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{
"metadata": {
"name": "Chapter4"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": "Quantum Physics"
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 4.1, Page number 117"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the energy and momentum of photon\n\n#importing modules\nimport math\n\n#Variable declaration\nc = 3*10**8; #velocity of light(m/sec)\nh = 6.62*10**-34; #planck's constant\nlamda = 1.2; #wavelength of photon(Angstrom)\ne = 1.6*10**-19; #conversion factor from J to eV\n\n#Calculation\nlamda = lamda*10**-10; #wavelength of photon(m)\nE = (h*c)/(lamda*e); #energy of photon(eV)\nE=math.ceil(E*10)/10; #rounding off to 1 decimal\np = h/lamda; #momentum of photon(kg m/s)\n\n#Result\nprint \"energy of the photon is\",E,\"eV\"\nprint \"momentum of the photon is\",p,\"kg m/s\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "energy of the photon is 10343.8 eV\nmomentum of the photon is 5.51666666667e-24 kg m/s\n"
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 4.2, Page number 117"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the number of photons emitted \n\n#importing modules\nimport math\n\n#Variable declaration\nh = 6.625*10**-34; #planck's constant\nnew = 900; #frequency(kHz)\nE1 = 10; #power radiated(kW)\n\n#Calculation\nE1 = E1*10**3; #power radiated(W)\nnew = new*10**3; #frequency(Hz)\nE = h*new; #energy of photon(J)\nN = E1/E; #number of photons emitted \n\n#Result\nprint \"number of photons emitted per second is\",N",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "number of photons emitted per second is 1.67714884696e+31\n"
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 4.3, Page number 118"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the number of photons emitted \n\n#importing modules\nimport math\n\n#Variable declaration\nc = 3*10**8; #velocity of light(m/sec)\nh = 6.63*10**-34; #planck's constant\nlamda = 5893; #wavelength of photon(Angstrom)\nE1 = 100; #power of lamp(W) \n\n#Calculation\nlamda = lamda*10**-10; #wavelength of photon(m)\nE = h*c/lamda; #energy of photon(J)\nN = E1/E; #number of photons emitted \n\n#Result\nprint \"number of photons emitted per second is\",N\nprint \"answer given in the book is wrong\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "number of photons emitted per second is 2.96279537456e+20\nanswer given in the book is wrong\n"
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 4.4, Page number 118"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the wavelength of scattered radiation\n\n#importing modules\nimport math\n\n#Variable declaration\nc = 3*10**8; #velocity of light(m/sec)\nh = 6.6*10**-34; #planck's constant\nm0 = 9.1*10**-31; #mass of photon(kg)\ntheta = 30; #viewing angle(degrees)\nlamda = 2.8*10**-10; #wavelength of photon(m)\n\n#Calculation\nx = math.pi/180; #conversion factor from degrees to radians\ntheta = theta*x; #viewing angle(radian) \nlamda_dash = (2*h*(math.sin(theta/2))**2/(m0*c))+lamda; #wavelength of scattered radiation(m)\nlamda_dash = lamda_dash*10**10; #wavelength of scattered radiation(Angstrom)\nlamda_dash=math.ceil(lamda_dash*10**5)/10**5; #rounding off to 5 decimals\n\n#Result\nprint \"wavelength of scattered radiation is\",lamda_dash,\"Angstrom\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "wavelength of scattered radiation is 2.80324 Angstrom\n"
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 4.5, Page number 119"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the deBroglie wavelength\n\n#importing modules\nimport math\n\n#Variable declaration\nh = 6.6*10**-34; #planck's constant\nm = 0.040; #mass of bullet(kg)\nv = 1; #speed of bullet(km/s)\n\n#Calculation\nv = v*10**3; #speed of bullet(m/s)\np = m*v; #momemtun of bullet(kg m/s)\nlamda = h/p; #deBroglie wavelength(m)\nlamda = lamda*10**10; #deBroglie wavelength(Angstrom)\n\n#Result\nprint \"deBroglie wavelength is\",lamda,\"Angstrom\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "deBroglie wavelength is 1.65e-25 Angstrom\n"
