{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# chapter 1: Atomic Nucleus"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.1;pg no:2"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.1, Page:2 \n",
" \n",
"\n",
"\n",
" The electric field in V/m is = 20000.0\n",
"\n",
" The force in N/C is = 20000.0\n",
"\n",
" The force on metal sphere in N is = 7.6e-05\n"
]
}
],
"source": [
"#cal of elelectric field and force\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.1, Page:2 \\n \\n\"\n",
"#Given:\n",
"v=1000# potential\n",
"d=0.05# distance\n",
"q=3.8*10**-9# charge\n",
"#solution:\n",
"e=v/d;#electric field\n",
"f=e;# force\n",
"f1=f*q;# force on metal sphere\n",
"print\"\\n The electric field in V/m is =\",e\n",
"print\"\\n The force in N/C is =\",f\n",
"print\"\\n The force on metal sphere in N is =\",f1"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.2;pg no:2"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.2, Page:2 \n",
" \n",
"\n",
"The potential in V is = 80.0\n"
]
}
],
"source": [
"#cal of potential\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.2, Page:2 \\n \\n\"\n",
"#Given:\n",
"energy=2*10**-6\n",
"c=2.5*10**-8# velocity of light\n",
"#solution:\n",
"v=energy/c# potential\n",
"print\"The potential in V is =\",v"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.3;pg no:3"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.3, Page:3 \n",
"\n",
"The wavelength in Angstroms is = 3.88\n",
"The photon wavelength in Angstroms is = 1242.38\n"
]
}
],
"source": [
"#cal of elecrtron and photon wavelength\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.3, Page:3 \\n\"\n",
"#Given:\n",
"energy=10 #in electron volts\n",
"m=9.1*10**-31# mass of electron in kg\n",
"h=6.626*10**-34# planck's constant J.s\n",
"c=3*10**8# speed of light in m/s\n",
"#solution (a):\n",
"energy1=energy*1.6*10**-19# energy in J\n",
"p=(2*m*energy1)**0.5# momentum\n",
"wavelength=h/p*(10)**10\n",
"print\"The wavelength in Angstroms is =\",round(wavelength,2)\n",
"#solution (b):\n",
"wavelength1=h*c/energy1*(10)**10;#photon wavelength\n",
"print\"The photon wavelength in Angstroms is =\",round(wavelength1,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.4;pg no:3"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.4, Page:3 \n",
" \n",
"\n",
"The energy in eV is = 150.77\n"
]
}
],
"source": [
"#cal of kinetic energy of an electron\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.4, Page:3 \\n \\n\"\n",
"#Given:\n",
"wavelength=10**-10\n",
"m=9.1*10**-31\n",
"h=6.626*10**-34\n",
"#solution:\n",
"p=h/wavelength\n",
"e=p*p/(2*m) # energy in J\n",
"e1=e/(1.6*10**-19)# energy in eV\n",
"print\"The energy in eV is =\",round(e1,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.5;pg no:3"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.5, Page:3 \n",
" \n",
"\n",
"The wavelength in 10^-5 Angstroms is = 0.66\n",
"The wavelength in 10^-5 Angstroms is = 0.65\n"
]
}
],
"source": [
"#cal of wavelength of oxygen and nitrogen nucleus\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.5, Page:3 \\n \\n\"\n",
"#Given:\n",
"m=1.66*10**-27# 1u=1.66*10^-27 kg\n",
"h=6.6262*10**-34#planck's constant in J.s\n",
"energy1=120# in Mev for oxygen\n",
"energy2=140# in MeV for nitrogen\n",
"#solution(a):\n",
"p=(2*m*16*energy1*(1.6022*10**-13))**0.5\n",
"wavelength1=h/p*(10)**15#wavelength in 10^-5 Angstroms\n",
"print\"The wavelength in 10^-5 Angstroms is =\",round(wavelength1,2)\n",
"#solution (b):\n",
"p=(2*m*14*energy2*(1.6022*10**-13))**0.5\n",
"wavelength2=h/p*(10)**15#wavelength in 10^-5 Angstroms\n",
"print\"The wavelength in 10^-5 Angstroms is =\",round(wavelength2,2)\n",
"# 1 Angstrom = 10^-10 m"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.6;pg no:3"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.6, Page:3 \n",
" \n",
"\n",
"The energy in eV is = 8275.0\n"
]
}
],
"source": [
"#cal of energy of a gamma photon\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.6, Page:3 \\n \\n\"\n",
"#Given:\n",
"wavelength=1.5*10**-10\n",
"h=6.62*10**-34\n",
"c=3*10**8\n",
"#solution:\n",
"e=(h*c)/wavelength# energy in J\n",
"e1=e/(1.6*10**-19)# energy in eV\n",
"print\"The energy in eV is =\",e1"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.7;pg no:4"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.7, Page:4 \n",
" \n",
"\n",
"\n",
" The threshold frequency in s^-1 is = 1.23634168427e+15\n",
"\n",
" The threshold wavelength in Angstroms is = 2426.51\n",
"\n",
" The energy of photoelectrone in eV is = 3.91\n"
]
}
],
"source": [
"#cal of threshold frequency,wavelength,energy of photoelectrone\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.7, Page:4 \\n \\n\"\n",
"#Given:\n",
"E=5.12*1.6*10**-19# energy in J\n",
"h=6.626*10**-34\n",
"c=3*10**8\n",
"wavelength=200*10**-9\n",
"w=2.3# in eV\n",
"#solution:\n",
"tf=E/h# (part a)\n",
"print\"\\n The threshold frequency in s^-1 is =\",round(tf,2)\n",
"tl=c/tf*10**10# (part b)\n",
"print\"\\n The threshold wavelength in Angstroms is =\",round(tl,2)\n",
"e=(h*c)/(wavelength*1.6*10**-19)# photon energy in eV (part c)\n",
"pe=e-w\n",
"print\"\\n The energy of photoelectrone in eV is =\",round(pe,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.8;pg no:4"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.8, Page:4 \n",
" \n",
"\n",
"\n",
" The velocity of alpha particles for 1 MeV in mega m/s is = 6.94\n",
"\n",
" The velocity of alpha particles for 2 MeV in mega m/s is = 9.82\n",
"\n",
" The velocity of deuteron particles for 1 MeV in mega m/s is = 9.82\n",
"\n",
" The velocity of deuteron particles for 2 MeV in mega m/s is = 13.88\n",
"\n",
" The velocity of proton particles for 1 MeV in mega m/s is = 13.88\n",
"\n",
" The velocity of proton particles for 2 MeV in mega m/s is = 19.63\n"
]
}
],
"source": [
"#cal of velocity of alpha particles,deuteron,proton\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.8, Page:4 \\n \\n\"\n",
"#Given:\n",
"e1=1 # in MeV\n",
"e2=2 # in MeV\n",
"ma=4 # in u(amu)\n",
"md=2 # in u(amu)\n",
"mp=1 # in u(amu)\n",
"# 1u = 1.6*10^-27 Kg\n",
"#solution: part a)For alpha particles\n",
"v1a=((2*e1*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n",
"print\"\\n The velocity of alpha particles for 1 MeV in mega m/s is =\",round(v1a/10**6,2)# For 1 MeV\n",
"v2a=((2*e2*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n",
"print\"\\n The velocity of alpha particles for 2 MeV in mega m/s is =\",round(v2a/10**6,2)# For 2 MeV\n",
"#solution: part b)For deuteron particles\n",
"v1b=((2*e1*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n",
"print\"\\n The velocity of deuteron particles for 1 MeV in mega m/s is =\",round(v1b/10**6,2) # For 1 MeV\n",
"v2b=((2*e2*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n",
"print\"\\n The velocity of deuteron particles for 2 MeV in mega m/s is =\",round(v2b/10**6,2) # For 2 MeV\n",
"#solution: part c)For proton particles\n",
