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Diffstat (limited to 'Nuclear_Physics')
| -rwxr-xr-x | Nuclear_Physics/Chapter12.ipynb | 1018 | ||||
| -rwxr-xr-x | Nuclear_Physics/README.txt | 10 |
2 files changed, 519 insertions, 509 deletions
diff --git a/Nuclear_Physics/Chapter12.ipynb b/Nuclear_Physics/Chapter12.ipynb index 7fbad62a..68299296 100755 --- a/Nuclear_Physics/Chapter12.ipynb +++ b/Nuclear_Physics/Chapter12.ipynb @@ -1,510 +1,510 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:4e35225448e1e6a4105db4a5e756370df804ed5cec1af259aab244cf2b566068"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Chapter12-Neutrons"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Exx1-pg573"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "## Exa12.1 : : Page-573 (2011)\n",
- "#calculate activity for Cu-63 and disintegrations\n",
- "import math \n",
- "N_0 = 6.23e+23; ## Avogadro's number, per mole\n",
- "m = 0.1; ## Mass of copper foil, Kg\n",
- "phi = 10**12; ## Neutron flux density, per square centimetre sec\n",
- "a_63 = 0.691; ## Abundance of Cu-63\n",
- "a_65 = 0.309; ## Abundance of Cu-65\n",
- "W_m = 63.57; ## Molecular weight, gram\n",
- "sigma_63 = 4.5e-24; ## Activation cross section for Cu-63, square centi metre\n",
- "sigma_65 = 2.3e-24; ## Activation cross section for Cu-65, square centi metre\n",
- "A_63 = phi*sigma_63*m*a_63/W_m*N_0; ## Activity for Cu-63, disintegrations per sec\n",
- "A_65 = phi*sigma_65*m*a_65/W_m*N_0; ## Activity for Cu-65, disintegrations per sec\n",
- "print'%s %.2e %s %.2e %s'%(\"\\nThe activity for Cu-63 is = \",A_63,\" disintegrations per sec\" and \"\\nThe activity for Cu-65 is = \",A_65,\" disintegrations per sec\");\n",
- "\n",
- "## Result\n",
- "## The activity for Cu-63 is = 3.047e+009 disintegrations per sec \n",
- "## The activity for Cu-65 is = 6.97e+008 disintegrations per sec "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "The activity for Cu-63 is = 3.05e+09 \n",
- "The activity for Cu-65 is = 6.97e+08 disintegrations per sec\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2-pg573"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "## Exa12.2 : : Page-573 (2011)\n",
- "import math \n",
- "#calculate enegy loss \n",
- "A_Be = 9.; ## Mass number of beryllium\n",
- "A_U = 238.; ## Mass number of uranium\n",
- "E_los_Be = (1-((A_Be-1)**2/(A_Be+1)**2))*100.; ## Energy loss for beryllium\n",
- "E_los_U = round((1-((A_U-1)**2/(A_U+1)**2))*100.); ## Energy loss for uranium\n",
- "print'%s %.2f %s %.2f %s '%(\"\\nThe energy loss for beryllium is = \",E_los_Be,\" percent\"and \" \\nThe energy loss for uranium is = \",E_los_U,\" percent\");\n",
- "\n",
- "## Check for greater energy loss !!!!\n",
- "if E_los_Be >= E_los_U :\n",
- " print(\"\\nThe energy loss is greater for beryllium\");\n",
- "else:\n",
- " print(\"\\nThe energy loss is greater for uranium\");\n",
- "\n",
- "\n",
- "## Result\n",
- "## The energy loss for beryllium is = 36 percent \n",
- "## The energy loss for uranium is = 2 percent\n",
- "## The energy loss is greater for beryllium \n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "The energy loss for beryllium is = 36.00 \n",
- "The energy loss for uranium is = 2.00 percent \n",
- "\n",
- "The energy loss is greater for beryllium\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3-pg574"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "## Exa12.3 : : Page-574 (2011)\n",
- "#calculate energy loss of neutron\n",
- "import math \n",