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 4.6, Page number 119"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the energy of particle\n\n#importing modules\nimport math\n\n#Variable declaration\nn = 1; #lowest energy state\na = 0.1; #width of box(nm)\nh = 6.625*10**-34; #planck's constant\ne = 1.602*10**-19; #conversion factor from J to eV\nm = 9.11*10**-31; #mass of particle(kg)\n\n#Calculation\na = a*10**-9; #width of box(m)\nE = (n**2)*(h**2)/(8*m*(a**2)); #energy of particle(J)\nE_eV = E/e; #energy of particle(eV)\nE_eV=math.ceil(E_eV*10)/10; #rounding off to 1 decimal\n\n#Result\nprint \"energy of particle is\",E,\"J or\",E_eV,\"eV\" ",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "energy of particle is 6.02231407794e-18 J or 37.6 eV\n"
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 4.7, Page number 120"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the minimum energy of an electron\n\n#importing modules\nimport math\n\n#Variable declaration\nn = 1; #lowest energy state\na = 4; #width of well(nm)\nh = 6.625*10**-34; #planck's constant\ne = 1.6025*10**-19; #conversion factor from J to eV\nm = 9.11*10**-31; #mass of electron(kg)\n\n#Calculation\na = a*10**-9; #width of box(m)\nE = (n**2)*(h**2)/(8*m*(a**2)); #energy of particle(J)\nE_eV = E/e; #energy of particle(eV)\nE_eV=math.ceil(E_eV*10**4)/10**4; #rounding off to 4 decimals\n\n#Result\nprint \"minimum energy of electron is\",E,\"J or\",E_eV,\"eV\" ",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "minimum energy of electron is 3.76394629871e-21 J or 0.0235 eV\n"
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 4.8, Page number 120"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the energy required to excite the electron\n\n#importing modules\nimport math\n\n#Variable declaration\nn1 = 1; #lowest energy state\nn2 = 6; #for 6th excited state\na = 0.1; #width of box(nm)\nh = 6.625*10**-34; #planck's constant\ne = 1.602*10**-19; #conversion factor from J to eV\nm = 9.11*10**-31; #mass of electron(kg)\n\n#Calculation\na = a*10**-9; #width of box(m)\nE1 = (n1**2)*(h**2)/(8*m*(a**2)); #energy of electron in ground state(J)\nE6 = (n2**2)*(h**2)/(8*m*(a**2)); #energy of electron in excited state(J)\nE = E6-E1; #energy required to excite the electron(J)\nE_eV = E/e; #energy required to excite the electron(eV)\nE_eV=math.ceil(E_eV*10)/10; #rounding off to 1 decimal\n\n#Result\nprint \"energy required to excite the electron is\",E,\"J or\",E_eV,\"eV\" \nprint \"answer for energy in eV given in the book is wrong\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "energy required to excite the electron is 2.10780992728e-16 J or 1315.8 eV\nanswer for energy in eV given in the book is wrong\n"
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 4.9, Page number 121"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the change in wavelength\n\n#importing modules\nimport math\n\n#Variable declaration\nh = 6.625*10**-34; #planck's constant\nc = 3*10**8; #velocity of light(m/sec)\nm0 = 9.11*10**-31; #rest mass of electron(kg)\nphi = 90; #angle of scattering(degrees)\nx = math.pi/180; #conversion factor from degrees to radians\n\n#Calculation\nphi = phi*x; ##angle of scattering(radian)\ndelta_lamda = h*(1-math.cos(phi))/(m0*c); #change in wavelength(m)\n\n#Result\nprint \"change in wavelength of X-ray photon is\",delta_lamda,\"m\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "change in wavelength of X-ray photon is 2.42407610684e-12 m\n"
}
],
"prompt_number": 12
},
{
"cell_type": "code",
"collapsed": false,
"input": "",
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
