{
"metadata": {
"name": "",
"signature": "sha256:18ac31f959977ef2080ed3a1b1a6990ce93e604dcfb0f72ab45c0c28a2428e0e"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Quantum Mechanics and Quantum Computing"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.1, Page number 41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration\n",
"c=3*10**8 #velocity of light in m/s\n",
"h=6.626*10**-34 #planks constant \n",
"m=1.67*10**-27 #mass of proton\n",
"\n",
"#Calculation\n",
"v=c/10 #velocity of proton\n",
"lamda=h/(m*v) #de Broglie wave length\n",
"\n",
"#Result\n",
"print(\"the de Broglie wavelength in m is \",lamda);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('the de Broglie wavelength in m is ', 1.3225548902195607e-14)\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.2, Page number 42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#importing modules\n",
"import math\n",
"\n",
"#Variable declaration\n",
"V=400; #potential in Volts\n",
"\n",
"#Calculation\n",
"lamda=12.56/math.sqrt(V); #de Broglie wavelength\n",
"\n",
"#Result\n",
"print(\"The de Broglie wavelength in Armstrong is\",lamda);\n",
"\n",
"#answer given in the book is wrong"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The de Broglie wavelength in Armstrong is', 0.628)\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.3, Page number 42\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#importing modules\n",
"import math\n",
"\n",
"#Variable declaration\n",
"m=1.674*10**(-27); #mass of neutron in kg\n",
"h=6.626*10**(-34);\n",
"E=0.025; #kinetic energy in eV\n",
"\n",
"#Calculation\n",
"Ej=E*1.6*10**-19; #kinetic energy in J\n",
"lamda=h/math.sqrt(2*m*Ej); #de Broglie wavelength\n",
"lamdaA=lamda*10**10; #converting wavelength from m to Armstrong\n",
"lamdaA=math.ceil(lamdaA*10**3)/10**3; #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print(\"The de Broglie wavelength in metres is\",lamda);\n",
"print(\"The de Broglie wavelength in Armstrong is\",lamdaA);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The de Broglie wavelength in metres is', 1.81062582829353e-10)\n",
"('The de Broglie wavelength in Armstrong is', 1.811)\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.4, Page number 42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#importing modules\n",
"import math\n",
"\n",
"#Variable declaration\n",
"V=1600; #potential in Volts\n",
"\n",
"#Calculation\n",
"lamda=12.56/math.sqrt(V); #de Broglie wavelength\n",
"lamda=math.ceil(lamda*10**2)/10**2; #rounding off to 2 decimals\n",
"\n",
"#Result\n",
"print(\"The de Broglie wavelength in Armstrong is\",lamda);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The de Broglie wavelength in Armstrong is', 0.32)\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.5, Page number 42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#importing modules\n",
"import math\n",
"\n",
"#Variable declaration\n",
"deltax=0.2; #distance in armstrong\n",
"h=6.626*10**(-34);\n",
"\n",
"#Calculation\n",
"delta_xm=deltax*10**-10; #distance in m\n",
"delta_p=h/(2*math.pi*delta_xm);\n",
"\n",
"#Result\n",
"print(\"The uncertainity in momentum of electron in kg m/sec is\",delta_p);"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The uncertainity in momentum of electron in kg m/sec is', 5.2728032646344916e-24)\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.6, Page number 43"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#importing modules\n",
"import math\n",
"\n",
"#Variable declaration\n",
"n1=1;\n",
"n2=1;\n",
"n3=1; #values in lowest energy\n",
"h=6.62*10**(-34);\n",
"M=9.1*10**-31; #mass in kg\n",
"L=0.1; #side in nm\n",
"\n",
