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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 11:Nuclear Structure"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:11.1,Page no:393"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration \n",
"u= 1.66*(10**(-27)) #atomic mass unit, kg\n",
"Mc= 12*u # atomic mass of Carbon-12, kg\n",
"R= 2.7 #radius of nucleus, fm\n",
"\n",
"#Calculation\n",
"import math\n",
"R=R*(10**(-15)) #converting to m\n",
"density= Mc/((4.0/3.0)*(math.pi)*(R**3)) # kg/m**3\n",
"\n",
"#Calculation\n",
"print\"Density of Carbon 12 nucleus is:%.2g\"%density,\"kg/m**3\"\n",
" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Density of Carbon 12 nucleus is:2.4e+17 kg/m**3\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:11.2,Page no:393"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"r= 2.4 #distance between centre of the protons, fm\n",
"r= r*(10**(-15)) #converting to m\n",
"e= 1.6*(10**(-19)) #charge of an electron, C\n",
"Po= 8.85*(10**(-12)) #Permittivity of free space, F/m\n",
"\n",
"#Calculation\n",
"K=1/(4*(math.pi)*Po) #constant, N.m**2/C**2\n",
"F= K*(e**2)/(r**2) #N\n",
"\n",
"#Result\n",
"print\"The repulsive force is: \",round(F),\"N\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The repulsive force is: 40.0 N\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:11.3,Page no:395"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"B= 1.0 #strength of magnetic field, T\n",
"Mneutron= 3.152*(10**(-8)) #Magnetic moment for neutron, eV/T\n",
"\n",
"#Calculation\n",
"#Part (a)\n",
"Mproton= 2.793*Mneutron #Magnetic moment for proton, eV/T\n",
"dE= 2*Mproton*B #eV\n",
"#Part (b)\n",
"h= 4.13*(10**(-15)) #Planck's constant, eV.s\n",
"Flarmor= dE/h #Hz\n",
"Flarmor= Flarmor/(10**6) #converting to MHz\n",
"\n",
"#Result\n",
"print\"(a).The energy difference is:%.4g\"%dE,\"eV\"\n",
"print\"(b).The Larmor frequency for a proton in the field is:\",round(Flarmor,1),\"MHz(APPROX)\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a).The energy difference is:1.761e-07 eV\n",
"(b).The Larmor frequency for a proton in the field is: 42.6 MHz(APPROX)\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:11.4,Page no:401"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration \n",
"Ebinding= 160.647 #binding nergy, MeV\n",
"Mh= 1.007825 #Mass of H1 atom, u\n",
"Mn= 1.008665 #Mass of neutron, u\n",
"Z=10 #number of protons\n",
"N=10 #number of neutrons\n",
"\n",
"#Calculation\n",
"Mneon= ((Z*Mh)+(N*Mn))-(Ebinding/931.49) #using Eqn 11.7\n",
"\n",
"#Result\n",
"print\"The atomic mass of Neon 10 isotope is: \",round(Mneon,3),\"u\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The atomic mass of Neon 10 isotope is: 19.992 u\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:11.5,Page no:402"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"\n",
"#Part(a)\n",
"Ca_42=41.958622 #[u] Mass of 42Ca20 \n",
"\n",
"#from table\n",
"M_neutron=1.008665 #[u] Mass of a free neutron\n",
"Ca_41=40.962278 #[u] Mass of 41Ca20 after removal of 1 neutron\n",
"#Part(b):\n",
"#from table\n",
"M_proton=1.007276466812 #[u] Mass of a free proton\n",
"K_19=40.96237 #[u] Mass of Potasium Isotope 41K19\n",
"\n",
"\n",
"#Calculation\n",
"#Part (a):\n",
"n_total=Ca_41+M_neutron\n",
"p_total=K_19+M_proton\n",
"neutron_BE=931.49*(n_total-Ca_42)#Binding energy in MeV\n",
"proton_BE=931.49*(p_total-Ca_42)\n",
"print \"(a).Binding energy of missing neutron is: \",round(neutron_BE,2),\"MeV\"\n",
"print\"(b).Binding energy for missing proton=\",round(proton_BE,2),\"MeV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a).Binding energy of missing neutron is: 11.48 MeV\n",
"(b).Binding energy for missing proton= 10.27 MeV\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:11.6,Page no:407"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration \n",
"Z= 30.0 #proton number\n",
"N=34.0 #Neutron number\n",
"\n",
"#Calculation\n",
"#Using Eqn 11.7\n",
"Mh= 1.007825 #Mass of H1 atom, u\n",
"Mn= 1.008665 #Mass of neutron, u\n",
"Mzinc= 63.929 #atomic mass of zinc, u\n",
"Ebinding= ((Z*Mh)+(N*Mn)-Mzinc)*931.49 #MeV\n",
"#Using semiempirical formula, Eqn 11.18, Page 407\n",
"a1= 14.1 #Mev\n",
"a2= 13.0 #MeV\n",
"a3= 0.595 #Mev\n",
"a4= 19.0 #MeV\n",
"a5= 33.5 #MeV\n",
"A= Z+N \n",
"E2= ((a1*A)-(a2*(A**(2.0/3.0)))-(a3*Z*(Z-1)/(A**(1.0/3.0)))-(a4*((A-2*Z)**2)/A)+(a5/(A**(3.0/4.0)))) #MeV\n",
"\n",
"#Result\n",
"print\"Binding energy of Zinc 64 isotope is: \",round(Ebinding,1),\"MeV\"\n",
"print\"The binding energy using semi-empirical formula, in MeV, is: \",round(E2,1),\"MeV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Binding energy of Zinc 64 isotope is: 559.2 MeV\n",
"The binding energy using semi-empirical formula, in MeV, is: 561.7 MeV\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:11.7,Page no:408"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"A=25\n",
"\n",
"#Calculation\n",
"#Derivation part\n",
"#dEb/dZ=-a3*(2*z-1)/A^(1/3)+4*a4*(A-2*Z)/A\n",
"#Z=a3*A**-1/3+4*a4/(2*a3*A^-1/3+8*a4*A**-1)\n",
"Z=(0.595*A**(-1.0/3.0)+76)/(1.19*A**(-1.0/3.0)+(152*A**-1))\n",
"print \"For A=25,Z=\",round(Z,1),\"\\nfor which we conclude that Z=\",round(Z),\"should be the atomic no. of most stable isobar of A=25\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"For A=25,Z= 11.7 \n",
"for which we conclude that Z= 12.0 should be the atomic no. of most stable isobar of A=25\n"
]
}
],
"prompt_number": 8
}
],
"metadata": {}
}
]
}
|
