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{
"metadata": {
"name": "",
"signature": "sha256:38e4520c8655f11fdcb5696673580e95d36ab2ce43843aea287c93b1f1a0b257"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"10: Quantum Physics and Schrodinger wave equation"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 10.1, Page number 258"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"me=9.1*10**-31; #mass of electron(kg)\n",
"h=6.625*10**-34; #planck's constant(Jsec)\n",
"deltax=10**-8; #uncertainity in position(m)\n",
"\n",
"#Calculation\n",
"deltap=(h/(2*math.pi*deltax)); #uncertainity principle(kgm/sec)\n",
"deltav=(deltap/me); #minimum uncertainity in velocity(m/sec)\n",
"\n",
"#Result\n",
"print \"minimum uncertainity in velocity is\",round(deltav/10**5,3),\"*10**5 m/sec\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"minimum uncertainity in velocity is 0.116 *10**5 m/sec\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 10.2, Page number 259"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"lamda=0.2865*10**-10; #wavelength(m)\n",
"mp=1.67*10**-27; #mass of proton(kg)\n",
"h=6.625*10**-34; #planck's constant(Jsec)\n",
"q=1.6*10**-19; #charge of proton(C)\n",
"\n",
"#Calculation\n",
"v=(h/(mp*lamda)); #velocity(m/sec)\n",
"KE=0.5*mp*(v**2); #kinetic energy of proton(J)\n",
"KE=KE/q; #kinetic energy of proton(eV)\n",
"\n",
"#Result\n",
"print \"kinetic energy of proton is\",int(KE),\"eV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"kinetic energy of proton is 1 eV\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 10.3, Page number 259"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"KE=0.025; #kinetic energy of neutron(eV)\n",
"q=1.6*10**-19; #charge of proton(C)\n",
"mn=1.676*10**-27; #mass of neutron(kg)\n",
"h=6.625*10**-34; #planck's constant(Jsec)\n",
"me=9.1*10**-31; #mass of electron(kg)\n",
"c=3*10**8; #velocity of light(m/s)\n",
"\n",
"#Calculation\n",
"KE=KE*q; #kinetic energy of neutron(J)\n",
"v=math.sqrt((2*KE)/mn); #velocity(m/s)\n",
"lamdan=h/(mn*v); #debroglie wavelength of neutron(m)\n",
"p=(h/lamdan); #momentum of electron and photon(kgm/s)\n",
"ve=(p/me); #velocity of electron(m/s)\n",
"Ee=0.5*p*ve; #energy of electron(J)\n",
"Ee=Ee/q; #energy of electron(eV)\n",
"Ep=h*c/lamdan; #energy of photon(J)\n",
"Ep=Ep/q; #energy of photon(eV)\n",
"\n",
"#Result\n",
"print \"wavelength of beam of neutron is\",round(lamdan*10**10,3),\"angstrom\"\n",
"print \"momentum of electron and photon is\",p,\"kgm/s\"\n",
"print \"energy of electron is\",round(Ee,2),\"eV\"\n",
"print \"energy of photon is\",round(Ep/10**3,2),\"*10**3 eV\"\n",
"print \"answers in the book vary due to rounding off errors\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"wavelength of beam of neutron is 1.809 angstrom\n",
"momentum of electron and photon is 3.66169359723e-24 kgm/s\n",
"energy of electron is 46.04 eV\n",
"energy of photon is 6.87 *10**3 eV\n",
"answers in the book vary due to rounding off errors\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 10.4, Page number 260"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"e=1.6*10**-19; #charge of electron(C)\n",
"V=200; #potential difference(V)\n",
"lamda=0.0202*10**-10; #debroglie wavelength(m)\n",
"h=6.625*10**-34; #planck's constant(Jsec)\n",
"\n",
"#Calculation\n",
"#eV=0.5*m*(v^2)\n",
"#mv=sqrt(2*m*eV)\n",
"m=((h**2)/(2*(lamda**2)*e*V)); #mass of particle(kg)\n",
"\n",
"#Result\n",
"print \"mass of particle is\",m,\"kg\"\n",
"print \"hence it is a proton\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"mass of particle is 1.68069555834e-27 kg\n",
"hence it is a proton\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 10.5, Page number 260"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration \n",
"mn=1.676*10**-27; #mass of neutron(kg)\n",
"e=1.6*10**-19; #charge of electron(C)\n",
"h=6.622*10**-34; #planck's constant(Jsec)\n",
"\n",
"#Calculation\n",
"E=e; #energy of neutron(J)\n",
"v=math.sqrt((2*E)/mn); #velocity of neutron(m/sec)\n",
"lamda=(h/(mn*v)); #de-broglie wavelength(m)\n",
"\n",
"#Result\n",
"print \"de-broglie wavelength of neutron is\",round(lamda*10**10,3),\"angstrom\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"de-broglie wavelength of neutron is 0.286 angstrom\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 10.6, Page number 261"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration \n",
"r=10**-14; #radius(m)\n",
"h=6.625*10**-34; #planck's constant(Jsec)\n",
"c=3*10**8; #velocity of light(m/s)\n",
"mo=9.1*10**-31; #rest mass of particle(kg)\n",
"q=1.6*10**-19; #charge of electron(C)\n",
"\n",
"\n",
"#Calculation\n",
"#acc. to uncertainity principle delx*delp >= (h/2*%pi)\n",
"deltax=2*r; #uncertainity in position(m)\n",
"deltap=(h/(2*math.pi*deltax)); ##uncertainity in momentum\n",
"#from einstein's relavistic relation E=mc2=KE+rest mass energy=0.5mv2+moc2\n",
"#when velocity of particle is very high\n",
"#m=(mo/sqrt(1-((v/c)^2))) where m-mass of particle with velocity v,mo-rest mass of particle, c-velocity of particle\n",
"p=deltap #assume\n",
"E=math.sqrt(((p*c)**2)+((mo*(c**2))**2)); #energy(J)\n",
"E=E/q; #energy(eV)\n",
"\n",
"#Result\n",
"print \"energy is\",round(E/10**6),\"MeV\"\n",
"print \"this value is much higher than experimentally obtained values of energy of electron of a radioactive nuclei i.e 4 Mev this proves that electron cannot reside within nucleus\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"energy is 10.0 MeV\n",
"this value is much higher than experimentally obtained values of energy of electron of a radioactive nuclei i.e 4 Mev this proves that electron cannot reside within nucleus\n"
]
}
],
"prompt_number": 19
}
],
"metadata": {}
}
]
}
|
