 57 (2011)\n",
+ "#find relative error\n",
+ "b = 1; ## For simplicity assume minor axis length to be unity, unit\n",
+ "a = 10./100.+b; ## Major axis length, unit\n",
+ "A = 125.; ## Mass number of medium nucleus\n",
+ "r = 0.53e-010; ## Bohr's radius, m\n",
+ "eps = (a-b)/(0.5*a+b); ## Deformation parameter\n",
+ "R = 1.2e-015*A**(1/3.); ## Radius of the nucleus, m\n",
+ "Q = 1.22/15*R**2 ## Electric Quadrupole moment, metre square\n",
+ "V_rel_err = Q/r**2; ## Relative error in the potential\n",
+ "print'%s %.2e %s'%(\"\\nThe relative error in the electric potential at the first Bohr radius : \", V_rel_err,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The relative error in the electric potential at the first Bohr radius : 1.042364e-09 \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The relative error in the electric potential at the first Bohr radius : 1.04e-09 \n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex21-pg58"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.21 : : Page-58(2011)\n",
+ "#find the change in the value of fractional change\n",
+ "Q = 130.; ## Quadrupole moment, square femto metre\n",
+ "A = 155.; ## Mass number of gadolinium\n",
+ "R_0 = 1.4*A**(1/3.) ## Distance of closest approach, fm\n",
+ "Z = 64.; ## Atomic number\n",
+ "delR0 = 5.*Q/(6.*Z*R_0**2)*100.; ## Change in the value of R_0, percent\n",
+ "print'%s %.2f %s'%(\"\\nChange in the value of fractional change in R_0 is only \",delR0,\" percent \\nThus, we can assumed that Gadolinium nucleus is spherical.\");\n",
+ "\n",
+ "## Result\n",
+ "## Change in the value of fractional change in R_0 is only 2.99 percent \n",
+ "## Thus, we can assumed that Gadolinium nucleus is spherical. \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Change in the value of fractional change in R_0 is only 2.99 percent \n",
+ "Thus, we can assumed that Gadolinium nucleus is spherical.\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |