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{
"metadata": {
"name": "",
"signature": "sha256:e2f7c7551d6e417829aadd6f90608da5e4f835298a7f6245d9b1f454185575b2"
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"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter8-Nuclear Forces"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3-pg349"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa8.3 : : Page-349 (2011)\n",
"#find The probability that the proton moves within the range of neutron\n",
"import math\n",
"b = 1.9e-15; ## Width of square well potential, metre\n",
"h_kt = 1.054571e-034; ## Reduced planck's constant, joule sec\n",
"c = 3e+08; ## Velocity of light, metre per sec\n",
"m_n = 1.67e-27; ## Mass of a nucleon , Kg\n",
"V_0 = 40*1.6e-13; ## Depth, metre\n",
"E_B = (V_0-(1/(m_n*c**2)*(math.pi*h_kt*c/(2*b))**2))/1.6e-13; ## Binding energy, mega electron volts\n",
"alpha = math.sqrt(m_n*c**2*E_B*1.6e-13)/(h_kt*c); ## scattering co efficient, per metre\n",
"P = (1+1/(alpha*b))**-1.; ## Probability\n",
"R_mean = math.sqrt (b**2./2.*(1./3.+4./math.pi**2.+2.5)); ## Mean square radius, metre\n",
"print'%s %.2f %s %.2e %s'%(\"\\nThe probability that the proton moves within the range of neutron = \",P,\" \\n\" \"The mean square radius of the deuteron = \",R_mean,\" metre\")\n",
"\n",
"\n",
"## Result\n",
"## The probability that the proton moves within the range of neutron = 0.50 \n",
"## The mean square radius of the deuteron = 2.42e-015 metre \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The probability that the proton moves within the range of neutron = 0.50 \n",
"The mean square radius of the deuteron = 2.42e-15 metre\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5-pg349"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa8.5 : : Page-349 (2011)\n",
"#find The total cross section for n-p scattering \n",
"import math\n",
"a_t = 5.38e-15;\n",
"a_s = -23.7e-15;\n",
"r_ot = 1.70e-15;\n",
"r_os = 2.40e-15;\n",
"m = 1.6748e-27;\n",
"E = 1.6e-13;\n",
"h_cut = 1.0549e-34;\n",
"K_sqr = m*E/h_cut**2;\n",
"sigma = 1/4.*(3.*4*math.pi*a_t**2./(a_t**2.*K_sqr+(1.-1/2.*K_sqr*a_t*r_ot)**2)+4*math.pi*a_s**2/(a_s**2*K_sqr+(1-1./2.*K_sqr*a_s*r_os)**2))*1e+028; ## Total cross-section for n-p scattering, barn\n",
"print'%s %.2f %s'%(\"\\nThe total cross section for n-p scattering = \",sigma,\" barn\");\n",
"\n",
"## Result\n",
"## The total cross section for n-p scattering = 2.911 barn \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The total cross section for n-p scattering = 2.91 barn\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8-pg351"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa8.8 : : Page-351 (2011)\n",
"#find The possible angular momentum states with their parities are as follows\n",
"import math\n",
"S = 1.; ## Spin angular momentum(s1+-s2), whereas s1 is the spin of proton and s2 is the spin of neutron.\n",
"m = 2.*S+1.; ## Spin multiplicity\n",
"j = 1.; ## Total angular momentum\n",
"print(\"\\nThe possible angular momentum states with their parities are as follows : \");\n",
"print'%s %.2f %s %.2f %s '%(\"\\n \",m, \" \" and \"S has even parity \",j,\"\");\n",
"print'%s %.2f %s %.2f %s '%(\"\\n \",m,\" \" and \"P has odd parity \", j,\"\");\n",
"print'%s %.2f %s %.2f %s'%(\"\\n \",m, \" \" and \"S has odd parity \",j,\"\"); \n",
"S = 0.;\n",
"m = 2.*S+1.\n",
"print(m)\n",
"print'%s %.2f %s %.2f %s '%(\"\\n \",m,\" \" and \"P has odd parity \", j,\"\");\n",
" \n",
"## Result \n",
"## The possible angular momentum states with their parities are as follows : \n",
"## 3S1 has even parity \n",
"## 3P1 has odd parity \n",
"## 3D1 has even parity\n",
"## 1P1 has odd parity "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The possible angular momentum states with their parities are as follows : \n",
"\n",
" 3.00 S has even parity 1.00 \n",
"\n",
" 3.00 P has odd parity 1.00 \n",
"\n",
" 3.00 S has odd parity 1.00 \n",
"1.0\n",
"\n",
" 1.00 P has odd parity 1.00 \n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex9-pg351"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Exa8.9 : : Page-351 (2011)\n",
"#find The possible states are\n",
"\n",
"print(\"\\nThe possible states are : \");\n",
"#For s = 0\n",
"s = 0; # Spin angular momentum\n",
"m = 2*s+1; # Spin multiplicity\n",
"for j in range(0,3): # Total angular momentum\n",
" l = j\n",
" if l == 0:\n",
" print\"%s %.1f %s %.d %s \"%(\"\",j,\"\"and \"S\",m,\"\")\n",
" elif l == 2:\n",
" print\"%s %.1f %s %.d %s \"%(\"\",j,\"\"and \"D\",m,\"\") \n",
" \n",
"\n",
"#For s = 1\n",
"s = 1;\n",
"m = 2*s+1;\n",
"l = 2\n",
"for j in range(0,3): \n",
" if j == 0:\n",
" print\"%s %.1f %s %.d %s \"%(\"\",j,\"\"and \"P\",m,\"\")\n",
" elif j ==1:\n",
" print\"%s %.d %s %.d %s \"%(\"\",j,\"\"and \"P\",m,\"\")\n",
" elif j ==2:\n",
" print\"%s %.d %s %.d %s \"%(\"\",j,\"\"and \"P\",m,\"\")\n",
" \n",
"\n",
"for j in range(2,3):\n",
" print\"%s %.d %s %.d %s \"%(\"\",j,\"\" and \"F\",m,\"\")\n",
"\n",
"\n",
"#Result\n",
"#Possible states are : \n",
"# The possible states are : \n",
"# 0S1, 2D1, 0P3, 1P3, 2P3 and 2F3 \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The possible states are : \n",
" 0.0 1 \n",
" 2.0 1 \n",
" 0.0 3 \n",
" 1 3 \n",
" 2 3 \n",
" 2 3 \n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex10-pg352"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa8.10 : : Page-352 (2011)\n",
"#find The kinetic energy of each nucleon and The total kinetic energy\n",
"import math\n",
"r = 2e-015; ## Range of nuclear force, metre\n",
"h_kt = 1.0546e-34; ## Reduced value of Planck's constant, joule sec\n",
"m = 1.674e-27; ## Mass of each nucleon, Kg\n",
"K = round (2*h_kt**2./(2*m*r**2*1.6023e-13)); ## Kinetic energy of each nucleon in centre of mass frame, mega electron volts\n",
"K_t = 2.*K; ## Total kinetic energy, mega electron volts\n",
"K_inc = 2.*K_t; ## Kinetic energy of the incident nucleon, mega electron volts\n",
"print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nThe kinetic energy of each nucleon = \",K,\" MeV\" and \"The total kinetic energy =\",K_t,\" MeV\"and \"The kinetic energy of the incident nucleon =\",K_inc,\" MeV\")\n",
"\n",
"\n",
"## Result\n",
"## "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The kinetic energy of each nucleon = 10.00 The total kinetic energy = 20.00 The kinetic energy of the incident nucleon = 40.00 MeV \n"
]
}
],
"prompt_number": 4
}
],
"metadata": {}
}
]
}
|