"#Calculation\n",
"L=L*10**-9; #side in m\n",
"n=(n1**2)+(n2**2)+(n3**2);\n",
"E1=(n*h**2)/(8*M*L**2); #energy in j\n",
"E1eV=E1/(1.6*10**-19); #energy in eV\n",
"E1eV=math.ceil(E1eV*10)/10; #rounding off to 1 decimals\n",
"\n",
"#Result\n",
"print(\"lowest energy of electron in Joule is\",E1);\n",
"print(\"lowest energy of electron is eV\",E1eV);\n",
"\n",
"#answer for lowest energy in eV given in the book is wrong"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('lowest energy of electron in Joule is', 1.8059505494505486e-17)\n",
"('lowest energy of electron is eV', 112.9)\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.7, Page number 43"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#importing modules\n",
"import math\n",
"\n",
"#Variable declaration\n",
"M=9.1*10**(-31); #mass of electron in kg\n",
"h=6.66*10**(-34);\n",
"E=2000; #kinetic energy in eV\n",
"\n",
"#Calculation\n",
"Ej=E*1.6*10**-19; #kinetic energy in J\n",
"lamda=h/math.sqrt(2*M*Ej); #de Broglie wavelength\n",
"lamdaA=lamda*10**9; #converting wavelength from m to nm\n",
"lamdaA=math.ceil(lamdaA*10**3)/10**3; #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print(\"The de Broglie wavelength in nm is\",lamdaA);"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The de Broglie wavelength in nm is', 0.028)\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.8, Page number 43"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#importing modules\n",
"import math\n",
"\n",
"#Variable declaration\n",
"n=1; #for minimum energy\n",
"h=6.626*10**(-34);\n",
"m=9.1*10**-31; #mass in kg\n",
"L=4*10**-10; #size in m\n",
"\n",
"#Calculation\n",
"E1=(n*h**2)/(8*m*L**2); #energy in j\n",
"\n",
"#Result\n",
"print(\"lowest energy of electron in Joule is\",E1);\n",
"\n",
"#answer given in the book is wrong"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('lowest energy of electron in Joule is', 3.7692201236263733e-19)\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.9, Page number 44"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"h=6.626*10**(-34);\n",
"m=9.1*10**-31; #mass in kg\n",
"lamda=1.66*10**-10; #wavelength in m\n",
"\n",
"#Calculation\n",
"v=h/(m*lamda); #velocity in m/sec\n",
"v_km=v*10**-3; #velocity in km/sec\n",
"E=(1/2)*m*v**2; #kinetic energy in joule\n",
"EeV=E/(1.6*10**-19); #energy in eV\n",
"EeV=math.ceil(EeV*10**3)/10**3; #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print(\"velocity of electron in m/sec is\",round(v));\n",
"print(\"velocity of electron in km/sec is\",round(v_km));\n",
"print(\"kinetic energy of electron in Joule is\",E);\n",
"print(\"kinetic energy of electron in eV is\",EeV);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('velocity of electron in m/sec is', 4386337.0)\n",
"('velocity of electron in km/sec is', 4386.0)\n",
"('kinetic energy of electron in Joule is', 8.754176510091736e-18)\n",
"('kinetic energy of electron in eV is', 54.714)\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.10, Page number 44"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#importing modules\n",
"import math\n",
"\n",
"#Variable decleration\n",
"V=15; #potential in kV\n",
"\n",
"#Calculation\n",
"v=V*10**3; #potential in V\n",
"lamda=12.26/math.sqrt(v); #de Broglie wavelength\n",
"lamda=math.ceil(lamda*10**2)/10**2 #rounding off to 2 decimals\n",
"\n",
"#result\n",
"print(\"The de Broglie wavelength in Armstrong is\",lamda);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The de Broglie wavelength in Armstrong is', 0.11)\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.11, Page number 44\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#importing modules\n",
