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| author | hardythe1 | 2015-06-03 15:27:17 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-06-03 15:27:17 +0530 |
| commit | 47d7279a724246ef7aa0f5359cf417992ed04449 (patch) | |
| tree | c613e5e4813d846d24d67f46507a6a69d1a42d87 /Nuclear_Physics/Chapter13.ipynb | |
| parent | 435840cef00c596d9e608f9eb2d96f522ea8505a (diff) | |
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diff --git a/Nuclear_Physics/Chapter13.ipynb b/Nuclear_Physics/Chapter13.ipynb new file mode 100755 index 00000000..c63c61c5 --- /dev/null +++ b/Nuclear_Physics/Chapter13.ipynb @@ -0,0 +1,430 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:e4b6100001a32fad047d14064effe0ad1c4e730aa6c2b9ae5ba17ef44d36b6f5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter13-Nuclear Fission and Fusion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg600"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.1 : : Page-600 (2011)\n",
+ "#calculate rate of fission and energy released\n",
+ "import math\n",
+ "E = 200.*1.6023e-13; ## Energy released per fission, joule\n",
+ "E_t = 2.; ## Total power produced, watt\n",
+ "R_fiss = E_t/E; ## Fission rate, fissions per sec\n",
+ "m = 0.5; ## Mass of uranium, Kg\n",
+ "M = 235.; ## Mass number of uranium\n",
+ "N_0 = 6.023e+26; ## Avogadro's number, per mole\n",
+ "N = m/M*N_0 ## Number of uranium nuclei\n",
+ "E_rel = N*E/4.08*10**-3; ## Energy released, kilocalories\n",
+ "print'%s %.2e %s %.2e %s'%(\"\\nThe rate of fission of U-235 = \",R_fiss,\" fissions per sec\"and\" \\nEnergy released =\",E_rel,\" kcal\");\n",
+ "\n",
+ "## Result\n",
+ "## The rate of fission of U-235 = 6.24e+010 fissions per sec \n",
+ "## Energy released = 1.006535e+010 kcal "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The rate of fission of U-235 = 6.24e+10 \n",
+ "Energy released = 1.01e+10 kcal\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg600"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.2 : : Page-600 (2011)\n",
+ "#calculate number of free neutrons in the reactor\n",
+ "import math\n",
+ "E = 200*1.6e-13; ## Energy released per fission, joules per neutron\n",
+ "t = 10**-3; ## Time, sec\n",
+ "P = E/t; ## Power produced by one free neutron, watt per neutron\n",
+ "P_l = 10**9; ## Power level, watt\n",
+ "N = P_l/P; ## Number of free neutrons in the reactor, neutrons\n",
+ "print'%s %.2e %s'%(\"\\nThe number of free neutrons in the reactor = \",N,\" neutrons\");\n",
+ "\n",
+ "## Result\n",
+ "## The number of free neutrons in the reactor = 3.125e+016 neutrons "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The number of free neutrons in the reactor = 3.12e+16 neutrons\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg600"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.3 : : Page-600 (2011)\n",
+ "#calculate average number of neutrons \n",
+ "import math\n",
+ "N_0_235 = 1.; ## Number of uranium 235 per 238 \n",
+ "N_0_238 = 20.; ## Number of uranium 238 for one uranium 235 \n",
+ "sigma_a_235 = 683.; ## Absorption cross section for uranium 235, barn\n",
+ "sigma_a_238 = 2.73; ## Absorption cross section for uranium 238, barn\n",
+ "sigma_f_235 = 583.; ## Fission cross section, barn\n",
+ "sigma_a = (N_0_235*sigma_a_235+N_0_238*sigma_a_238)/(N_0_235+N_0_238); ##Asorption cross sec, barn\n",
+ "sigma_f = N_0_235*sigma_f_235/(N_0_235+N_0_238); ## Fisssion cross section \n",
+ "v = 2.43;\n",
+ "eta = v*sigma_f/sigma_a; ## Average number of neutron released per absorption\n",