- "A = 12.; ## Mass number of Carbon\n",
- "alpha = (A-1)**2/(A+1)**2; ## Scattering coefficient\n",
- "E_loss = 1/2.*(1-alpha)*100.; ## Energy loss of neutron\n",
- "print'%s %.2f %s'%(\"\\nThe energy loss of neutron = \",E_loss,\" percent\")\n",
- "\n",
- "## Result\n",
- "## The energy loss of neutron = 14.201 percent \n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "The energy loss of neutron = 14.20 percent\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4-pg574"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "## Exa12.4 : : Page-574 (2011)\n",
- "#calculate number of collisions of neutrons \n",
- "import math \n",
- "zeta = 0.209; ## Moderated assembly\n",
- "E_change = 100./1.; ## Change in energy of the neutron\n",
- "E_thermal = 0.025; ## Thermal energy of the neutron, electron volts\n",
- "E_n = 2*10**6; ## Energy of the neutron, electron volts\n",
- "n = 1/zeta*math.log(E_change); ## Number of collisions of neutrons to loss 99 percent of their energies \n",
- "n_thermal = 1/zeta*math.log(E_n/E_thermal); ## Number of collisions of neutrons to reach thermal energies\n",
- "print'%s %.2f %s %.2f %s'%(\"\\nThe number of collisions of neutrons to loss 99 percent of their energies = \",n,\" \\nThe number of collisions of neutrons to reach thermal energies = \",n_thermal,\"\")\n",
- "\n",
- "## Result\n",
- "## The number of collisions of neutrons to loss 99 percent of their energies = 22 \n",
- "## The number of collisions of neutrons to reach thermal energies = 87 \n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "The number of collisions of neutrons to loss 99 percent of their energies = 22.03 \n",
- "The number of collisions of neutrons to reach thermal energies = 87.07 \n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5-pg574"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "## Exa12.5 : : Page-574 (2011)\n",
- "#calculate average distance travelled by the neutron\n",
- "import math\n",
- "import scipy\n",
- "from scipy import integrate\n",
- "L = 1.; ## For simplicity assume thermal diffusion length to be unity, unit\n",
- "def fun(x):\n",
- " y=x*math.exp(-x/L)\n",
- " return y\n",
- "x_b = scipy.integrate.quad(fun, 0, 100); ## Average distance travelled by the neutron, unit\n",
- "x_b1=x_b[0]\n",
- "def fun2(x): \n",
- " y1=x**2*math.exp(-x/L)\n",
- " return y1\n",
- "X=scipy.integrate.quad(fun2, 0, 100)\n",
- "x_rms = math.sqrt(X[0]); ## Root mean square of the distance trvelled by the neutron, unit\n",
- "print'%s %.2f %s'%(\"\\nThe average distance travelled by the neutron = \", x_b1,\"*L\");\n",
- "print'%s %.2f %s %.2f %s '%(\"\\nThe root mean square distance travelled by the neutron = \",x_rms,\"\"and \"\",x_rms,\"x_bar\")\n",
- "\n",
- "## Result\n",
- "## The average distance travelled by the neutron = 1*L\n",
- "## The root mean square distance travelled by the neutron = 1.414L = 1.414x_bar \n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "The average distance travelled by the neutron = 1.00 *L\n",
- "\n",
- "The root mean square distance travelled by the neutron = 1.41 1.41 x_bar \n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex6-pg574"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "## Exa12.6 : : Page-574 (2011)\n",
- "#calculate neutron flux through water \n",
- "import math\n",
- "Q = 5e+08; ## Rate at which neutrons produce, neutrons per sec\n",
- "r = 20.; ## Distance from the source, centi metre\n",
- "## For water\n",