"import math\n",
"\n",
"#Calculation\n",
"m=1.675*10**-27; #mass of neutron in kg\n",
"h=6.626*10**-34;\n",
"E=10; #kinetic energy in keV\n",
"\n",
"#Calculation\n",
"EeV=E*10**3; #Energy in eV\n",
"Ej=EeV*1.6*10**-19; #kinetic energy in J\n",
"v=math.sqrt(2*Ej/m); #velocity in m/s\n",
"lamda=h/(m*v); #de broglie wavelength in m\n",
"lamda_A=lamda*10**10; #de broglie wavelength in armstrong\n",
"lamda_A=math.ceil(lamda_A*10**4)/10**4 #rounding off to 4 decimals\n",
"\n",
"#Result\n",
"print(\"The velocity in m/sec is\",round(v));\n",
"print(\"The de Broglie wavelength in metres is\",lamda);\n",
"print(\"The de Broglie wavelength in Armstrong is\",lamda_A);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The velocity in m/sec is', 1382189.0)\n",
"('The de Broglie wavelength in metres is', 2.861996093951046e-13)\n",
"('The de Broglie wavelength in Armstrong is', 0.0029)\n"
]
}
],
"prompt_number": 28
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.12, Page number 45"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#importing modules\n",
"import math\n",
"\n",
"#Variable decleration\n",
"m=9.1*10**-31; #mass of electron in kg\n",
"h=6.6*10**-34;\n",
"E=2; #kinetic energy in keV\n",
"\n",
"#Calculation\n",
"EeV=E*10**3; #Energy in eV\n",
"Ej=EeV*1.6*10**-19; #kinetic energy in J\n",
"p=math.sqrt(2*m*Ej); #momentum\n",
"lamda=h/p; #de broglie wavelength in m\n",
"lamda_A=lamda*10**10; #de broglie wavelength in armstrong\n",
"lamda_A=math.ceil(lamda_A*10**4)/10**4 #rounding off to 4 decimals\n",
"\n",
"#Result\n",
"print(\"The de Broglie wavelength in metres is\",lamda);\n",
"print(\"The de Broglie wavelength in Armstrong is\",lamda_A);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The de Broglie wavelength in metres is', 2.7348483695436575e-11)\n",
"('The de Broglie wavelength in Armstrong is', 0.2735)\n"
]
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.13, Page number 45"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#import module\n",
"import math\n",
"\n",
"#Variable decleration\n",
"m=1.676*10**-27; #mass of neutron in kg\n",
"h=6.62*10**-34;\n",
"E=0.025; #kinetic energy in eV\n",
"\n",
"#Calculation\n",
"Ej=E*1.6*10**-19; #kinetic energy in J\n",
"v=math.sqrt(2*Ej/m); #velocity in m/s\n",
"lamda=h/(m*v); #wavelength in m\n",
"lamda_A=lamda*10**10; #de broglie wavelength in armstrong\n",
"lamda_A=math.ceil(lamda_A*10**5)/10**5 #rounding off to 5 decimals\n",
"\n",
"#Result\n",
"print(\"The neutrons wavelength in metres is\",lamda);\n",
"print(\"The wavelength in Armstrong is\",lamda_A);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The neutrons wavelength in metres is', 1.8079065940980725e-10)\n",
"('The wavelength in Armstrong is', 1.80791)\n"
]
}
],
"prompt_number": 30
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.14, Page number 45"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#import module\n",
"import math\n",
"\n",
"#Variable decleration\n",
"V=10; #potential in kV\n",
"\n",
"#Calculation\n",
"V=V*10**3; #potential in V\n",
"lamda=12.26/math.sqrt(V); #wavelength\n",
"\n",
"#Result\n",
"print(\"The wavelength in Armstrong is\",lamda);"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The wavelength in Armstrong is', 0.1226)\n"
]
}
],
"prompt_number": 31
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.15, Page number 45"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#import module\n",
"import math\n",
"\n",
"#Varialble decleration\n",
"h=6.626*10**-34;\n",
"m=9.1*10**-31; #mass in kg\n",
"l=1; #width in armstrong\n",
"\n",
"#Calculation\n",
"L=l*10**-10; #width in m\n",
"#permitted electron energies En=(n**2*h**2)/(8*m*L**2)\n",
"#let X = h**2/(8*m*L**2)\n",
"X = h**2/(8*m*L**2); #energy in J\n",
"XeV=X/(1.6*10**-19); #energy in eV\n",
"#in the 1st level n1=1\n",
"n1=1;\n",
"E1=(n1**2)*XeV; #energy in eV\n",
"\n",
"#in second level n2=2\n",
"n2=2;\n",
"E2=(n2**2)*XeV; #energy in eV\n",
"#in third level n3=\n",
"n3=3;\n",
"E3=(n3**2)*XeV; #energy in eV\n",
"\n",
"#Result\n",
"print(\"minimum energy the electron can have in eV is\",round(E1));\n",
"print(\"other values of energy are in eV and in eV\",round(E2),round(E3));\n",
"\n",
"#answers given in the book are wrong"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('minimum energy the electron can have in eV is', 38.0)\n",
"('other values of energy are in eV and in eV', 151.0, 339.0)\n"
]
}
],
"prompt_number": 32
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.16, Page number 46\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#import module\n",
"import math\n",
"\n",
"#Variable decleration\n",
"n=1; #lowest state\n",
"L=10; #width in armstrong\n",
"\n",
"#Calculation\n",
"L=L*10**-10; #width in m\n",
"x=L/2;\n",
"delta_x=1; #interval in armstrong\n",
"delta_x=delta_x*10**-10; #interval in m\n",
"psi1=(math.sqrt(2/L))*math.sin(math.pi*x/L);\n",
"A=psi1**2;\n",
"p=A*delta_x;\n",
"p=math.ceil(p*10)/10; #de broglie wavelength in armstrong\n",
"\n",
"#Result\n",
"print(\"probability of finding the particle is \",p);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('probability of finding the particle is ', 0.2)\n"
]
}
],
"prompt_number": 33
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.17, Page number 46"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#import module\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable decleration\n",
"d=970; #density of Na in kg/m^3\n",
"n=6.02*10**26;\n",
"h=6.62*10**(-34);\n",
"m=9.1*10**-31; #mass in kg\n",
"w=23; #atomic weight\n",
"\n",
"#Calculation\n",
"N=(d*n)/w; #number of atoms per m^3\n",
"A=(h**2)/(8*m);\n",
"B=(3*N)/math.pi;\n",
"Ef=A*B**(2/3);\n",
"EfeV=Ef/(1.6*10**-19);\n",
"EfeV=math.ceil(EfeV*10**2)/10**2 #rounding of to 2 decimals\n",
"\n",
"#Result\n",
"print(\"fermi energy of Na in eV is\",EfeV);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('fermi energy of Na in eV is', 3.16)\n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.18, Page number 46"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#import module\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable decleration\n",
"n1=1;\n",
"n2=1;\n",
"n3=1; #values in lowest energy\n",
"h=6.62*10**(-34);\n",
"m=9.1*10**-31; #mass in kg\n",
"L=0.1; #side in nm\n",
"\n",
"#Calculation\n",
"L=L*10**-9; #side in m\n",
"n=(n1**2)+(n2**2)+(n3**2);\n",
"E1=(n*h**2)/(8*m*L**2); #energy in j\n",
"E1eV=E1/(1.6*10**-19); #energy in eV\n",
"E1eV=math.ceil(E1eV*10**1)/10**1 #rounding off to 2 decimals\n",
"\n",
"#Result\n",
"print(\"lowest energy of electron in Joule is\",E1);\n",
"print(\"lowest energy of electron in eV is\",E1eV);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('lowest energy of electron in Joule is', 1.8059505494505486e-17)\n",
"('lowest energy of electron in eV is', 112.9)\n"
]
}
],
"prompt_number": 35
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.19, Page number 47"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#import module\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable decleration\n",
"mn=1.676*10**-27; #mass of neutron in kg\n",
"me=9.1*10**-31; #mass of electron in kg\n",
"h=6.62*10**-34;\n",
"c=3*10**8; #velocity of light in m/sec\n",
"\n",
"#Calculation\n",
"En=2*me*c**2;\n",
"lamda=h/math.sqrt(2*mn*En); #wavelength in m\n",
"lamda_A=lamda*10**10; #converting lamda from m to A\n",