+ "print'%s %.2f %s'%(\"\\nThe average number of neutrons released per absorption = \", eta,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The average number of neutrons released per absorption = 1.921 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The average number of neutrons released per absorption = 1.92 \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg600"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.4 : : Page-600(2011)\n",
+ "#find excitaion energy for uranum 236 and 239 and 235\n",
+ "import math\n",
+ "a_v = 14.0; ## Volume binding energy constant, mega electron volts\n",
+ "a_s = 13.0; ## Surface binding energy constant, mega electron volts\n",
+ "a_c = 0.583; ## Coulomb constant, mega electron volts\n",
+ "a_a = 19.3; ## Asymmetric constant, mega electron volts\n",
+ "a_p = 33.5; ## Pairing energy constant, mega electron volts\n",
+ "Z = 92.; ## Atomic number \n",
+ "## For U-236\n",
+ "A = 235.; ## Mass number\n",
+ "E_exc_236 = a_v*(A+1-A)-a_s*((A+1)**(2./3.)-A**(2./3.))-a_c*(Z**2/(A+1.)**(1./3.)-Z**2/A**(1/3.))-a_a*((A+1-2*Z)**2/(A+1)-(A-2*Z)**2/A)+a_p*(A+1)**(-3./4.); ## Excitation energy for uranium 236, mega electron volts\n",
+ "## For U-239\n",
+ "A = 238.; ## Mass number\n",
+ "E_exc_239 = a_v*(A+1-A)-a_s*((A+1)**(2./3.)-A**(2./3.))-a_c*(Z**2/(A+1)**(1/3.)-Z**2/A**(1./3.))-a_a*((A+1.-2.*Z)**2/(A+1)-(A-2*Z)**2/A)+a_p*((A+1)**(-3/4.)-A**(-3/4.)); ## Excitation energy for uranium 239\n",
+ "## Now calculate the rate of spontaneous fissioning for U-235\n",
+ "N_0 = 6.02214e+23; ## Avogadro's constant, per mole\n",
+ "M = 235.; ## Mass number\n",
+ "t_half = 3e+17*3.15e+7; ## Half life, years \n",
+ "D = 0.693/t_half; ## Decay constant, per year\n",
+ "N = N_0/M; ## Mass of uranium 235, Kg\n",
+ "dN_dt = N*D*3600; ## Rate of spontaneous fissioning of uranium 235, per hour\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nThe excitation energy for uranium 236 =\",E_exc_236,\" MeV\"and \"\\nThe excitation energy for uranium 239 = \",E_exc_239,\" MeV\"and\"\\nThe rate of spontaneous fissioning of uranium 235 = \",dN_dt,\" per hour\");\n",
+ "\n",
+ "## Result\n",
+ "## The excitation energy for uranium 236 = 6.8 MeV\n",
+ "## The excitation energy for uranium 239 = 5.9 MeV\n",
+ "## The rate of spontaneous fissioning of uranium 235 = 0.68 per hour \n",
+ "\n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The excitation energy for uranium 236 = 6.77 \n",
+ "The excitation energy for uranium 239 = 5.90 \n",
+ "The rate of spontaneous fissioning of uranium 235 = 0.68 per hour \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg601"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.5 : : Page-601 (2011)\n",
+ "#calculate total energy released during fusion\n",
+ "import math \n",
+ "a = 10**5; ## Area of the lake, square mile\n",
+ "d = 1/20.; ## Depth of the lake, mile\n",
+ "V = a*d*(1.6e+03)**3; ## Volume of the lake, cubic metre\n",
+ "rho = 10**3; ## Density of water, kg per cubic metre\n",
+ "M_water = V*rho; ## Total mass of water in the lake, Kg\n",
+ "N_0 = 6.02214e+26; ## Avogadro's constant, per mole\n",
+ "A = 18.; ## Milecular mass of water\n",
+ "N = M_water*N_0/A; ## Number of molecules of water, molecules\n",
+ "abund_det = 0.0156e-02; ## Abundance of deterium\n",
+ "N_d = N*2*abund_det; ## Number of deterium atoms\n",
+ "E_per_det = 43/6.; ## Energy released per deterium atom, mega electron volts\n",
+ "E_t = N_d*E_per_det; ## Total energy released during fusion, mega electron volt\n",
+ "print'%s %.2e %s'%(\"\\nThe total energy released during fusion = \",E_t,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## Total energy released during fusion = 1.53e+039 MeV\n",
+ "\n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The total energy released during fusion = 1.53e+39 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg601"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.6 : : Page-601 (2011)\n",
