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{
"metadata": {
"name": "MP-13"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": "Nuclear Reaction and Application"
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 13.1 Page 417"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#initiation of variable\nv=1*1.0*10**-6.0*10**2; p=7.9; m=p*v;Na=6.023*10**23 #given values and various constants in suitable units\nM=56.0;N=m*Na/M; #number of atoms\ni=3.0*10**-6;\nq=1.6*10**-19;\n\n#calculation\nIo=i/q; #intensity\ns=0.6*10**-24;S=1; #given values in suitable units\nR=N*s*Io/S; #rate of neutrons\n\n#result\nprint\"The rate of neutrons emitted from the target in particles per second is %.1e\" %round(R,3);",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The rate of neutrons emitted from the target in particles per second is 9.6e+07\n"
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 13.2 Page 419"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#initiation of variable\nA=197.0; m=30*10**-3;phi=3.0*10**12; #given values and various constants taken in suitable units\nAr=99.0*10**-24; Na=6.023*10**23\n\n#calculation\nR=(phi*Na*Ar*m/A); #rate or production of gold\nt=2.7*24*60 # time of decay\nAct=R*(0.693/t); #activity /sec\nActCi=Act/(3.7*10**4); # in terms of curie(Ci)\n\n#result\nprint\"The activity is found out to be %.1e\" %round(Act,3),\"/sec i.e \" ,ActCi,\"Ci\";\n",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The activity is found out to be 4.9e+06 /sec i.e 131.228932295 Ci\n"
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 13.3 Page 423"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#initiation of variable\nfrom math import exp\nv=1.5*1.5*2.5*(10**-6)*10**2; #volume in cm3\np=8.9; #density in g/cm3\nm=p*v;Na=6.023*10**23 #mass and Avagadro's number\nM=58.9; #Given values\n\n#calculation\nN=m*Na/M;\ni=12*10**-6; #thickness of beam\nq=1.6*10**-19;\nIo=i/(2*q); #intensity\ns=0.64*10**-24; #Given values\nS=1.5*1.5;\nR=N*s*Io/S; #rate of production of 61Cu\n\n#result\nprint \"The rate of neutrons emitted from the target in particles/second is %.1e\" %round(R,3);\n\n#part b\nact=R*(1-(exp((0.693)*(-2/3.41)))); #activity\n\n#result\nprint\"The activity after 2.0h in /sec is %.1e\" %round(act,3);\n",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The rate of neutrons emitted from the target in particles/second is 546058064.516\nThe activity after 2.0h in /sec is 182378303.69\n"
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 13.4 Page 425"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#initiation of variable\nm2H=2.014102; #mass of various particles\nmn=1.008665;m63Cu=62.929599;\nm64Zn=63.929145;c2=931.5; #c^2=931.5 MeV\nQ=(m2H+m63Cu-mn-m64Zn)*c2; #Q of the reaction\n\n#result\nprint\"The value of Q is in MeV\",round(Q,3);\n\n\n#part b\nKx=12.00;Ky=16.85;\nKy=Q+Kx-Ky #kinetic energy of 64Zn\n\n#result\nprint\"The value of Ky was found out to be in MeV\",round(Ky,3);",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The value of Q is in MeV 5.487\nThe value of Ky was found out to be in MeV 0.637\n"
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 13.5 Page 425"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#initiation of variable\nmp=1.007825;m3H=3.016049; #mass of the particle\nm2H=2.014102;c2=931.5; #constant\nQ=(mp+m3H-(2*m2H))*c2; #Q of thereaction\n\n#result\nprint\"The value of q was found out to be in MeV\",round(Q,3);\n\n#partb\nKth1= -Q*(1+(mp/m3H)); #threshold energy of kinetic energy\nKth2=-Q*(1+(m3H/mp)); #threshold kinetic energy in case2\n\n#result\nprint\"The threshold kinetic energy in case-1 in MeV\",round(Kth1,3);\nprint\"The threshold kinetic energy in case-2 in MeV\",round(Kth2,3);",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The value of q was found out to be in MeV -4.033\nThe threshold kinetic energy in case-1 in MeV 5.381\nThe threshold kinetic energy in case-2 in MeV 16.104\n"
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
}
|