"lamda_A=math.ceil(lamda_A*10**6)/10**6 #rounding off to 6 decimals\n",
"\n",
"#Result\n",
"print(\"The de broglie wavelength in Angstrom is\",lamda_A);"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The de broglie wavelength in Angstrom is', 0.000283)\n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.20, Page number 47 ***************************************************************************"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#import module\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable decleration\n",
"n2=2; #second quantum state\n",
"n4=4; #fourth quantum state\n",
"h=6.626*10**-34;\n",
"m=9.1*10**-31; #mass in kg\n",
"a=2; #potential box length in armstrong\n",
"\n",
"#Calculation\n",
"a=a*10**-10; #length in m\n",
"A=n2**2*h**2;\n",
"B=8*m*a**2;\n",
"E2=A/B; #energy in j\n",
"E2eV=E2/(1.6*10**-19); #energy in eV\n",
"C=n4**2*h**2;\n",
"E4=C/B; #energy in j\n",
"E4eV=E4/(1.6*10**-19); #energy in eV\n",
"\n",
"#Result\n",
"print(\"energy corresponding to second quantum state in Joule is\",E2);\n",
"print(\"energy corresponding to second quantum state in eV is\",E2eV);\n",
"print(\"energy corresponding to fourth quantum state in Joule is\",E4);\n",
"print(\"energy corresponding to fourth quantum state in eV is\",E4eV);\n",
"\n",
"\n",
"#answers given in the book are wrong"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('energy corresponding to second quantum state in Joule is', 6.030752197802197e-18)\n",
"('energy corresponding to second quantum state in eV is', 37.69220123626373)\n",
"('energy corresponding to fourth quantum state in Joule is', 2.412300879120879e-17)\n",
"('energy corresponding to fourth quantum state in eV is', 150.7688049450549)\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.21, Page number 48 ***********"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#import module\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable decleration\n",
"V=344; #accelerated voltage in V\n",
"n=1; #first reflection\n",
"theta=60; #glancing angle in degrees\n",
"\n",
"#Calculation\n",
"lamda=12.27/math.sqrt(V);\n",
"d=(n*lamda)/(2*math.sin(theta));\n",
"\n",
"#Result\n",
"print(\"The spacing of the crystal in Angstrom is\",lamda);"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The spacing of the crystal in Angstrom is', 0.6615540636030947)\n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.22, Page number 49 *************"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#import module\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable decleration\n",
"n2=2; #second quantum state\n",
"n3=3; #fourth quantum state\n",
"h=6.626*10**-34;\n",
"m=9.1*10**-31; #mass in kg\n",
"a=1*10**-10; #width of potential well in m\n",
"\n",
"#Calculation\n",
"B=8*m*a**2;\n",
"E1=h**2/B; #ground state energy\n",
"E1eV=E1/(1.6*10**-19); #energy in eV\n",
"A=n2**2*h**2;\n",
"E2=A/B; #energy in j\n",
"E2eV=E2/(1.6*10**-19); #energy in eV\n",
"C=n3**2*h**2;\n",
"E3=C/B; #energy in j\n",
"E3eV=E3/(1.6*10**-19); #energy in eV\n",
"E1=math.ceil(E1*10**3)/10**3 #rounding off to 3 decimals\n",
"E1eV=math.ceil(E1eV*10**3)/10**3 #rounding off to 3 decimals\n",
"E2eV=math.ceil(E2eV*10**3)/10**3 #rounding off to 3 decimals\n",
"E3eV=math.ceil(E3eV*10**3)/10**3 #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print(\"ground state energy in Joule is\",E1);\n",
"print(\"ground state energy in eV is\",E1eV);\n",
"print(\"first energy state in eV is\",E2eV);\n",
"print(\"second energy state in eV is\",E3eV);\n",
"\n",
"#answers given in the book are wrong by one decimal"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('ground state energy in Joule is', 0.001)\n",