+ "#calculate temperature attained by thermonuclear device\n",
+ "import math \n",
+ "r = 1/2.; ## Radius of the tube, metre\n",
+ "a = math.pi*r**2; ## Area of the torus, square metre\n",
+ "V = 3*math.pi*a; ## Volume of the torus, cubic metre\n",
+ "P = 10**-5*13.6e+3*9.81; ## Pressure of the gas, newton per square metre\n",
+ "C = 1200e-6; ## Capacitance, farad\n",
+ "v = 4e+4; ## potential, volts\n",
+ "T_room = 293; ## Room temperature, kelvin\n",
+ "N_k = P*V/T_room; ## From gas equation\n",
+ "E = 1/2.*C*v**2; ## Energy stored, joules\n",
+ "T_k = 1/6.*E/(N_k*10.); ## Temperature attained by thermonuclear device, kelvin\n",
+ "print'%s %.2e %s'%(\"\\nThe temperature attained by thermonuclear device =\",T_k,\" K\");\n",
+ "\n",
+ "## Result\n",
+ "## The temperature attained by thermonuclear device = 4.75e+005 K \n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The temperature attained by thermonuclear device = 4.75e+05 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg601"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.7 : : Page-601 (2011)\n",
+ "#calculate energy released by the sun temperautre of the sun\n",
+ "import math \n",
+ "G = 6.67e-11; ## Gravitational constant, newton square m per square kg\n",
+ "r = 7e+08; ## Radius of the sun, metre\n",
+ "M_0 = 2e+30; ## Mass of the sun, kg\n",
+ "E_rel = 3/5.*G*M_0**2/r; ## Energy released by the sun, joule\n",
+ "E_dia_shrink_10 = E_rel/9.; ## Energy released when sun diameter shrink by 10 percent, joule\n",
+ "R = 8.314; ## Universal gas constant, joule per kelvin per kelvin per mole\n",
+ "T = E_rel/(M_0*R); ## Temperature of the sun, kelvin\n",
+ "print'%s %.2e %s %.2e %s %.2e %s '%(\"\\nThe energy released by the sun = \",E_rel,\" joule\"and\" \\nThe energy released when sun diameter is shrinked by 10 percent = \",E_dia_shrink_10,\" joule\"and\" \\nThe temperature of the sun = \",T,\" kelvin \");\n",
+ "\n",
+ "## Result\n",
+ "## The energy released by the sun = 2.29e+041 joule \n",
+ "## The energy released when sun diameter is shrinked by 10 percent = 2.54e+040 joule \n",
+ "## The temperature of the sun = 1.38e+010 kelvin \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy released by the sun = 2.29e+41 \n",
+ "The energy released when sun diameter is shrinked by 10 percent = 2.54e+40 \n",
+ "The temperature of the sun = 1.38e+10 kelvin \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg602"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.8 : : Page-602 (2011)\n",
+ "# estimated Q-value\n",
+ "import math\n",
+ "A_0 = 240.; ## Mass number of parent nucleus\n",
+ "A_1 = 120.; ## Mass number of daughter nucleus\n",
+ "B_120 = 8.5; ## Binding energy of daughter nucleus\n",
+ "B_240 = 7.6; ## Binding energy of parent nucleus\n",
+ "Q = 2*A_1*B_120-A_0*B_240; ## Estimated Q-value, mega electron volts\n",
+ "print'%s %.2f %s'%(\"\\nThe estimated Q-value is = \",Q,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The estimated Q-value is = 216 MeV \n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The estimated Q-value is = 216.00 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg602"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.9 : : Page-602 (2011)\n",
+ "#find The asymmetric binding energy\n",
+ "import math\n",
+ "E = 31.7; ## Energy, MeV\n",
+ "a_a = 5/9.*2**(-2/3.)*E; ## Asymmetric binding energy term, mega electron volts\n",
+ "print'%s %.2f %s'%(\"\\nThe asymmetric binding energy term = \",a_a,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The asymmetric binding energy term = 11.1 MeV \n",
+ "\n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The asymmetric binding energy term = 11.09 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |