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| author | kinitrupti | 2017-05-12 18:53:46 +0530 |
|---|---|---|
| committer | kinitrupti | 2017-05-12 18:53:46 +0530 |
| commit | 6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d (patch) | |
| tree | 22789c9dbe468dae6697dcd12d8e97de4bcf94a2 /Nuclear_Physics_by_D._C._Tayal/Chapter5.ipynb | |
| parent | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (diff) | |
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diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter5.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter5.ipynb new file mode 100755 index 00000000..8e2edd81 --- /dev/null +++ b/Nuclear_Physics_by_D._C._Tayal/Chapter5.ipynb @@ -0,0 +1,347 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:5bd75b768cbef982b8c705dbf643c0d6d3b446f1e0997915eea25d6d1e7b9a2f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter5-Alpha particles"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.1 : : Page 203 (2011)\n",
+ "#find Disintegration energy and Barrier height for alpha-particle\n",
+ "import math\n",
+ "E_a = 8.766; ## Energy of the alpha particle, MeV\n",
+ "A = 212.; ## Atomic mass of Po-212, amu\n",
+ "M_a = 4.; ## Atomic mass of alpha particle, amu\n",
+ "e = 1.6e-019; ## Charge of an electron, coulomb\n",
+ "Z = 82.; ## Atomic number of Po-212\n",
+ "R_0 = 1.4e-015; ## Distance of closest approach,metre\n",
+ "K = 8.99e+09; ## Coulomb constant\n",
+ "E = E_a*A/(A-M_a); ## Disintegration energy, mega electron volts\n",
+ "B_H = 2.*Z*e**2.*K/(R_0*A**(1./3.)*1.6*10**-13.); ## Barrier height for an alpha particle within the nucleus, MeV\n",
+ "print\"%s %.3f %s %.3f %s \"%(\"Disintegration energy :\",E,\" MeV Barrier height for alpha-particle: \",B_H,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## Disintegration energy : 8.935 MeV \n",
+ "## Barrier height for alpha-particle: 28.26 MeV "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Disintegration energy : 8.935 MeV Barrier height for alpha-particle: 28.259 MeV \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.2 : : Page 203 (2011)\n",
+ "## We have to make calculation for alpha particle and for proton\n",
+ "import math\n",
+ "E_a = 8.766; ## Energy of the alpha particle, mega electron volts\n",
+ "A_Bi = 209.; ## Atomic mass of Bi-209, atomic mass unit\n",
+ "A_a = 4.; ## Atomic mass of alpha particle, atomic mass unit\n",
+ "A_p = 1.; ## Atomic mass of proton, atomic mass unit\n",
+ "e = 1.6e-019; ## Charge of an electron, coulomb\n",
+ "Z = 83.; ## Atomic number of bismuth\n",
+ "R_0 = 1.4e-015; ## Distance of closest approach,metre\n",
+ "K = 8.99e+09; ## Coulomb constant\n",
+ "B_H_a = 2.*Z*e**2.*K/(R_0*1.6e-013*(A_Bi**(1/3.)+A_a**(1/3.))); ## Barrier height for an alpha particle, mega electron volts\n",
+ "B_H_p = 1.*Z*e**2.*K/(R_0*1.6e-013*(A_Bi**(1/3.)+A_p**(1/3.))); ## Barrier height for proton, mega electron volts\n",
+ "print'%s %.2f %s %.2f %s'%(\"Barrier height for the alpha particle =\",B_H_a,\" MeV Barrier height for the proton = \",B_H_p,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## Barrier height for the alpha particle = 22.67 MeV \n",
+ "## Barrier height for the proton = 12.30 MeV "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Barrier height for the alpha particle = 22.67 MeV Barrier height for the proton = 12.30 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.3 : : Page 203 (2011)\n",
+ "## We have also calculate the value of magnetic field in a particular orbit. \n",
+ "import math\n",
+ "C = 3e+08; ## Velocity of light, m/S\n",
+ "M_0 = 6.644e-027*(C)**2/(1.60218e-013); ## Rest mass of alpha particle, MeV\n",
+ "T = 5.998; ## Kinetic energy of alpha particle emitted by Po-218\n",
+ "q = 2*1.60218e-019; ## Charge of alpha particle, C\n",
+ "V = math.sqrt(C**2*T*(T+2*M_0)/(T+M_0)**2); ## Velocity of alpha particle,metre per sec\n",
+ "B_r = V*M_0*(1.60218e-013)/(C**2.*q*math.sqrt(1-V**2./C**2.)); ## magnetic field in a particular orbit, Web per mtere\n",
+ "print'%s %.2e %s %.2f %s '%(\"The velocity of alpha particle : \",V,\" m/sThe magnetic field in a particular orbit :\",B_r,\" Wb/m\");\n",
+ "\n",
+ "## Result\n",
+ "## The velocity of alpha particle : 1.699e+007 m/s\n",
+ "## The magnetic field in a particular orbit : 0.3528 Wb/m"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of alpha particle : 1.70e+07 m/sThe magnetic field in a particular orbit : 0.35 Wb/m \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.4: : Page 204 (2011)\n",
+ "#find The probability of leakage of alpha-particle through potential barrier\n",
+ "import math\n",
+ "a = 10**-14; ## Width of the potential barrier, m\n",
+ "E = 5*1.60218e-013; ## Energy of the alpha particle, joule\n",
+ "V = 10*1.60218e-013; ## Potential height, joule\n",
+ "M_0 = 6.644e-027; ## Rest mass of the alpha particle, joule\n",
+ "h_red = 1.05457e-034; ## Reduced value of Planck's constant,joule sec \n",
+ "T = 4*math.exp(-2*a*math.sqrt(2*M_0*(V-E)/h_red**2)); ## Probability of leakage through through potential barrier\n",
+ "print'%s %.2e %s'%(\"The probability of leakage of alpha-particle through potential barrier = \",T,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The probability of leakage of alpha-particle through potential barrier = 1.271e-008 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The probability of leakage of alpha-particle through potential barrier = 1.27e-08 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.6: : Page 204 (2011)\n",
+ "#find The disintegration constant increases \n",
+ "import math\n",
+ "Z_D = 82.; ## Atomic number of Po\n",
+ "E_Po210 = 5.3; ## Alpha-source for Po210, MeV\n",
+ "E_Po214 = 7.7; ## Alpha-source for Po214, MeV\n",
+ "log_lambda_Po210 = -1*1.72*Z_D*E_Po210**(-1/2.); \n",
+ "log_lambda_Po214 = -1*1.72*Z_D*E_Po214**(-1/2.); \n",
+ "delta_OM_t = log_lambda_Po214 - log_lambda_Po210; ## Difference in order of magnitude of life times of Po214 and Po210\n",
+ "print'%s %.2d %s'%(\"The disintegration constant increases by a factor of some 10^ \" ,delta_OM_t,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The disintegration constant increases by a factor of some 10^10 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The disintegration constant increases by a factor of some 10^ 10 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg205"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.8: : Page 205 (2011)\n",
+ "#find The half life of Pu \n",
+ "import math\n",
+ "N = 120.1*6.023e+023/239.; ## Number of Pu nuclei\n",
+ "P_rel = 0.231; ## Power released, watt\n",
+ "E_rel = 5.323*1.6026e-13; ## Energy released, joule\n",
+ "decay_rate = P_rel/E_rel; ## Decay rate of Pu239, per hour\n",
+ "t_half = N*math.log(2)/(decay_rate*365.*86400.); ## Half life of Po239, sec\n",
+ "print\"%s %.2e %s\"%(\"The half life of Pu = \",t_half,\" yr\");\n",
+ "\n",
+ "## Result\n",
+ "## The half life of Pu = 2.46e+004 yr "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The half life of Pu = 2.46e+04 yr\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg205"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.9 : : Page 205(2011)\n",
+ "#find The slope of alpha decay energy versus atomic number\n",
+ "import math\n",
+ "a_v = 14.; ## Volume energy constant, MeV\n",
+ "a_s = 13.; ## Surface energy constant, MeV\n",
+ "a_c = 0.60; ## Coulomb energy constant, MeV\n",
+ "a_a = 19.; ## Asymmetric energy constant, MeV\n",
+ "A = 202.; ## Mass number\n",
+ "Z = 82; ## Atomic number \n",
+ "dE_by_dN = -8/9*a_s/A**(4/3.)-4/3.*a_c*Z/A**(4/3.)*(1-4.*Z/(3.*A))-16.*a_a*Z/A**2.*(1.-2.*Z/A); ## Slope, mega electron volts per nucleon\n",
+ "print'%s %.2f %s'%(\"The slope of alpha decay energy versus atomic number = \",dE_by_dN,\" MeV/nucleon\");\n",
+ "\n",
+ "## Result\n",
+ "## The slope of alpha decay energy versus atomic number = -0.15007 MeV/nucleon "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The slope of alpha decay energy versus atomic number = -0.15 MeV/nucleon\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg206"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.10 : : Page 206 (2011)\n",
+ "#find The hindrance factor for alpha particle\n",
+ "import math\n",
+ "h_kt = 1.05457e-34; ## Reduced Planck's constant, joule sec\n",
+ "e = 1.60218e-19; ## Charge of an electron, coulomb\n",
+ "l = 2.; ## Orbital angular momentum\n",
+ "eps_0 = 8.5542e-12; ## Absolute permittivity of free space, coulomb square per newton per metre square\n",
+ "Z_D = 90.; ## Atomic number of daughter nucleus\n",
+ "m = 6.644e-27; ## Mass of alpha particle, Kg\n",
+ "R = 8.627e-15; ## Radius of daughter nucleus, metre\n",
+ "T1_by_T0 = math.exp(2.*l*(l+1.)*h_kt/e*math.sqrt(math.pi*eps_0/(Z_D*m*R))); ## Hindrance factor\n",
+ "print'%s %.2f %s'%(\"The hindrance factor for alpha particle = \" ,T1_by_T0,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The hindrance factor for alpha particle = 1.768 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hindrance factor for alpha particle = 1.77 \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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