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{
"metadata": {
"name": "Chapter13",
"signature": "sha256:3612dc1ad0c2617d4c1e1da21ea7791334391562bfe93ddc9d0f98037d4fe15d"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 13: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\n",
"v=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\n",
"M=56.0;N=m*Na/M; #number of atoms\n",
"i=3.0*10**-6;\n",
"q=1.6*10**-19;\n",
"\n",
"#calculation\n",
"Io=i/q; #intensity\n",
"s=0.6*10**-24;S=1; #given values in suitable units\n",
"R=N*s*Io/S; #rate of neutrons\n",
"\n",
"#result\n",
"print\"The rate of neutrons emitted from the target in particles per second is %.1e\" %round(R,3);\n",
"print\"Slight difference in answer due to approximation error\""
],
"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",
"slight difference due to approximation error\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 13.2, Page 419"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initiation of variable\n",
"A=197.0; m=30*10**-3;phi=3.0*10**12; #given values and various constants taken in suitable units\n",
"Ar=99.0*10**-24; Na=6.023*10**23\n",
"\n",
"#calculation\n",
"R=(phi*Na*Ar*m/A); #rate or production of gold\n",
"t=2.7*24*60 # time of decay\n",
"Act=R*(0.693/t); #activity /sec\n",
"ActCi=Act/(3.7*10**4); # in terms of curie(Ci)\n",
"\n",
"#result\n",
"print\"The activity is found out to be %.1e\" %round(Act,3),\"/sec i.e \" ,round(ActCi,3),\"muCi\"\n",
"print\"Slight difference in answer due to approximation error\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The activity is found out to be 4.9e+06 /sec i.e 131.229 muCi\n",
"slight difference due to approximation error\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 13.3, Page 423"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initiation of variable\n",
"from math import exp\n",
"v=1.5*1.5*2.5*(10**-6)*10**2; #volume in cm3\n",
"p=8.9; #density in g/cm3\n",
"m=p*v;Na=6.023*10**23 #mass and Avagadro's number\n",
"M=58.9; #Given values\n",
"\n",
"#calculation\n",
"N=m*Na/M;\n",
"i=12*10**-6; #thickness of beam\n",
"q=1.6*10**-19;\n",
"Io=i/(2*q); #intensity\n",
"s=0.64*10**-24; #Given values\n",
"S=1.5*1.5;\n",
"R=N*s*Io/S; #rate of production of 61Cu\n",
"\n",
"#result\n",
"print \"The rate of neutrons emitted from the target in particles/second is %.1e\" %round(R,3);\n",
"\n",
"#part b\n",
"act=R*(1-(exp((0.693)*(-2/3.41)))); #activity\n",
"\n",
"#result\n",
"print\"The activity after 2.0h in /sec is %.1e\" %round(act,3),\"=4.9mCi\";\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The rate of neutrons emitted from the target in particles/second is 5.5e+08\n",
"The activity after 2.0h in /sec is 1.8e+08 =4.9mCi\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 13.4, Page 425"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initiation of variable\n",
"m2H=2.014102; #mass of various particles\n",
"mn=1.008665;m63Cu=62.929599;\n",
"m64Zn=63.929145;c2=931.5; #c^2=931.5 MeV\n",
"Q=(m2H+m63Cu-mn-m64Zn)*c2; #Q of the reaction\n",
"\n",
"#result\n",
"print\"The value of Q is in MeV\",round(Q,3);\n",
"\n",
"\n",
"#part b\n",
"Kx=12.00;Ky=16.85;\n",
"Ky=Q+Kx-Ky #kinetic energy of 64Zn\n",
"\n",
"#result\n",
"print\"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\n",
"The 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\n",
"mp=1.007825;m3H=3.016049; #mass of the particle\n",
"m2H=2.014102;c2=931.5; #constant\n",
"Q=(mp+m3H-(2*m2H))*c2; #Q of the reaction\n",
"\n",
"#result\n",
"print\"The value of q was found out to be in MeV\",round(Q,3);\n",
"\n",
"#partb\n",
"Kth1= -Q*(1+(mp/m3H)); #threshold energy of kinetic energy\n",
"Kth2=-Q*(1+(m3H/mp)); #threshold kinetic energy in case2\n",
"\n",
"#result\n",
"print\"The threshold kinetic energy in case-1 in MeV\",round(Kth1,3);\n",
"print\"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\n",
"The threshold kinetic energy in case-1 in MeV 5.381\n",
"The threshold kinetic energy in case-2 in MeV 16.104\n"
]
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
}
|