- "lambda_wtr = 0.45; ## Transport mean free path, centi metre\n",
- "L_wtr = 2.73; ## Thermal diffusion length, centi metre\n",
- "phi_wtr = 3*Q/(4.*math.pi*lambda_wtr*r)*math.exp(-r/L_wtr); ## Neutron flux for water, neutrons per square centimetre per sec\n",
- "## For heavy water\n",
- "lambda_h_wtr = 2.40; ## Transport mean free path, centi metre\n",
- "L_h_wtr = 171.; ## Thermal diffusion length, centi metre\n",
- "phi_h_wtr = 3*Q/(4.*math.pi*lambda_h_wtr*r)*math.exp(-r/L_h_wtr); ## Neutron flux for heavy water, neutrons per square centimetre per sec\n",
- "print'%s %.2e %s %.2e %s '%(\"\\nThe neutron flux through water = \",phi_wtr,\" neutrons per square cm per sec\"and \"\\nThe neutron flux through heavy water = \",phi_h_wtr,\" neutrons per square cm per sec\")\n",
- "\n",
- "## Result\n",
- "## The neutron flux through water = 8.730e+003 neutrons per square cm per sec \n",
- "## The neutron flux through heavy water = 2.212e+006 neutrons per square cm per sec \n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "The neutron flux through water = 8.73e+03 \n",
- "The neutron flux through heavy water = 2.21e+06 neutrons per square cm per sec \n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7-pg575"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "## Exa12.7 : : Page-575 (2011)\n",
- "#calculate neutron flux and diffusion length\n",
- "import math\n",
- "k = 1.38e-23; ## Boltzmann constant, joules per kelvin\n",
- "T = 323.; ## Temperature, kelvin\n",
- "E = (k*T)/1.6e-19; ## Thermal energy, joules\n",
- "sigma_0 = 13.2e-28; ## Cross section, square metre\n",
- "E_0 = 0.025; ## Energy of the neutron, electron volts\n",
- "sigma_a = sigma_0*math.sqrt(E_0/E); ## Absorption cross section, square metre\n",
- "t_half = 2.25; ## Half life, hours\n",
- "D= 0.69/t_half; ## Decay constant, per hour\n",
- "N_0 = 6.023e+026; ## Avogadro's number, per \n",
- "m_Mn = 55.; ## Mass number of mangnese\n",
- "w = 0.1e-03; ## Weight of mangnese foil, Kg\n",
- "A = 200.; ## Activity, disintegrations per sec\n",
- "N = N_0*w/m_Mn; ## Number of mangnese nuclei in the foil\n",
- "x1 = 1.5; ## Base, metre\n",
- "x2 = 2.0; ## Height, metre\n",
- "phi = A/(N*sigma_a*0.416); ## Neutron flux, neutrons per square metre per sec\n",
- "phi1 = 1.; ## For simplicity assume initial neutron flux to be unity, neutrons/Sq.m-sec\n",
- "phi2 = 1/2.*phi1; ## Given neutron flux, neutrons/Sq.m-sec\n",
- "L1 = 1/math.log(phi1/phi2)/(x2-x1); ## Thermal diffusion length for given neutron flux, m\n",
- "L = math.sqrt(1./((1./L1)**2+(math.pi/x1)**2+(math.pi/x2)**2)); ## Diffusion length, metre\n",
- "print'%s %.2e %s %.2f %s '%(\"\\nThe neutron flux = \",phi,\" neutrons per square metre per sec\"and \" \\nThe diffusion length = \",L,\" metre\");\n",
- "\n",
- "## Result\n",
- "## The neutron flux = 3.51e+008 neutrons per square metre per sec \n",
- "## The diffusion length = 0.38 metre\n",
- "## Note: the difussion length is solved wrongly in the testbook\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "The neutron flux = 3.51e+08 \n",
- "The diffusion length = 0.38 metre \n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8-pg575"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "## Exa12.8 : : Page-575(2011)\n",
- "#find diffusion length for thermal neutron\n",
- "import math\n",
- "N_0 = 6.023e+026; ## Avogadro's number, per mole\n",
- "rho = 1.62e+03; ## Density, kg per cubic metre\n",
- "sigma_a = 3.2e-31; ## Absorption cross section, square metre\n",
