{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 4:Atomic Structure"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:4.1,Page no:125"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration \n",
"E= -13.6; #Energy required to separate electron and proton, eV\n",
"e= 1.6*(10**(-19)); #charge of an electron, C\n",
"E= E*e; #converting to J\n",
"Po= 8.85*(10**(-12)); #Permittivity of free space, F/m\n",
"\n",
"#Calculation\n",
"import math\n",
"r= e**2/(8*(math.pi)*Po*E); #radius, m\n",
"r= -r;\n",
"m= 9.1*(10**(-31)); #mass of electron, kg\n",
"v=e/math.sqrt(4*(math.pi)*Po*m*r); #velocity, m/s\n",
"\n",
"#Result\n",
"print\"The orbital radius of the electron is:%.2g\"%r,\"m\"\n",
"print\"The velocity of electron is:%.2g\"%v,\"m/s\"\n",
" \n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The orbital radius of the electron is:5.3e-11 m\n",
"The velocity of electron is:2.2e+06 m/s\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:4.2,Page no:135"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration \n",
"n1=1.0; #initial state\n",
"n2=3.0; #final state\n",
"E= -13.6; #energy in ground state, eV\n",
"\n",
"#Calculation\n",
"dE= E*((1/n2**2)-(1/n1**2)); #Change in energy, eV\n",
"\n",
"#Result\n",
"print\"The energy change of Hydrogen atom is: \",round(dE,1),\"eV\"\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The energy change of Hydrogen atom is: 12.1 eV\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:4.3,Page no:135"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"#Part(a)\n",
"Rn= 10.0**(-5); #radius of Rydberg atom, m\n",
"Ao= 5.29*(10**(-11)); #Bohr radius, m\n",
"\n",
"#Calculation\n",
"n= math.sqrt(Rn/Ao); #Quantum number\n",
"E1= -13.6; #Ground state energy level, eV\n",
"En= E1/n**2.0; #Nth state energy level, eV\n",
"\n",
"#Result\n",
"print\"(a).The quantum number of the Rydberg atom is: \",round(n)\n",
"print\"(b).The energy ofthe rydberg atom is:%.3g\"%En,\"eV\"\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a).The quantum number of the Rydberg atom is: 435.0\n",
"(b).The energy ofthe rydberg atom is:-7.19e-05 eV\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:4.4,Page no:138"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"n1= 3.0; #initial state\n",
"n2= 2.0; #final state\n",
"R= 1.097*(10**7); #Rydberg's constant, m**(-1)\n",
"\n",
"#Calculation\n",
"k= (1/n2**2)-(1/n1**2);\n",
"l= 1/(k*R); #longest wavelength, m\n",
"l= l*(10**9); #converting to nm\n",
"\n",
"#Result\n",
"print\"The longest in Balmer series of Hydrogen, in nm, is: \",round(l),\"nm\"\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The longest in Balmer series of Hydrogen, in nm, is: 656.0 nm\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:4.5,Page no:139"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"n1=1.0; #initial state\n",
"n2=2.0; #final state\n",
"E1= 2.18*(10**(-18)); #Rydberg's constant, J\n",
"h= 6.63*(10**(-34)); #Planck's constant, J.s\n",
"\n",
"#Calculation\n",
"f1= (E1/h)*(2.0/n1**3); #Frequency for first orbit, rev/s\n",
"f2= (E1/h)*(2.0/n2**3); #Frequency for second orbit, rev/s\n",
"print\"Ans (A):Frequency of revolution for orbit n=1 is,f1: %.3g\"%f1,\"rev/s\"\n",
"print\"Frequency of revolution for orbit n=2 is,f2:%.2e\"%f2,\"rev/s\"\n",
"print\"which is equivalent to 0.823*10**15 rev/s 'without' any scientific notation\\n\"\n",
"#Part (b)\n",
"n1=2.0; #initial orbit\n",
"n2=1.0; #final orbit\n",
"f= (E1/(h))*((1.0/(n2**2))-(1.0/n1**3)); #frequency, Hz\n",
"print\"Ans(B):Frequency of emitted photon is: %.3g\"%f,\"Hz\\n\"\n",
"#Part (c)\n",
"n= 2.0; #orbit\n",
"f= f2; #from part (a)\n",
"dt= 10.0**(-8); # time duration, s\n",
"N= f*dt; #Number of revolutions\n",
"#Result\n",
"print\"Ans(C):Number of revolutions the electron makes is:%.3g\"%N\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Ans (A):Frequency of revolution for orbit n=1 is,f1: 6.58e+15 rev/s\n",
"Frequency of revolution for orbit n=2 is,f2:8.22e+14 rev/s\n",
"which is equivalent to 0.823*10**15 rev/s 'without' any scientific notation\n",
"\n",
"Ans(B):Frequency of emitted photon is: 2.88e+15 Hz\n",
"\n",
"Ans(C):Number of revolutions the electron makes is:8.22e+06\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:4.7,Page no:142"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"#Part (a) \n",
"Me= 9.1*(10**(-31)); #mass of electron, kg\n",
"m= 207*Me; #mass of muon, kg\n",
"\n",
"#Calculation\n",
"Mp= 1836*Me; #mass of proton, kg\n",
"Mreduced= (m*Mp)/(m+Mp); #reduced mass, kg\n",
"Ao= 5.29*(10**(-11)); #Bohr's orbit for n=1, m\n",
"r1= Ao; #expected orbit for atom, m\n",
"r2= (Me/Mreduced)*r1; #reduced radius of orbit, m\n",
"#Part (b)\n",
"E=-13.6; # energy for elctron in n=1, eV\n",
"Ereduced= (Mreduced/Me)*E; #energy for eectron in mounic atom, eV\n",
"Ereduced= Ereduced/(10**3);#converting to keV\n",
"\n",
"#Result\n",
"print\"(A)Radius of the mounic atom formed, in m, is:%.3g\"%r2,\"m\"\n",
"print\"(B)Ionisation energy for the muonic atom is: \",round(Ereduced,2),\"keV\"\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(A)Radius of the mounic atom formed, in m, is:2.84e-13 m\n",
"(B)Ionisation energy for the muonic atom is: -2.53 keV\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:4.8,Page no:156"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration \n",
"I= 7.7; #Intensity of beam, MeV\n",
"Dgold= 1.93*(10**4); #density of gold foil used, kg/m**3\n",
"u= 1.66*(10**(-27)); #atomic mass unit, kg\n",
"Mgold= 197*u; #atomic mass of gold, per atom\n",
"\n",
"#Calculation\n",
"n= Dgold/Mgold; #number of atoms per unit volume, atoms/m**3\n",
"Zgold= 79; #atomic number of gold\n",
"e= 1.6*(10**(-19)); #electronis charge, C\n",
"KE= (I*e)/(10**(-6)); #converting to J\n",
"angle= 45; #degree\n",
"p=1/math.tan(math.radians(angle/2));\n",
"Po= 8.85*(10**(-12)); #Permittivity of free space, F/m\n",
"t= 3*(10**(-7)); #thickness of foil, m\n",
"f= (math.pi)*n*t*(((Zgold*(e**2))/(4*(math.pi)*Po*KE))**2)*(p**2) #using Rutherford scattering formula\n",
"\n",
"#Result\n",
"print\"f=%.g\"%f\n",
"print\"Fraction of the beam scattered through 45 degree or more is: \",round(f*100,3),\"%\"\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"f=7e-05\n",
"Fraction of the beam scattered through 45 degree or more is: 0.007 %\n"
]
}
],
"prompt_number": 8
}
],
"metadata": {}
}
]
}