"v1p=((2*e1*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n",
"print\"\\n The velocity of proton particles for 1 MeV in mega m/s is =\",round(v1p/10**6,2) # For 1 MeV\n",
"v2p=((2*e2*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n",
"print\"\\n The velocity of proton particles for 2 MeV in mega m/s is =\",round(v2p/10**6,2) # For 2 MeV"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.9;pg no:5"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.9, Page:5 \n",
" \n",
"\n",
"The energy in MeV is = 934.0\n"
]
}
],
"source": [
"#cal of The energy equivalence\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.9, Page:5 \\n \\n\"\n",
"#Given:\n",
"m=1./(6.023*10**23)#mass of 1 atom in g\n",
"m1=m*10**-3#mass of 1 atom in Kg\n",
"c=3.*10**8# velocity in m/s\n",
"#solution:\n",
"e=m1*c*c; # energy in J\n",
"e1=e/(1.6*10**-13)# energy in MeV\n",
"print\"The energy in MeV is =\",round(e1)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.10;pg no:5"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.10, Page:5 \n",
" \n",
"\n",
"The energy in eV is = 13.26\n"
]
}
],
"source": [
"#cal of The energy of formation\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.10, Page:5 \\n \\n\"\n",
"#Given:\n",
"enthalpy=1278 # enthalpy of combustion in kJ/mol\n",
"#solution:\n",
"energy=(enthalpy*1000)/(6.022*10**23*1.6*10**-19)\n",
"print\"The energy in eV is =\",round(energy,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.11;pg no:5"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.11, Page:5 \n",
" \n",
"\n",
"\n",
" The mean binding energy of helium atom in MeV is = 7.07\n",
"\n",
" The mean binding energy of oxygen atom in MeV is = 7.98\n"
]
}
],
"source": [
"#cal of mean binding energy of helium and oxygen\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.11, Page:5 \\n \\n\"\n",
"#Given:\n",
"mh=1.0078\n",
"mn=1.0087\n",
"ma=4.0026\n",
"mo=15.9949\n",
"Ah=4.0026 # atomic mass of helium\n",
"Ao=15.9949 # atomic mass of oxygen\n",
"#solution:\n",
"# part (a)\n",
"B1=(2*mh+2*mn-ma)*931 # in MeV\n",
"Bh=B1/Ah\n",
"print\"\\n The mean binding energy of helium atom in MeV is =\",round(Bh,2)\n",
"# part (b)\n",
"B2=(8*mh+8*mn-mo)*931 # in MeV\n",
"Bo=B2/Ao\n",
"print\"\\n The mean binding energy of oxygen atom in MeV is =\",round(Bo,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.12;pg no:6"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.12, Page:6 \n",
"\n",
"\n",
" The mean binding energy of Be atom in MeV is = 7.059\n",
"From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\n"
]
}
],
"source": [
"#cal of mean binding energy of Be atom\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.12, Page:6 \\n\"\n",
"#Given:\n",
"mh=1.0078;\n",
"mn=1.0087;\n",
"ABe=8.0053; # atomic mass of beryllium\n",
"#solution:\n",
"B1=(4*mh+4*mn-ABe)*931; # in MeV\n",
"Bh=B1/ABe;\n",
"print\"\\n The mean binding energy of Be atom in MeV is =\",round(Bh,3)\n",
"print\"From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.13;pg no:6"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.13, Page:6 \n",
"\n",
"The amount of coal required in Kg is = 2.5\n"
]
}
],
"source": [
"#cal of amount of coal\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.13, Page:6 \\n\"\n",
"#Given:\n",
"e=200; # in Mev\n",
"m=0.235; # weight of uranium atom in Kg\n",
"enthalpy=393.5; # in KJ/mol\n",
"Na=6.02*10**23;\n",
"#solution:\n",
"e1=e*1.6*10**-19*10**6;\n",
"atoms=Na/m;\n",
"e2=atoms*e1;#energy released in J\n",
"m1=(e2*12)/(393.5*1000*1000);# in Kg\n",
"m2=m1/1000;# in tons\n",
"print\"The amount of coal required in Kg is =\", round(m2/1000,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.14;pg no:7"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.14, Page:7 \n",
" \n",
"\n",
"The energy release in part (a) in eV/molecule is = 2.51\n",
"The energy release in part (b) in eV/molecule is = 9.23\n"
]
}
],
"source": [
"#cal of The energy releases\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.14, Page:7 \\n \\n\"\n",
"#Given:\n",
"H1=241.8; # in KJ/mol\n",
"H2=887.2; # in KJ/mol\n",
"# 1 KJ/mol = 0.0104 eV/atom\n",
"#solution: part (a)\n",
"e1=H1*0.0104;\n",
"print\"The energy release in part (a) in eV/molecule is =\",round(e1,2)\n",
"#solution: part (b)\n",
"e2=H2*0.0104;\n",
"print\"The energy release in part (b) in eV/molecule is =\",round(e2,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.15;pg no:7"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.15, Page:7 \n",
"\n",
"The energy release in part (a) in KJ/mol of carbondioxide is = 394.9\n",
"The energy release in part (b) in KJ/mol of alumina is = 1676.0\n",
"The energy release in part (c) in MJ/atom of U(235) is = 19.264\n"
]
}
],
"source": [
"#cal of The energy releases\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.15, Page:7 \\n\"\n",
"#Given:\n",
"H1=4.1; # in eV/molecule\n",
"H2=17.4; # in eV/molecule\n",
"H3=200;# in MeV/atom of U\n",
"# 1 eV/atom = 96.32 KJ/mol\n",
"#solution: part (a)\n",
"e1=H1*96.32;\n",
"print\"The energy release in part (a) in KJ/mol of carbondioxide is =\",round(e1,1)\n",
"#solution: part (b)\n",
"e2=H2*96.32;\n",
"print\"The energy release in part (b) in KJ/mol of alumina is =\",round(e2,1)\n",
"#solution: part (c)\n",
"e3=H3*1000*96.32;# in MJ/atom of U(235)\n",
"print\"The energy release in part (c) in MJ/atom of U(235) is =\",round(e3/10**6,3)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.16;pg no:7"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.16, Page:7 \n",
" \n",
"\n",
"\n",
" The rate of energy release in MW is= 949.25\n"
]
}
],
"source": [
"#cal of The rate of energy release\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.16, Page:7 \\n \\n\"\n",
"#Given:\n",
"e=200.; #MeV/ atom of U\n",
"# 1 eV = 1.6*10^-19 J\n",
"Na=6.023*10**23;\n",
"M=0.235; # mass in Kg\n",
"#solution:\n",
"e1=e*1.6*10**-19*10**6;\n",
"A=Na/M;\n",
"e2=A*e1; # energy released in MJ/day\n",
"e3=e2/(24.*3600.);\n",
"print\"\\n The rate of energy release in MW is=\",round(e3/10**6,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.17;pg no:8"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.17, Page:8 \n",
" \n",
"\n",
"\n",
" The mass loss in 10^-27 Kg/He formed is = 0.0464\n"
]
}
],
"source": [
"#cal of The mass loss\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.17, Page:8 \\n \\n\"\n",
"#Given:\n",
"e=26.03; # in MeV\n",
"#solution:\n",
"loss=e/931; #in atomic mass units (u)\n",
"# 1 u = 1.66*10^-27 Kg\n",
"m=(loss*1.66*10**-27)/(1*10**-27);\n",
"print\"\\n The mass loss in 10^-27 Kg/He formed is =\",round(m,4)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.18;pg no:8"
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.18, Page:8 \n",
"\n",
"\n",
" The energy loss in MeV is = -4.0312\n"
]
}
],
"source": [
"#cal of The energy loss\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.18, Page:8 \\n\"\n",
"#Given:\n",
"mh=1.007825;\n",