- "sigma_s = 4.8e-28; ## Scattered cross section, square metre\n",
- "A = 12.; ## Mass number\n",
- "lambda_a = A/(N_0*rho*sigma_a); ## Absorption mean free path, metre\n",
- "lambda_tr = A/(N_0*rho*sigma_s*(1.-2./(3.*A))); ## Transport mean free path, metre\n",
- "L = math.sqrt(lambda_a*lambda_tr/3.); ## Diffusion length for thermal neutron\n",
- "print'%s %.2f %s'%(\"\\nThe diffusion length for thermal neutron = \",L,\" metre \")\n",
- "\n",
- "## Result\n",
- "## The diffusion length for thermal neutron = 0.590 metre \n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "The diffusion length for thermal neutron = 0.59 metre \n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex9-pg575"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "## Exa12.9 : : Page-575 (2011)\n",
- "#calculate graphite and neutron age and slowing down length and same as berylliums\n",
- "import math\n",
- "E_0 = 2e+06; ## Average energy of the neutron, electron volts\n",
- "E = 0.025; ## Thermal energy of the neutron, electron volts\n",
- "## For graphite\n",
- "A = 12. ## Mass number\n",
- "sigma_g = 33.5; ## The value of sigma for graphite\n",
- "tau_0 = 1./(6.*sigma_g**2)*(A+2./3.)/(1.-2./(3.*A))*math.log(E_0/E); ## Age of neutron for graphite, Sq.m\n",
- "L_f = math.sqrt(tau_0); ## Slowing down length of neutron through graphite, m\n",
- "print'%s %.2f %s'%(\"\\nFor Graphite, A = \", A,\"\");\n",
- "print'%s %.2f %s'%(\"\\nNeutron age = \",tau_0*1e+004,\" Sq.cm\");\n",
- "print'%s %.2f %s'%(\"\\nSlowing down length =\",L_f,\" m\");\n",
- "## For beryllium\n",
- "A = 9. ## Mass number\n",
- "sigma_b = 57.; ## The value of sigma for beryllium\n",
- "tau_0 = 1/(6.*sigma_b**2)*(A+2./3.)/(1.-2./(3.*A))*math.log(E_0/E); ## Age of neutron for beryllium, Sq.m\n",
- "L_f = math.sqrt(tau_0); ## Slowing down length of neutron through graphite, m\n",
- "print'%s %.2f %s'%(\"\\n\\nFor Beryllium, A = \", A,\"\");\n",
- "print'%s %.2f %s'%(\"\\nNeutron age = \",tau_0*1e+004,\" Sq.cm\");\n",
- "print'%s %.2e %s'%(\"\\nSlowing down length = \",L_f,\" m\");\n",
- "\n",
- "## Result\n",
- "## For Graphite, A = 12\n",
- "## Neutron age = 362 Sq.cm\n",
- "## Slowing down length = 0.190 m\n",
- "\n",
- "## For Beryllium, A = 9\n",
- "## Neutron age = 97 Sq.cm\n",
- "## Slowing down length = 9.9e-002 m "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "For Graphite, A = 12.00 \n",
- "\n",
- "Neutron age = 362.46 Sq.cm\n",
- "\n",
- "Slowing down length = 0.19 m\n",
- "\n",
- "\n",
- "For Beryllium, A = 9.00 \n",
- "\n",
- "Neutron age = 97.46 Sq.cm\n",
- "\n",
- "Slowing down length = 9.87e-02 m\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex10-pg576"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "## Exa12.10 : : Page-576 (2011)\n",
- "#find enegy of the neutrons\n",
- "import math\n",
- "theta = 3.5*math.pi/180.; ## Reflection angle, radian\n",
- "d = 2.3e-10; ## Lattice spacing, metre\n",
- "n = 1.; ## For first order\n",
- "h = 6.6256e-34; ## Planck's constant, joule sec\n",
- "m = 1.6748e-27; ## Mass of the neutron, Kg\n",
- "E = n**2*h**2/(8.*m*d**2*math.sin(theta)**2*1.6023e-19); ## Energy of the neutrons, electron volts\n",
- "print'%s %.2f %s'%(\"\\nThe energy of the neutrons = \",E,\" eV\");\n",
- "\n",
- "## Result\n",
- "## The energy of the neutrons = 1.04 eV \n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "The energy of the neutrons = 1.04 eV\n"
- ]
- }
- ],
- "prompt_number": 14
- }
- ],
- "metadata": {}
- }
- ]
+{ + "metadata": { + "name": "", + "signature": "sha256:7bae67184aff3b5018323a5a7fd8d73277287467e3722f9114aa02be5e49dd72" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter12-Neutrons" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exx1-pg573" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "## Exa12.1 : : Page-573 (2011)\n", + "#calculate activity for Cu-63 and disintegrations\n", + "import math \n", + "N_0 = 6.23e+23; ## Avogadro's number, per mole\n", + "m = 0.1; ## Mass of copper foil, Kg\n", + "phi = 10**12; ## Neutron flux density, per square centimetre sec\n", + "a_63 = 0.691; ## Abundance of Cu-63\n", + "a_65 = 0.309; ## Abundance of Cu-65\n", + "W_m = 63.57; ## Molecular weight, gram\n", + "sigma_63 = 4.5e-24; ## Activation cross section for Cu-63, square centi metre\n", + "sigma_65 = 2.3e-24; ## Activation cross section for Cu-65, square centi metre\n", + "A_63 = phi*sigma_63*m*a_63/W_m*N_0; ## Activity for Cu-63, disintegrations per sec\n", + "A_65 = phi*sigma_65*m*a_65/W_m*N_0; ## Activity for Cu-65, disintegrations per sec\n", + "print'%s %.2e %s %.2e %s'%(\"\\nThe activity for Cu-63 is = \",A_63,\" disintegrations per sec\" and \"\\nThe activity for Cu-65 is = \",A_65,\" disintegrations per sec\");\n", + "\n", + "## Result\n", + "## The activity for Cu-63 is = 3.047e+009 disintegrations per sec \n", + "## The activity for Cu-65 is = 6.97e+008 disintegrations per sec " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The activity for Cu-63 is = 3.05e+09 \n", + "The activity for Cu-65 is = 6.97e+08 disintegrations per sec\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2-pg573" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "## Exa12.2 : : Page-573 (2011)\n", + "import math \n", + "#calculate enegy loss \n", + "A_Be = 9.; ## Mass number of beryllium\n", + "A_U = 238.; ## Mass number of uranium\n", + "E_los_Be = (1-((A_Be-1)**2/(A_Be+1)**2))*100.; ## Energy loss for beryllium\n", + "E_los_U = round((1-((A_U-1)**2/(A_U+1)**2))*100.); ## Energy loss for uranium\n", + "print'%s %.2f %s %.2f %s '%(\"\\nThe energy loss for beryllium is = \",E_los_Be,\" percent\"and \" \\nThe energy loss for uranium is = \",E_los_U,\" percent\");\n", + "\n", + "## Check for greater energy loss !!!!\n", + "if E_los_Be >= E_los_U :\n", + " print(\"\\nThe energy loss is greater for beryllium\");\n", + "else:\n", + " print(\"\\nThe energy loss is greater for uranium\");\n", + "\n", + "\n", + "## Result\n", + "## The energy loss for beryllium is = 36 percent \n", + "## The energy loss for uranium is = 2 percent\n", + "## The energy loss is greater for beryllium \n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The energy loss for beryllium is = 36.00 \n", + "The energy loss for uranium is = 2.00 percent \n", + "\n", + "The energy loss is greater for beryllium\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex3-pg574" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "## Exa12.3 : : Page-574 (2011)\n", + "#calculate energy loss of neutron\n", + "import math \n", + "A = 12.; ## Mass number of Carbon\n", + "alpha = (A-1)**2/(A+1)**2; ## Scattering coefficient\n", + "E_loss = 1/2.