"('ground state energy in eV is', 37.693)\n",
"('first energy state in eV is', 150.769)\n",
"('second energy state in eV is', 339.23)\n"
]
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.23, Page number 49"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#import module\n",
"import math\n",
"\n",
"#Variable decleration\n",
"n3=3; #fourth quantum state\n",
"h=6.626*10**-34;\n",
"m=9.1*10**-31; #mass in kg\n",
"\n",
"\n",
"#ground state energy E1 = h**2/(8*m*a**2)\n",
"#second excited state E3 = (9*h**2)/(8*m*a**2)\n",
"#required energy E = E3-E1\n",
"#E = (9*h**2)/(8*m*a**2) - h**2/(8*m*a**2)\n",
"#E = (h**2/(8*m*a**2))*(9-1)\n",
"#therefore E = (8*h**2)/(8*m*a**2)\n",
"#hence E = (h**2)/(m*a**2)\n",
"\n",
"#Result \n",
"# the required energy is E = (h**2)/(m*a**2)"
],
"language": "python",
"metadata": {},
"outputs": [],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.24, Page number 50"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#import module\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable decleration\n",
"delta_x=10**-8; #length of box in m\n",
"h=6.626*10**-34;\n",
"m=9.1*10**-31; #mass in kg\n",
"\n",
"#Calculation\n",
"delta_v=h/(m*delta_x); #uncertainity in m/sec\n",
"delta_vk=delta_v*10**-3; #uncertainity in km/sec\n",
"delta_vk=math.ceil(delta_vk*10**2)/10**2 #rounding off to 2 decimals\n",
"\n",
"#Result\n",
"print(\"minimum uncertainity in velocity in m/sec is\",round(delta_v));\n",
"print(\"minimum uncertainity in velocity in km/sec is\",delta_vk);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('minimum uncertainity in velocity in m/sec is', 72813.0)\n",
"('minimum uncertainity in velocity in km/sec is', 72.82)\n"
]
}
],
"prompt_number": 40
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.25, Page number 50"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#import module\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable decleration\n",
"mp=1.6*10**-27; #mass of proton in kg\n",
"me=9.1*10**-31; #mass of electron in kg\n",
"h=6.626*10**(-34);\n",
"c=3*10**10; #velocity of light in m/sec\n",
"\n",
"#Calculation\n",
"Ep=me*c**2;\n",
"lamda=h/math.sqrt(2*mp*Ep); #wavelength in m\n",
"lamda_A=lamda*10**10; #converting lamda from m to A\n",
"\n",
"#Result\n",
"print(\"The de broglie wavelength in Angstrom is\",lamda_A);"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The de broglie wavelength in Angstrom is', 4.092931643497047e-06)\n"
]
}
],
"prompt_number": 41
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.26, Page number 51 *************************************************"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#import module\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable decleration\n",
"m=1.675*10**(-27); #mass of neutron in kg\n",
"h=6.626*10**(-34);\n",
"n=1; #diffractive order\n",
"d=0.314; #spacing in nm\n",
"E=0.04; #kinetic energy in eV\n",
"\n",
"#Calculation\n",
"d=d*10**-9; #spacing in m\n",
"Ej=E*1.6*10**-19; #kinetic energy in J\n",
"lamda=h/math.sqrt(2*m*Ej); #de Broglie wavelength\n",
"lamdaA=lamda*10**9; #converting wavelength from m to nm\n",
"theta=math.asin((n*lamda)/(2*d));\n",
"print(\"The de Broglie wavelength in metres is\",lamda);\n",
"print(\"The de Broglie wavelength in nm is\",lamdaA);\n",
"print(\"glancing angle in degrees is\",theta);\n",
"\n",
"#answer given in the book is wrong"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"('The de Broglie wavelength in metres is', 1.4309980469755228e-10)\n",
"('The de Broglie wavelength in nm is', 0.1430998046975523)\n",
"('glancing angle in degrees is', 0.2298853909391574)\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}