"mt=3.016049;\n",
"md=2.014102;\n",
"#solution:\n",
"m1=(mh+mt-2*md);\n",
"e=(-m1)*931; # in MeV\n",
"print\"\\n The energy loss in MeV is =\",round(-e,4)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.19;pg no:8"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.19, Page:8 \n",
"\n",
"The mean binding energy of tritium atom in MeV is = 2.811\n",
"The mean binding energy of nickel atom in MeV is = 8.716\n"
]
}
],
"source": [
"#cal of mean binding energy of tritium and nickel atom\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.19, Page:8 \\n\"\n",
"#Given:\n",
"mh=1.007825;\n",
"mn=1.008665;\n",
"mt=3.016049; # atomic mass of Tritium\n",
"mNi=59.93528; # atomic mass of Nickel\n",
"#solution:\n",
"# part (a)\n",
"B1=(1*mh+2*mn-mt)*931; # in MeV\n",
"Bh=B1/mt;\n",
"print\"The mean binding energy of tritium atom in MeV is =\",round(Bh,3)\n",
"# part (b)\n",
"B2=(28*mh+32*mn-mNi)*931; # in MeV\n",
"Bo=B2/mNi;\n",
"print\"The mean binding energy of nickel atom in MeV is =\",round(Bo,3)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.20;pg no:9"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.20, Page:9 \n",
"\n",
"The mean binding energy of Cl (35) atom in MeV is = 8.5281\n",
"The mean binding energy of Cl (37) atom in MeV is = 8.5784\n",
"The increase in mean binding energy of Cl atom in MeV is = 0.05\n"
]
}
],
"source": [
"#cal of mean binding energy of Cl\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.20, Page:9 \\n\"\n",
"#Given:\n",
"mh=1.00783;\n",
"mn=1.00867;\n",
"m35=34.96885; # atomic mass of Cl (35)\n",
"m37=36.96590; # atomic mass of Cl (37)\n",
"#solution:\n",
"B1=(17*mh+18*mn-m35)*931; # in MeV\n",
"Bh=B1/m35;\n",
"print\"The mean binding energy of Cl (35) atom in MeV is =\",round(Bh,4)\n",
"B2=(17*mh+20*mn-m37)*931; # in MeV\n",
"Bo=B2/m37;\n",
"print\"The mean binding energy of Cl (37) atom in MeV is =\",round(Bo,4)\n",
"Bi=Bo-Bh;\n",
"print\"The increase in mean binding energy of Cl atom in MeV is =\",round(Bi,2)\n",
"# NOTE: The answer depends upon how much precise value you take for atomic masses."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##example 1.21;pg no:9"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 1.21, Page:9 \n",
" \n",
"\n",
"\n",
" The mean binding energy of Na(22) in MeV is = 7.9236\n",
"\n",
" The mean binding energy of Na(23)in MeV is = 8.1154\n",
"\n",
" The mean binding energy of Na(24) in MeV is = 8.0717\n"
]
}
],
"source": [
"#cal of mean binding energy of Na\n",
"#intiation of all variables\n",
"# Chapter 1\n",
"print\"Example 1.21, Page:9 \\n \\n\"\n",
"#Given:\n",
"mh=1.0078;\n",
"mn=1.0087;\n",
"m22=21.99431;# atomic mass of Na 22\n",
"m23=22.9898;# atomic mass of Na 23\n",
"m24=23.9909;# atomic mass of Na 24\n",
"#solution:\n",
"# part (a)\n",
"B1=((11*mh+11*mn)-m22)*931; # in MeV\n",
"Bh=B1/m22;\n",
"print\"\\n The mean binding energy of Na(22) in MeV is =\",round(Bh,4)\n",
"# part (b)\n",
"B2=((11*mh+12*mn)-m23)*931; # in MeV\n",
"Bo=B2/m23;\n",
"print\"\\n The mean binding energy of Na(23)in MeV is =\",round(Bo,4)\n",
"# part (c)\n",
"B3=((11*mh+13*mn)-m24)*931; # in MeV\n",
"Bs=B3/m24;\n",
"print\"\\n The mean binding energy of Na(24) in MeV is =\",round(Bs,4)"
]
},
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"cell_type": "code",
"execution_count": null,
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"collapsed": true
},
"outputs": [],
"source": []
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