*(1-alpha)*100.; ## Energy loss of neutron\n", + "print'%s %.2f %s'%(\"\\nThe energy loss of neutron = \",E_loss,\" percent\")\n", + "\n", + "## Result\n", + "## The energy loss of neutron = 14.201 percent \n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The energy loss of neutron = 14.20 percent\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex4-pg574" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "## Exa12.4 : : Page-574 (2011)\n", + "#calculate number of collisions of neutrons \n", + "import math \n", + "zeta = 0.209; ## Moderated assembly\n", + "E_change = 100./1.; ## Change in energy of the neutron\n", + "E_thermal = 0.025; ## Thermal energy of the neutron, electron volts\n", + "E_n = 2*10**6; ## Energy of the neutron, electron volts\n", + "n = 1/zeta*math.log(E_change); ## Number of collisions of neutrons to loss 99 percent of their energies \n", + "n_thermal = 1/zeta*math.log(E_n/E_thermal); ## Number of collisions of neutrons to reach thermal energies\n", + "print'%s %.2f %s %.2f %s'%(\"\\nThe number of collisions of neutrons to loss 99 percent of their energies = \",n,\" \\nThe number of collisions of neutrons to reach thermal energies = \",n_thermal,\"\")\n", + "\n", + "## Result\n", + "## The number of collisions of neutrons to loss 99 percent of their energies = 22 \n", + "## The number of collisions of neutrons to reach thermal energies = 87 \n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The number of collisions of neutrons to loss 99 percent of their energies = 22.03 \n", + "The number of collisions of neutrons to reach thermal energies = 87.07 \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex5-pg574" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "## Exa12.5 : : Page-574 (2011)\n", + "#calculate average distance travelled by the neutron\n", + "import math\n", + "import scipy\n", + "from scipy import integrate\n", + "L = 1.; ## For simplicity assume thermal diffusion length to be unity, unit\n", + "def fun(x):\n", + " y=x*math.exp(-x/L)\n", + " return y\n", + "x_b = scipy.integrate.quad(fun, 0, 100); ## Average distance travelled by the neutron, unit\n", + "x_b1=x_b[0]\n", + "def fun2(x): \n", + " y1=x**2*math.exp(-x/L)\n", + " return y1\n", + "X=scipy.integrate.quad(fun2, 0, 100)\n", + "x_rms = math.sqrt(X[0]); ## Root mean square of the distance trvelled by the neutron, unit\n", + "print'%s %.2f %s'%(\"\\nThe average distance travelled by the neutron = \", x_b1,\"*L\");\n", + "print'%s %.2f %s %.2f %s '%(\"\\nThe root mean square distance travelled by the neutron = \",x_rms,\"\"and \"\",x_rms,\"x_bar\")\n", + "\n", + "## Result\n", + "## The average distance travelled by the neutron = 1*L\n", + "## The root mean square distance travelled by the neutron = 1.414L = 1.414x_bar \n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The average distance travelled by the neutron = 1.00 *L\n", + "\n", + "The root mean square distance travelled by the neutron = 1.41 1.41 x_bar \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6-pg574" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "## Exa12.6 : : Page-574 (2011)\n", + "#calculate neutron flux through water \n", + "import math\n", + "Q = 5e+08; ## Rate at which neutrons produce, neutrons per sec\n", + "r = 20.; ## Distance from the source, centi metre\n", + "## For water\n", + "lambda_wtr = 0.45; ## Transport mean free path, centi metre\n", + "L_wtr = 2.73; ## Thermal diffusion length, centi metre\n", + "phi_wtr = 3*Q/(4.*math.pi*lambda_wtr*r)*math.exp(-r/L_wtr); ## Neutron flux for water, neutrons per square centimetre per sec\n", + "## For heavy water\n", + "lambda_h_wtr = 2.40; ## Transport mean free path, centi metre\n", + "L_h_wtr = 171.; ## Thermal diffusion length, centi metre\n", + "phi_h_wtr = 3*Q/(4.