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| author | hardythe1 | 2015-06-03 15:27:17 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-06-03 15:27:17 +0530 |
| commit | df60071cf1d1c18822d34f943ab8f412a8946b69 (patch) | |
| tree | ab059cf19bad4a1233a464ccf5d72cf8b3fb323c /Nuclear_Physics/Chapter3.ipynb | |
| parent | fba055ce5aa0955e22bac2413c33493b10ae6532 (diff) | |
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diff --git a/Nuclear_Physics/Chapter3.ipynb b/Nuclear_Physics/Chapter3.ipynb new file mode 100755 index 00000000..3abe9ff1 --- /dev/null +++ b/Nuclear_Physics/Chapter3.ipynb @@ -0,0 +1,342 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0f490ccea3b13c889d4ee8d26efe8421ceedda6e18896bd5d0ed1b94aaa5c5de"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter3-Interaction of Nuclear radiations with matter"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg123"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.1 : : Page-123 (2011)\n",
+ "#find range of an alpha particle and metre the thickness\n",
+ "E = 9.; ## Energy of the alpha particle, MeV\n",
+ "S = 1700.; ## Stopping power of Al\n",
+ "D = 2700.; ## Density of Al, Kg per cubic metre\n",
+ "R_air = 0.00318*E**(3/2.); ## Range of an alpha particle in air,metre\n",
+ "R_Al = R_air/S; ## Range of an alpha particle in Al, metre\n",
+ "T = D*1./S; ## Thickness in Al of 1m air, Kg per square metre\n",
+ "print\"%s %.2e %s %.2f %s \"%(\"The range of an alpha particle = \",R_Al,\" metre The thickness in Al of 1 m air =\",T,\" Kg per square metre\");\n",
+ "\n",
+ "## Result\n",
+ "## The range of an alpha particle = 5.05e-05 metre \n",
+ "## The thickness in Al of 1 m air = 1.59 Kg per square metre \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The range of an alpha particle = 5.05e-05 metre The thickness in Al of 1 m air = 1.59 Kg per square metre \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.4: : Page-124 (2011)\n",
+ "#find the haif value thickness for beta absorptions\n",
+ "import math\n",
+ "E_max = 1.17; ## Maximum energy of the beta particle, mega electron volts\n",
+ "D = 2.7; ## Density of Al,gram per cubic metre\n",
+ "u_m = 22./E_max; ## Mass absorption coefficient,centimetre square per gram\n",
+ "x_h = math.log(2.)/(u_m*D); ## Half value thickness for beta absorption, cm\n",
+ "print'%s %.2f %s'%(\"The Half value thickness for beta absorption = \",x_h,\" cm\"); \n",
+ "\n",
+ "## Result \n",
+ "## The Half value thickness for beta absorption = 0.014 cm \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Half value thickness for beta absorption = 0.01 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.7: : Page 125(2011)\n",
+ "#calculate ratio of raditon loss to ionisation\n",
+ "Z = 82.; ## Atomic number\n",
+ "E = 1.; ## Energy of the beta paricle, MeV\n",
+ "I_l = 800.; ## Ionisation loss, MeV\n",
+ "R = Z*E/I_l; ## Ratio of radiation loss to ionisation loss\n",
+ "E_1 = I_l/Z; ## Energy of the beta particle when radiation radiation loss is equal to ionisation loss, MeV\n",
+ "\n",
+ "print'%s %.2e %s %.2f %s '%(\"The ratio of radiation loss to ionisation loss =\",R,\" The energy of the beta particle = \",E_1,\" MeV \");\n",
+ "\n",
+ "## Result\n",
+ "## The ratio of radiation loss to ionisation loss = 1.025e-01 \n",
+ "## The energy of the beta particle = 9.76 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of radiation loss to ionisation loss = 1.02e-01 The energy of the beta particle = 9.76 MeV \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.8 : : Page 125(2011)\n",
+ "#find The half value thickness of Al and The mass absorption coefficient\n",
+ "import math\n",
+ "x = 0.25; ## Thickness of Al, metre\n",
+ "U_l = 1./x*math.log(50.); ## Linear absorption coefficient\n",
+ "d = 2700.; ## density of the Al, Kg per cubic centimetre \n",
+ "x_h = math.log(2.)/U_l; ## Half value thickness of Al, metre\n",
+ "U_m = U_l/d; ## Mass absorption coefficient, square metre per Kg\n",
+ "print'%s %.2f %s %.3f %s'%(\"The half value thickness of Al = \",x_h,\" Kg per cubic metre The mass absorption coefficient = \",U_m,\" square metre per Kg \");\n",
+ "\n",
+ "## Result\n",
+ "## The half value thickness of Al = 0.0443 Kg per cubic metre \n",
+ "## The mass absorption coefficient = 0.00580 square metre per Kg \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The half value thickness of Al = 0.04 Kg per cubic metre The mass absorption coefficient = 0.006 square metre per Kg \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.9 : : Page-125(2011)\n",
+ "#find The energy of the compton recoil electrons\n",
+ "import math\n",
+ "E_g = 2.19*1.6e-013; ## Energy of the gamma rays, joule\n",
+ "m_e = 9.10939e-031; ## Mass of the electron, Kg\n",
+ "C = 3e+08; ## Velocity of light, m/s\n",
+ "E_max = (E_g/(1.+(m_e*C**2)/(2.*E_g)))/(1.6e-013); ## Energy of the compton recoil electron, MeV\n",
+ "print\"%s %.2f %s\"%(\"The energy of the compton recoil electrons = \",E_max,\" MeV\"); \n",
+ "\n",
+ "## Result \n",
+ "## The energy of the compton recoil electrons = 1.961 MeV\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy of the compton recoil electrons = 1.96 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.10 : : Page-125(2011)\n",
+ "#find The average energy of the positron \n",
+ "m_e = 9.1e-31; ## Mass of the positron, Kg\n",
+ "e = 1.6e-19; ## Charge of the positron, coulomb\n",
+ "c = 3e+08; ## Velocity of the light, metre per sec\n",
+ "eps = 8.85e-12; ## Absolute permittivity of free space, per N per metre-square per coulomb square\n",
+ "h = 6.6e-34; ## Planck's constant, joule sec\n",
+ "E = e**2.*m_e*c/(eps*h*1.6e-13); ## Average energy of the positron, mega electron volts\n",
+ "print'%s %.2f %s'%(\"The average energy of the positron = \",E,\"Z MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The average energy of the positron = 0.0075Z MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The average energy of the positron = 0.01 Z MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.11 : : Page-125(2011)\n",
+ "#find The refractive index of the gas and The angle at which Cerenkov radiation is emitted \n",
+ "import math\n",
+ "P = 1.; ## Momentum of the proton, GeV/c\n",
+ "M_0 = 0.94; ## Rest mass of the proton, GeV/c-square\n",
+ "G = math.sqrt((P/M_0)**2.+1.) ## Lorentz factor\n",
+ "V = math.sqrt(1.-1./G**2.); ## Minimum velocity of the electron, m/s\n",
+ "u = 1/V; ## Refractive index of the gas\n",
+ "print'%s %.2f %s'%(\"The refractive index of the gas =\", u,\"\"); \n",
+ "u = 1.6; ## Refractive index\n",
+ "theta = round (math.acos(1/(u*V))*180/3.14); ## Angle at which cerenkov radiatin is emitted,degree\n",
+ "print'%s %.2f %s'%(\"The angle at which Cerenkov radiation is emitted =\",theta,\" degree\") \n",
+ "\n",
+ "## Result \n",
+ "## The refractive index of the gas = 1.37\n",
+ "## The angle at which Cerenkov radiation is emitted = 31 degree \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The refractive index of the gas = 1.37 \n",
+ "The angle at which Cerenkov radiation is emitted = 31.00 degree"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.12 : : Page-126(2011)\n",
+ "#find The minimum kinetic energy required to electron to emit cerenkov radiation\n",
+ "import math\n",
+ "n = 1+1.35e-04; ## Refractive index of the medium\n",
+ "V_min = 1./n; ## Minimum velocity of the electron, m/s\n",
+ "p = (1.+V_min)*(1.-V_min); ## It is nothing but just to take the product \n",
+ "G_min = 1./math.sqrt(p); ## Lorentz factor\n",
+ "m_e = 9.10939e-031; ## Mass of the electron, Kg\n",
+ "C = 3e+08; ## Velocity of light, metre per sec\n",
+ "T_min = ((G_min-1.)*m_e*C**2.)/(1.602e-013); ## Minimum kinetic energy required by an electro to emit cerenkov radiation, mega electron volts\n",
+ "print'%s %.2f %s'%(\"The minimum kinetic energy required to electron to emit cerenkov radiation = \",T_min,\" MeV\");\n",
+ " \n",
+ "## Result \n",
+ "## The minimum kinetic energy required to electron to emit cerenkov radiation = 30.64 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum kinetic energy required to electron to emit cerenkov radiation = 30.64 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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