*math.pi*lambda_h_wtr*r)*math.exp(-r/L_h_wtr); ## Neutron flux for heavy water, neutrons per square centimetre per sec\n", + "print'%s %.2e %s %.2e %s '%(\"\\nThe neutron flux through water = \",phi_wtr,\" neutrons per square cm per sec\"and \"\\nThe neutron flux through heavy water = \",phi_h_wtr,\" neutrons per square cm per sec\")\n", + "\n", + "## Result\n", + "## The neutron flux through water = 8.730e+003 neutrons per square cm per sec \n", + "## The neutron flux through heavy water = 2.212e+006 neutrons per square cm per sec \n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The neutron flux through water = 8.73e+03 \n", + "The neutron flux through heavy water = 2.21e+06 neutrons per square cm per sec \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex7-pg575" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "## Exa12.7 : : Page-575 (2011)\n", + "#calculate neutron flux and diffusion length\n", + "import math\n", + "k = 1.38e-23; ## Boltzmann constant, joules per kelvin\n", + "T = 323.; ## Temperature, kelvin\n", + "E = (k*T)/1.6e-19; ## Thermal energy, joules\n", + "sigma_0 = 13.2e-28; ## Cross section, square metre\n", + "E_0 = 0.025; ## Energy of the neutron, electron volts\n", + "sigma_a = sigma_0*math.sqrt(E_0/E); ## Absorption cross section, square metre\n", + "t_half = 2.25; ## Half life, hours\n", + "D= 0.69/t_half; ## Decay constant, per hour\n", + "N_0 = 6.023e+026; ## Avogadro's number, per \n", + "m_Mn = 55.; ## Mass number of mangnese\n", + "w = 0.1e-03; ## Weight of mangnese foil, Kg\n", + "A = 200.; ## Activity, disintegrations per sec\n", + "N = N_0*w/m_Mn; ## Number of mangnese nuclei in the foil\n", + "x1 = 1.5; ## Base, metre\n", + "x2 = 2.0; ## Height, metre\n", + "phi = A/(N*sigma_a*0.416); ## Neutron flux, neutrons per square metre per sec\n", + "phi1 = 1.; ## For simplicity assume initial neutron flux to be unity, neutrons/Sq.m-sec\n", + "phi2 = 1/2.*phi1; ## Given neutron flux, neutrons/Sq.m-sec\n", + "L1 = 1/math.log(phi1/phi2)/(x2-x1); ## Thermal diffusion length for given neutron flux, m\n", + "L = math.sqrt(1./((1./L1)**2+(math.pi/x1)**2+(math.pi/x2)**2)); ## Diffusion length, metre\n", + "print'%s %.2e %s %.2f %s '%(\"\\nThe neutron flux = \",phi,\" neutrons per square metre per sec\"and \" \\nThe diffusion length = \",L,\" metre\");\n", + "\n", + "## Result\n", + "## The neutron flux = 3.51e+008 neutrons per square metre per sec \n", + "## The diffusion length = 0.38 metre\n", + "## Note: the difussion length is solved wrongly in the testbook\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The neutron flux = 3.51e+08 \n", + "The diffusion length = 0.38 metre \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex8-pg575" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "## Exa12.8 : : Page-575(2011)\n", + "#find diffusion length for thermal neutron\n", + "import math\n", + "N_0 = 6.023e+026; ## Avogadro's number, per mole\n", + "rho = 1.62e+03; ## Density, kg per cubic metre\n", + "sigma_a = 3.2e-31; ## Absorption cross section, square metre\n", + "sigma_s = 4.8e-28; ## Scattered cross section, square metre\n", + "A = 12.; ## Mass number\n", + "lambda_a = A/(N_0*rho*sigma_a); ## Absorption mean free path, metre\n", + "lambda_tr = A/(N_0*rho*sigma_s*(1.-2./(3.*A))); ## Transport mean free path, metre\n", + "L = math.sqrt(lambda_a*lambda_tr/3.); ## Diffusion length for thermal neutron\n", + "print'%s %.2f %s'%(\"\\nThe diffusion length for thermal neutron = \",L,\" metre \")\n", + "\n", + "## Result\n", + "## The diffusion length for thermal neutron = 0.590 metre \n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The diffusion length for thermal neutron = 0.59 metre \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex9-pg575" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "## Exa12.9 : : Page-575 (2011)\n", + "#calculate graphite and neutron age and slowing down length and same as berylliums\n", + "import math\n", + "E_0 = 2e+06; ## Average energy of the neutron, electron volts\n", + "E = 0.025; ## Thermal energy of the neutron, electron volts\n", + "## For graphite\n", + "A = 12. ## Mass number\n", + "sigma_g = 33.5; ## The value of sigma for graphite\n", + "tau_0 = 1./(6.*sigma_g**2)*(A+2./3.)/(1.-2./(3.*A))*math.log(E_0/E); ## Age of neutron for graphite, Sq.m\n", + "L_f = math.sqrt(tau_0); ## Slowing down length of neutron through graphite, m\n", + "print'%s %.2f %s'%(\"\\nFor Graphite, A = \", A,\"\");\n", + "print'%s %.2f %s'%(\"\\nNeutron age = \",tau_0*1e+004,\" Sq.cm\");\n", + "print'%s %.2f %s'%(\"\\nSlowing down length =\",L_f,\" m\");\n", + "## For beryllium\n", + "A = 9. ## Mass number\n", + "sigma_b = 57.; ## The value of sigma for beryllium\n", + "tau_0 = 1/(6.*sigma_b**2)*(A+2./3.)/(1.-2./(3.*A))*math.log(E_0/E); ## Age of neutron for beryllium, Sq.m\n", + "L_f = math.sqrt(tau_0); ## Slowing down length of neutron through graphite, m\n", + "print'%s %.2f %s'%(\"\\n\\nFor Beryllium, A = \", A,\"\");\n", + "print'%s %.2f %s'%(\"\\nNeutron age = \",tau_0*1e+004,\" Sq.cm\");\n", + "print'%s %.2e %s'%(\"\\nSlowing down length = \",L_f,\" m\");\n", + "\n", + "## Result\n", + "## For Graphite, A = 12\n", + "## Neutron age = 362 Sq.cm\n", + "## Slowing down length = 0.190 m\n", + "\n", + "## For Beryllium, A = 9\n", + "## Neutron age = 97 Sq.cm\n", + "## Slowing down length = 9.9e-002 m " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "For Graphite, A = 12.00 \n", + "\n", + "Neutron age = 362.46 Sq.cm\n", + "\n", + "Slowing down length = 0.19 m\n", + "\n", + "\n", + "For Beryllium, A = 9.00 \n", + "\n", + "Neutron age = 97.46 Sq.cm\n", + "\n", + "Slowing down length = 9.87e-02 m\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex10-pg576" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "## Exa12.10 : : Page-576 (2011)\n", + "#find enegy of the neutrons\n", + "import math\n", + "theta = 3.5*math.pi/180.; ## Reflection angle, radian\n", + "d = 2.3e-10; ## Lattice spacing, metre\n", + "n = 1.; ## For first order\n", + "h = 6.6256e-34; ## Planck's constant, joule sec\n", + "m = 1.6748e-27; ## Mass of the neutron, Kg\n", + "E = n**2*h**2/(8.*m*d**2*math.sin(theta)**2*1.6023e-19); ## Energy of the neutrons, electron volts\n", + "print'%s %.2f %s'%(\"\\nThe energy of the neutrons = \",E,\" eV\");\n", + "\n", + "## Result\n", + "## The energy of the neutrons = 1.04 eV \n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The energy of the neutrons = 1.04 eV\n" + ] + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] }
\ No newline at end of file diff --git a/Nuclear_Physics/README.txt b/Nuclear_Physics/README.txt new file mode 100755 index 00000000..cb19ed24 --- /dev/null +++ b/Nuclear_Physics/README.txt @@ -0,0 +1,10 @@ +Contributed By: harsha vardhan +Course: msc +College/Institute/Organization: iitbombay +Department/Designation: msc chem +Book Title: Nuclear Physics +Author: D. C. Tayal +Publisher: Himalaya Publishing House, Mumbai +Year of publication: 2011 +Isbn: 978-93-5024-743-3 +Edition: 5
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