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Diffstat (limited to 'Quantum_mechanics_by_M.C.Jain')
| -rwxr-xr-x | Quantum_mechanics_by_M.C.Jain/chapter13.ipynb | 62 | ||||
| -rwxr-xr-x | Quantum_mechanics_by_M.C.Jain/chapter2.ipynb | 710 | ||||
| -rwxr-xr-x | Quantum_mechanics_by_M.C.Jain/chapter3.ipynb | 519 | ||||
| -rwxr-xr-x | Quantum_mechanics_by_M.C.Jain/chapter4.ipynb | 329 | ||||
| -rwxr-xr-x | Quantum_mechanics_by_M.C.Jain/chapter5.ipynb | 294 | ||||
| -rwxr-xr-x | Quantum_mechanics_by_M.C.Jain/chapter7.ipynb | 238 | ||||
| -rwxr-xr-x | Quantum_mechanics_by_M.C.Jain/chapter8.ipynb | 269 |
7 files changed, 0 insertions, 2421 deletions
diff --git a/Quantum_mechanics_by_M.C.Jain/chapter13.ipynb b/Quantum_mechanics_by_M.C.Jain/chapter13.ipynb deleted file mode 100755 index 9d30125f..00000000 --- a/Quantum_mechanics_by_M.C.Jain/chapter13.ipynb +++ /dev/null @@ -1,62 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:7906c3ed6cf316425e6f8055a34bac268639221c2739520f565ed9645c9cd808"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13 Atomic structure I: One- Electron Atoms"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.1 Page no 252"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "d=0.1 #m\n",
- "v=10.0**3 #m/s\n",
- "a=50 #gradient of a magnet field Wb/m**2/m\n",
- "b=9.274*10**-27 #J/Wb/m**2\n",
- "h=1.6605*10**-27\n",
- "\n",
- "#Calculation\n",
- "M=107.868*h\n",
- "z=(b/M)*a*(d**2/v**2)\n",
- "\n",
- "#Result\n",
- "print\"seperation between the two component \",round(z*10**8,1),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "seperation between the two component 2.6 mm\n"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Quantum_mechanics_by_M.C.Jain/chapter2.ipynb b/Quantum_mechanics_by_M.C.Jain/chapter2.ipynb deleted file mode 100755 index 3d83df64..00000000 --- a/Quantum_mechanics_by_M.C.Jain/chapter2.ipynb +++ /dev/null @@ -1,710 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:46bc70330d4213802afb03e252b2ad32eb9319ed4cc2a32fe2c16df97a5f1978"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 2 Particle nature of Radiation; The origin of Quantum theory"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2 Page no-12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=40 #W\n",
- "lembda=6000*10**-10 #m\n",
- "h=6.63*10**-34 #Js\n",
- "c=3*10**8 #m/s\n",
- "\n",
- "#Calculation\n",
- "n=(E*lembda)/(h*c)\n",
- "\n",
- "#Result\n",
- "print\"No. of photons emitted per second are given by \",round(n*10**-19,2),\"*10**19\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "No. of photons emitted per second are given by 12.07 *10**19\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.3 Page no-12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=3.2 #ev\n",
- "energy=3.8 #ev\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "c=energy-a\n",
- "Energy=c*e\n",
- "\n",
- "#Result\n",
- "print\"Kinetic energy of the photoelectron is given by \",Energy,\"Joule\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Kinetic energy of the photoelectron is given by 9.6e-20 Joule\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.4 Page no-12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "W=3.45 #ev\n",
- "h=6.63*10**-34 #Js\n",
- "c=3*10**8 #m/s\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "lembda=(h*c)/(W*e)\n",
- "\n",
- "#Result\n",
- "print\"Maximum wavelength of photon is \",round(lembda*10**10,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum wavelength of photon is 3603.0 A\n"
- ]
- }
- ],
- "prompt_number": 193
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.5 Page no-12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "W=3 #ev\n",
- "h=6.63*10**-34\n",
- "e=1.6*10**-19\n",
- "lembda=3.0*10**-7 #m\n",
- "c=3*10**8 #m/s\n",
- "\n",
- "#Calculation\n",
- "v0=(W*e)/h\n",
- "v=c/lembda\n",
- "E=h*(v-v0)\n",
- "E1=(h*(v-v0))/(1.6*10**-19)\n",
- "V0=E/e\n",
- "\n",
- "#Result\n",
- "print\"(a) Threshold frequency \",round(v0*10**-15,2),\"*10**15 HZ\"\n",
- "print\"(b) Maximum energy of photoelectron \",round(E1,2),\"eV\"\n",
- "print\"(c) Stopping potential \",round(V0,2),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Threshold frequency 0.72 *10**15 HZ\n",
- "(b) Maximum energy of photoelectron 1.14 eV\n",
- "(c) Stopping potential 1.14 V\n"
- ]
- }
- ],
- "prompt_number": 197
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.6 Page no-13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "v0=6*10**14 #s**-1\n",
- "h=6.63*10**-34\n",
- "e=1.6*10**-19\n",
- "V0=3\n",
- "\n",
- "#Calculaton\n",
- "W=h*v0\n",
- "W0=(h*v0)/e\n",
- "V=(e*V0+h*v0)/h\n",
- "\n",
- "#Result \n",
- "print\"work function is given by \",round(W0,3),\"ev\"\n",
- "print\"frequency is given by \",round(V*10**-15,2),\"*10**15 s-1\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "work function is given by 2.486 ev\n",
- "frequency is given by 1.32 *10**15 s-1\n"
- ]
- }
- ],
- "prompt_number": 88
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.7 Page no 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "lembda=6800.0*10**-10 #m\n",
- "h=6.6*10**-34\n",
- "W=2.3 #ev\n",
- "c=3*10**8 #m/s\n",
- "\n",
- "#Calculation\n",
- "E=((h*c)/lembda)/1.6*10**-19\n",
- "\n",
- "#Result\n",
- "print\"Energy is \",round(E*10**38,2),\"ev\"\n",
- "print\"since the energy of incident photon is less then the work function of Na, photoelecrticemession is not possible with the given light.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy is 1.82 ev\n",
- "since the energy of incident photon is less then the work function of Na, photoelecrticemession is not possible with the given light.\n"
- ]
- }
- ],
- "prompt_number": 200
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.8 Page no 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "lembda=3500*10**-10 #m\n",
- "h=6.6*10**-34\n",
- "c=3*10**8 #m/s\n",
- "\n",
- "#calculation \n",
- "E=((h*c)/lembda)/1.6*10**-19\n",
- "\n",
- "#Result\n",
- "print\"Energy is \" ,round(E*10**38,2),\"ev\"\n",
- "print\"1.9 ev < E < 4.2 ev,only metal B will yield photoelectrons\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy is 3.54 ev\n",
- "1.9 ev < E < 4.2 ev,only metal B will yield photoelectrons\n"
- ]
- }
- ],
- "prompt_number": 201
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.9 Page no 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "lembda=6.2*10**-6\n",
- "W=0.1 #ev\n",
- "h=6.6*10**-34 #Js\n",
- "c=3*10**8 #m/s\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "E=((h*c)/(lembda*e))-W\n",
- "\n",
- "#Result\n",
- "print\"Maximum kinetic energy of photoelectron \",round(E,1),\"ev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum kinetic energy of photoelectron 0.1 ev\n"
- ]
- }
- ],
- "prompt_number": 112
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.10 Page no 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#given\n",
- "e=1.60*10**-19 #C\n",
- "slope=4.12*10**-15 #Vs\n",
- "\n",
- "#Calculation\n",
- "h=slope*e\n",
- "\n",
- "#Result\n",
- "print\"Value of plank's constant \",h,\"Js\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Value of plank's constant 6.592e-34 Js\n"
- ]
- }
- ],
- "prompt_number": 114
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.11 Page no 15"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "W=2.26*1.6*10**-19 #ev\n",
- "v=10**6 #m/s\n",
- "m=9*10**-31\n",
- "\n",
- "#Calculation\n",
- "V=((1/2.0)*m*v**2+W)/h\n",
- "\n",
- "#Result\n",
- "print\"frequency of incident radiation \",round(V*10**-15,2),\"*10**15 HZ\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "frequency of incident radiation 1.23 *10**15 HZ\n"
- ]
- }
- ],
- "prompt_number": 118
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.12 Page no 15"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#given\n",
- "V1=.82 #volts\n",
- "V2=1.85 #volts\n",
- "lembda1=4.0*10**-7 #m\n",
- "lembda2=3.0*10**-7\n",
- "e=1.6*10**-19\n",
- "c=3.0*10**8 #m/s\n",
- "\n",
- "#Calculation\n",
- "lembda=(1/lembda2)-(1/lembda1)\n",
- "h=(e*(V2-V1))/(c*lembda)\n",
- "\n",
- "#Result\n",
- "print\"(a) plank's constant \",h,\"Js\"\n",
- "print\"(b) no, because the stopping potentialdepends only on the wavelength of light and not on its intensity.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) plank's constant 6.592e-34 Js\n",
- "(b) no, because the stopping potentialdepends only on the wavelength of light and not on its intensity.\n"
- ]
- }
- ],
- "prompt_number": 202
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.13 Page no 16"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "\n"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#given\n",
- "h=6.62*10**-34 #Js\n",
- "c=3*10**8 #m/s\n",
- "lembda=4560.0*10**-10 #m\n",
- "p=1*10**-3 #W\n",
- "a=0.5/100\n",
- "e=1.6*10**-19\n",
- "\n",
- "#calculation\n",
- "E=(h*c)/lembda\n",
- "N=p/E #Number of photons incedent on the surface\n",
- "n=N*a\n",
- "I=n*e\n",
- "\n",
- "#result\n",
- "print\"Photoelectric current \",round(I*10**6,2),\"*10**-6 A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Photoelectric current 1.84 *10**-6 A\n"
- ]
- }
- ],
- "prompt_number": 131
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.14 Page no 22"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#given\n",
- "m0=9.1*10**-31 #Kg\n",
- "c=3*10**8 #m/s\n",
- "h=6.6*10**-34 #Js\n",
- "v1=2.0*10**-10 #m\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "v= (h/(m0*c))*(1-(math.cos(90))*3.14/180.0)\n",
- "v2=v+v1\n",
- "v0=v2-v1\n",
- "E=(h*c*(v0))/(v1*v2)\n",
- "b=(1/(math.sin(90)*3.14/180.0))*((v2*10**-10/v1)-math.cos(90)*3.14/180.0)\n",
- "angle=3.14/2.0-math.atan(b)\n",
- "\n",
- "#Result\n",
- "print \"(a) the wavelength of scattered photon is \",round(v2*10**10,3),\"A\"\n",
- "print\"(b) The energy of recoil electron is \",round(E*10**17,2),\"*10**-17 J\"\n",
- "print\"(c) angle at which the recoil electron appears \",round(angle,2),\"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) the wavelength of scattered photon is 2.024 A\n",
- "(b) The energy of recoil electron is 1.19 *10**-17 J\n",
- "(c) angle at which the recoil electron appears 1.11 degree\n"
- ]
- }
- ],
- "prompt_number": 278
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.15 Page no 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given \n",
- "E=0.9 #Mev\n",
- "a=120 #degree\n",
- "m=9.1*10**-31 #Kg\n",
- "c=3*10**8 #m/s\n",
- "\n",
- "#calculation\n",
- "b=((m*c**2)/1.6*10**-19)*10**32\n",
- "energy=E/(1+2*(E/b)*(3/4.0))\n",
- "\n",
- "#Result\n",
- "print \"energy of scattered photon \",round(energy,3),\"Mev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "energy of scattered photon 0.247 Mev\n"
- ]
- }
- ],
- "prompt_number": 142
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.16 Page no 24"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "v1=2.000*10**-10 #m\n",
- "v2=2.048*10**-10 #m\n",
- "a=180 #degree\n",
- "a1=60 #degree\n",
- "h=6.6*10**-34\n",
- "c=3*10**8\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "b=(v2-v1)/(1-math.cos(a*3.14/180.0))\n",
- "V=v1+b*(1-math.cos(60*3.14/180.0))\n",
- "E=(h*c*(V-v1))/(V*v1)\n",
- "\n",
- "#Result\n",
- "print\"(a) wavelength of radiation scattered at an angle of 60 degree \",round(V*10**10,3),\"A\"\n",
- "print \"(b) Energy of the recoiul electron is \",round(E*10**18,2),\"*10**-18 J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) wavelength of radiation scattered at an angle of 60 degree 2.012 A\n",
- "(b) Energy of the recoiul electron is 5.9 *10**-18 J\n"
- ]
- }
- ],
- "prompt_number": 277
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.17 Page no 24"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=4*10**3*1.6*10**-19\n",
- "m0=9.1*10**-31\n",
- "b=6.4*10**-16\n",
- "d=102.39*10**-16\n",
- "h=6.3*10**-34\n",
- "c=3*10**8\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "p=math.sqrt(2*m0*E)\n",
- "d=b+d\n",
- "lembda=(2*h*c)/d\n",
- "\n",
- "#Result\n",
- "print\"Wavelength of incident photon is \", round(lembda*10**10,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Wavelength of incident photon is 0.35 A\n"
- ]
- }
- ],
- "prompt_number": 233
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.19 Page no 26"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=1.02 #Mev\n",
- "b=0.51\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "alpha=E/b\n",
- "a=1/(math.sqrt(2*(alpha+2)))\n",
- "angle=2*(math.asin(a)*180/3.14)\n",
- "e=E/(1.0+alpha*(1-(math.cos(angle*3.14/180.0))))\n",
- "\n",
- "#Result\n",
- "print\"(a) Angle for symmetric scattering is \", round(angle,1),\"degree\"\n",
- "print \"(b) energy of the scattered photon is \",round(e,2),\"Mev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Angle for symmetric scattering is 41.4 degree\n",
- "(b) energy of the scattered photon is 0.68 Mev\n"
- ]
- }
- ],
- "prompt_number": 263
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Quantum_mechanics_by_M.C.Jain/chapter3.ipynb b/Quantum_mechanics_by_M.C.Jain/chapter3.ipynb deleted file mode 100755 index 7ef8f5c9..00000000 --- a/Quantum_mechanics_by_M.C.Jain/chapter3.ipynb +++ /dev/null @@ -1,519 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:8da78b3914613cbb3785b48315fd900bd210e4e84b50eb2e4aa86b6821a0f0e1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 3 Atoms and the Bohr model"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.1 Page no 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=-3.4 #ev\n",
- "h=6.63*10**-34 #Js\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "n=math.sqrt(-13.6/E)\n",
- "M=(n*h)/(2.0*math.pi)\n",
- "\n",
- "#Result\n",
- "print\"Angular momentum of electron is given by \",round(M,36),\" Js\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angular momentum of electron is given by 2.11e-34 Js\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.2 Page no 40"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=13.6 #ev\n",
- "n1=4\n",
- "n2=2\n",
- "\n",
- "#Calculation\n",
- "energy=E*((1/2.0**2)-(1/4.0**2))\n",
- "\n",
- "#Result\n",
- "print\"Energy of photon emitted in the transition is \",energy,\"ev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy of photon emitted in the transition is 2.55 ev\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3 Page no 40"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n1=3\n",
- "n2=2\n",
- "E1=-1.5 #ev\n",
- "E2=-3.4 #ev\n",
- "h=6.63*10**-34 #Js\n",
- "c=3*10**8 #m/s\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "v=(h*c)/((E1-E2)*e)\n",
- "\n",
- "#Result\n",
- "print\"Wavelength is \",round(v,10),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Wavelength is 6.543e-07 m\n"
- ]
- }
- ],
- "prompt_number": 131
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.4 Page no 40"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "v=1200 #A\n",
- "R=1.097*10**7 #m-1\n",
- "n1=2.0\n",
- "n2=3.0\n",
- "\n",
- "#Calculation\n",
- "v1=(R*(1-(1/n1**2)))\n",
- "v2=(R*(1-(1/n2**2)))\n",
- "V=v1/v2\n",
- "V1=V*v\n",
- "\n",
- "#Result\n",
- "print\"Wavelength of the second line is \", V1,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Wavelength of the second line is 1012.5 A\n"
- ]
- }
- ],
- "prompt_number": 136
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.5 Page no 41"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=1.097*10**7 #m-1\n",
- "n=2\n",
- "\n",
- "#Calculation\n",
- "v=n**2/(3.0*R)\n",
- "v1=1/R # for n=infinite\n",
- "\n",
- "#Result\n",
- "print\"longest wavelength is \",round(v*10**10,0),\"A\"\n",
- "print\"shortest wavelength is \",round(v1*10**10,1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "longest wavelength is 1215.0 A\n",
- "shortest wavelength is 911.6 A\n"
- ]
- }
- ],
- "prompt_number": 138
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.6 Page no 41"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=47.2 # 3ev\n",
- "n1=2\n",
- "n2 =3\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Z=math.sqrt(E/(13.6*((1/2.0**2)-(1/3.0**2))))\n",
- "\n",
- "#Result\n",
- "print\"Atomic number of the atom is \",round(Z,0)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Atomic number of the atom is 5.0\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.7 Page no 42"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Z=1.0\n",
- "n=1.0 #for the ground state of hydrogen\n",
- "Z1=4 #for Be++\n",
- "n1=2.0\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "n1=math.sqrt((n**2/Z)*Z1)\n",
- "r=(Z1**2/n1**2)/(Z**2/n**2) #Ratio of two energies\n",
- "\n",
- "#Result\n",
- "print\"nBe++= \", n1\n",
- "print\"comparison is \",r"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "nBe++= 2.0\n",
- "comparison is 4.0\n"
- ]
- }
- ],
- "prompt_number": 143
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.8 Page no 42"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Z=3.0\n",
- "n=3 #for Li++\n",
- "Z1=1.0\n",
- "n1=1 #for hydrogen\n",
- "\n",
- "#Calculation\n",
- "r=(n**2/Z)/(n1**2/Z1)\n",
- "\n",
- "#Result\n",
- "print\"orbital ratio of two states \",r"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "orbital ratio of two states 3.0\n"
- ]
- }
- ],
- "prompt_number": 144
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.9 Page no 42"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "v=970.6 #A\n",
- "h=6.63*10**-34 #Js \n",
- "c=3*10**8 #m/s\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "E=((h*c)/(v*e))*10**10\n",
- "En=-13.6+E\n",
- "n=math.sqrt(-13.6/En)\n",
- "E3=-13.6/(3.0**2)\n",
- "vmax=(h*c)/((-E3+En)*(1.6*10**-19))\n",
- "\n",
- "#Result\n",
- "print\"Longest wavelength is \",round(vmax*10**10),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Longest wavelength is 17292.0 A\n"
- ]
- }
- ],
- "prompt_number": 159
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.10 Page no 43"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Z=2\n",
- "E=13.6 #ev\n",
- "E0=10.04 #ev\n",
- "\n",
- "#Calculation\n",
- "Ei=Z**2*E\n",
- "E1=-Ei\n",
- "E3=E1/(3.0**2)\n",
- "Ee=E0+E3\n",
- "\n",
- "#Result\n",
- "print\"Required stopping potential is \", round(Ee,0),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Required stopping potential is 4.0 V\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.11 Page no 44"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Ei=4*2.2*10**-18 #Joule\n",
- "h=6.6*10**-34 #Js\n",
- "c=3*10**8 #m/s\n",
- "\n",
- "#Calculation\n",
- "E1=-Ei\n",
- "E2=E1/(2.0**2)\n",
- "v=(h*c)/(Ei+E2)\n",
- "\n",
- "#Result\n",
- "print\"Wavelength is \", round(v*10**10,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Wavelength is 300.0 A\n"
- ]
- }
- ],
- "prompt_number": 173
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.12 Page no 44"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n1=3\n",
- "n2 =1\n",
- "E=13.6 #ev\n",
- "\n",
- "#Calculation\n",
- "E1=E/(3.0**2) #Binding energy of the atom in n=3 state\n",
- "energy=E-E1 #Energy required for the atomic electron to jump from n=1 to n=3 state\n",
- "\n",
- "#Result\n",
- "print\"The electron beam must, therefore be accelerated through a potential difference of \",round(energy,2),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The electron beam must, therefore be accelerated through a potential difference of 12.09 V\n"
- ]
- }
- ],
- "prompt_number": 58
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.13 Page no 46"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Rh=1.09678*10**7 #m-1\n",
- "Rhe=1.09722*10**7 #m-1\n",
- "\n",
- "#Calculation\n",
- "Mr=(Rhe-Rh)/(Rh-(Rhe/4.0)) #ratio of electron mass\n",
- "\n",
- "#Result\n",
- "print\"Ratio of the electron mas to the proton mass \",round(Mr*10**4,2),\"*10**-4\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ratio of the electron mas to the proton mass 5.35 *10**-4\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Quantum_mechanics_by_M.C.Jain/chapter4.ipynb b/Quantum_mechanics_by_M.C.Jain/chapter4.ipynb deleted file mode 100755 index 214047e7..00000000 --- a/Quantum_mechanics_by_M.C.Jain/chapter4.ipynb +++ /dev/null @@ -1,329 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:b24ac6268abeb9e8ceafeb2b8fcac357229b810faa15684921ddb633803d3d99"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 4 Wave nature of matter and the need for a wave function"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.1 Page no 58"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given \n",
- "V=100 #volts\n",
- "\n",
- "#Calculation \n",
- "import math\n",
- "wavelength=12.3/(math.sqrt(V))\n",
- "\n",
- "#Result\n",
- "print\"de Broglie wavelength of electrons \", wavelength,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "de Broglie wavelength of electrons 1.23 A\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.2 Page no 58"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "K=100 #ev\n",
- "h=6.63*10**-34\n",
- "m=9.1*10**-31\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "v=h/(math.sqrt(2*m*K*e))\n",
- "\n",
- "#Result\n",
- "print\"de broglie wavelength of electrons \",round(v*10**10,1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "de broglie wavelength of electrons 1.2 A\n"
- ]
- }
- ],
- "prompt_number": 109
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.3 Page no 58"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=1.675*10**-27 #mass of neutron in kg\n",
- "v=1.4*10**-10 #de broglie wavelength in m\n",
- "h=6.63*10**-34 #Js\n",
- "\n",
- "#Calculation \n",
- "K=(h**2/(2*m*(v**2)))/(1.6*10**-19)\n",
- "\n",
- "#Result\n",
- "print\"Kinetic energy of neutron is \", round(K*10**2,2),\"*10**-2 ev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Kinetic energy of neutron is 4.18 *10**-2 ev\n"
- ]
- }
- ],
- "prompt_number": 113
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4 Page no 58"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=-3.4 #total energy in ev\n",
- "h=6.63*10**-34 #Js\n",
- "m=9.1*10**-31\n",
- "e=1.6*10**-19 \n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "K=-E\n",
- "v=h/(math.sqrt(2*m*K*e))\n",
- "\n",
- "#Result \n",
- "print\"(a) Kinetic energy \",K,\"ev\"\n",
- "print\"(b) de broglie wavelength of the electron is \",round(v*10**10,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Kinetic energy 3.4 ev\n",
- "(b) de broglie wavelength of the electron is 6.663 A\n"
- ]
- }
- ],
- "prompt_number": 64
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.5 Page no 59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=1.672*10**-27 #mass of neutron in kg\n",
- "h=6.60*10**-34 #Js\n",
- "v=1.0*10**-10 #de broglie wavelength in m\n",
- "\n",
- "#Calculation\n",
- "K=(h**2/(2.0*m*v**2))/(1.6*10**-19)\n",
- "\n",
- "#Result\n",
- "print\"Kinetic energy of a neutron is \", round(K*10**2,2),\"*10**-2 ev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Kinetic energy of a neutron is 8.14 *10**-2 ev\n"
- ]
- }
- ],
- "prompt_number": 116
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.6 Page no 59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=10*10**-3 #mass of a ball in kg \n",
- "v=1 #Speed in m/s\n",
- "h=6.63*10**-34 #Js\n",
- "\n",
- "#Calculation\n",
- "V=h/(m*v) #Wavelength\n",
- "\n",
- "#Result\n",
- "print \"de broglie wavelength is \",V,\"m\"\n",
- "print\"This wavelength is negligible compared to the dimensions of the ball. therefore its effect can not be observed.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "de broglie wavelength is 6.63e-32 m\n",
- "This wavelength is negligible compared to the dimensions of the ball. therefore its effect can not be observed.\n"
- ]
- }
- ],
- "prompt_number": 87
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.7 Page no 59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "T=27 #temperature in degree c\n",
- "K=1.38*10**-23 #boltzmann constant in J/K\n",
- "h=6.63*10**-34 #Js\n",
- "m=1.67*10**-27\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "T1=T+273\n",
- "v=h/(math.sqrt(2*m*K*T1))\n",
- "\n",
- "#Result\n",
- "print\"de broglie wavelength is \",round(v*10**10,2),\"A\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "de broglie wavelength is 1.78 A\n"
- ]
- }
- ],
- "prompt_number": 94
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.10 Page no 64"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=100 #ev\n",
- "a=10 #degree\n",
- "n=1\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "v=12.3/(math.sqrt(V)) #De broglie wavelength\n",
- "d=v/(2*math.sin(a*3.14/180.0))\n",
- "n=(2*d)/v\n",
- "\n",
- "#Result\n",
- "print\"(a) Spacing between the crystal plane is \", round(d,2),\"A\"\n",
- "print\"(b) Peaks in the interference pattern is \",round(n,2)\n",
- "print\"the largest possible value of n is 5\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Spacing between the crystal plane is 3.54 A\n",
- "(b) Peaks in the interference pattern is 5.76\n",
- "the largest possible value of n is 5\n"
- ]
- }
- ],
- "prompt_number": 117
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Quantum_mechanics_by_M.C.Jain/chapter5.ipynb b/Quantum_mechanics_by_M.C.Jain/chapter5.ipynb deleted file mode 100755 index 04b23240..00000000 --- a/Quantum_mechanics_by_M.C.Jain/chapter5.ipynb +++ /dev/null @@ -1,294 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:b28e13437e72a8d6bfe6e7e88f944bafb4762aecc71f80de95734c39265374bd"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 5 Wave packets and the uncertainty principles"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.2 Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "r=10.0**-14 #m\n",
- "h=1.054*10**-34 #Js\n",
- "m=1.67*10**-27\n",
- "\n",
- "#Calculation\n",
- "p=h/r\n",
- "E=(h**2/(2*m*(r**2)))/(1.6*10**-13)\n",
- "\n",
- "#Result\n",
- "print\"Kinetic energy \",round(E,2),\"Mev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Kinetic energy 0.21 Mev\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.3 Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=100 #ev\n",
- "m=9.1*10**-31\n",
- "e=1.6*10**-19\n",
- "x=10.0**-6 #m\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "p=math.sqrt(2*m*E*e)\n",
- "p1=h/x\n",
- "theta=p1/p\n",
- "\n",
- "#Result\n",
- "print\"uncertainty in the angle of emergence \", round(theta*10**4,1),\"*10**-4 radians\"\n",
- "print \"4 seconds of arc\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "uncertainty in the angle of emergence 0.2 *10**-4 radians\n",
- "4 seconds of arc\n"
- ]
- }
- ],
- "prompt_number": 44
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.4 Page no 90"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "p=0.2*10**-3*10 #Kg m/s\n",
- "h=1.054*10**-34\n",
- "x=1*10**-2 #m\n",
- "\n",
- "#Calculation\n",
- "p1=h/x\n",
- "a=p1/p\n",
- "\n",
- "#Result\n",
- "print\"uncertainty in the angle of emergence \",a,\"radians\"\n",
- "print \"1.1*10**-24 seconds of arc\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "uncertainty in the angle of emergence 5.27e-30 radians\n",
- "1.1*10**-24 seconds of arc\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5 Page no 90"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=50*10**-3 #kgram\n",
- "accuracy=0.01\n",
- "v=300 #m/s\n",
- "h=1.054*10**-34\n",
- "\n",
- "#Calculation\n",
- "p=m*(v*accuracy)/100.0\n",
- "x=h/p\n",
- "\n",
- "#Result\n",
- "print\"position of the bullet \",round(x,32),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "position of the bullet 7e-32 m\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.6 Page no 91"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "t=10.0**-12 #s\n",
- "h1=1.054*10**-34\n",
- "h=6.625*10**-34\n",
- "\n",
- "#Calculation\n",
- "E=h1/t\n",
- "v=E/h\n",
- "\n",
- "#Result\n",
- "print\"uncertainity in energy is \",E,\"J\"\n",
- "print\"uncertainity in frequency is \",round(v,-9),\"Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "uncertainity in energy is 1.054e-22 J\n",
- "uncertainity in frequency is 1.59e+11 Hz\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.8 Page no 92"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "r=5*10**-15 #m\n",
- "h=1.05*10**-34\n",
- "m=1.67*10**-27\n",
- "e=1.6*10**-13\n",
- "\n",
- "#Calculation\n",
- "xmax=2*r #maximum uncertainity in the position of the nucleon\n",
- "pmin=h/xmax #minimum uncertainity in the momentum of particle\n",
- "Kmin=pmin**2/(2.0*m*e)\n",
- "\n",
- "#Result\n",
- "print\"minimum kinetic energy is\",round(Kmin,1),\"Mev\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "minimum kinetic energy is 0.2 Mev\n"
- ]
- }
- ],
- "prompt_number": 70
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.10 Page no 92"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "v=6000*10**-10 #Wavelength in m\n",
- "t=10**-8 #s\n",
- "c=3*10**8\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "v1=v**2/(2.0*math.pi*c*t)\n",
- "\n",
- "#Result\n",
- "print\"width of a line \",round(v1,15),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "width of a line 1.9e-14 m\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Quantum_mechanics_by_M.C.Jain/chapter7.ipynb b/Quantum_mechanics_by_M.C.Jain/chapter7.ipynb deleted file mode 100755 index 50969726..00000000 --- a/Quantum_mechanics_by_M.C.Jain/chapter7.ipynb +++ /dev/null @@ -1,238 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:df09a01a18ea5b59bfa81431af3bf540f8785c24b7f1fddb025b95df17f9d0c1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 7 particle in a potential well"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.4 Page no 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=9.1*10**-31 #Kg\n",
- "h=1.05*10**-34 #Js\n",
- "ev=1.6*10**-19\n",
- "n1=1\n",
- "n2=2 \n",
- "n3=3\n",
- "a=10**-10 #m\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "E1=((n1**2*math.pi**2*h**2)/(8.0*m*a**2))/(1.6*10**-19) #ev\n",
- "E2=n2**2*E1\n",
- "E3=n3**2*E1\n",
- "\n",
- "#Result \n",
- "print\"three lowest energy levels are \",round(E1,1),\"ev,\", round(E2,1),\"ev and \",round(E3,2),\"ev\"\n",
- "print \"their eigenfunctions are 1/10**-5*cos(pie*x/2*10**-10),1/10**-5*sin(pie*x/10**-10) and 1/10**-5*cos(3*pie*x/2*10**-10)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "three lowest energy levels are 9.3 ev, 37.4 ev and 84.08 ev\n",
- "their eigenfunctions are 1/10**-5*cos(pie*x/2*10**-10),1/10**-5*sin(pie*x/10**-10) and 1/10**-5*cos(3*pie*x/2*10**-10)\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.5 Page no 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=10.0*10**-3 #kgram\n",
- "l= 10.0*10**-2 #Length in m\n",
- "h=1.054*10**-34\n",
- "n1=1\n",
- "n2=2 \n",
- "n3=3\n",
- "\n",
- "#Calculation\n",
- "E1=(((3.14*h*n1)**2)/(2.0*m*(l**2)))/(1.6*10**-19)\n",
- "E2=(((3.14*h*n2)**2)/(2.0*m*(l**2)))/(1.6*10**-19)\n",
- "E3=(((3.14*h*n3)**2)/(2.0*m*(l**2)))/(1.6*10**-19)\n",
- "\n",
- "#Result\n",
- "print\"energies are \", round(E1,46),\"ev,\",round(E2,46),\"ev,\",round(E3,45),\"ev\"\n",
- "print\"these energies are extremely small and close together and hence can't be measured\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "energies are 3.4e-45 ev, 1.37e-44 ev, 3.1e-44 ev\n",
- "these energies are extremely small and close together and hence can't be measured\n"
- ]
- }
- ],
- "prompt_number": 84
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7 Page no 136"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=10**-9 #Width in m\n",
- "v=9.0*10**-9\n",
- "h=1.054*10**-34 #Js\n",
- "c=3*10**8 #m/s\n",
- "m=9.1*10**-31\n",
- "v1=(9.0+1)*10**-9\n",
- "v2=(9.0-1)*10**-9\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "n=math.sqrt((4*c*m*(L**2))/(v*math.pi*h))\n",
- "n1=math.sqrt((4*c*m*(L**2))/(v1*math.pi*h))\n",
- "n2=math.sqrt((4*c*m*(L**2))/(v2*math.pi*h))\n",
- "\n",
- "#Result\n",
- "print\"value of n is \",round(n,0),\", When + sign is taken \",round(n2,0),\", when -ve sign is taken \",round(n1,0)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "value of n is 19.0 , When + sign is taken 20.0 , when -ve sign is taken 18.0\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.8 Page no 136"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L1=0.4\n",
- "L2=0.6\n",
- "L=1 #Say\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "dx=(L2-L1)*L\n",
- "#for ground state\n",
- "P1=2/L*(math.sin(math.pi*L/2.0*L))**2*dx\n",
- "#for first excited state\n",
- "P2=2/L*(math.sin(2*math.pi*L/2.0*L))**2*dx\n",
- "#for second excited state\n",
- "P3=2/L*(math.sin(3*math.pi*L/2.0*L))**2*dx\n",
- "\n",
- "#Result\n",
- "print\"(a) probability for ground state \", P1\n",
- "print\"(b) probability for first excited state \",round(P2,1)\n",
- "print\"(c) Probability for second excited state \", P3"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) probability for ground state 0.4\n",
- "(b) probability for first excited state 0.0\n",
- "(c) Probability for second excited state 0.4\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.9 Page no 140"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=10.0**-14 #m\n",
- "m=1.6*10**-27 #mass of a nucleon in kg\n",
- "h=1.054*10**-34 #Js\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Emin=((3*(math.pi**2)*(h**2))/(2.0*m*(a**2)))/(1.6*10**-19)\n",
- "\n",
- "#Result\n",
- "print\"minimum energy of a nucleon is \", round(Emin*10**-6,1),\"Mev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "minimum energy of a nucleon is 6.4 Mev\n"
- ]
- }
- ],
- "prompt_number": 17
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Quantum_mechanics_by_M.C.Jain/chapter8.ipynb b/Quantum_mechanics_by_M.C.Jain/chapter8.ipynb deleted file mode 100755 index 8c294517..00000000 --- a/Quantum_mechanics_by_M.C.Jain/chapter8.ipynb +++ /dev/null @@ -1,269 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:52aec0ea98abde6e8f04f9a7bf644564cf443f3e3280ece147b509c27e4e9a1b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 8 Scattering of psrticles by barriers and wells"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.2 page no 155 "
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "b=-32\n",
- "a=32.0\n",
- "c=1\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "r=(-b+(math.sqrt(b**2-(4*a*c))))/(2.0*a)\n",
- "\n",
- "#Result\n",
- "print\"The ratio of E/V0 = \", round(r*10**0)\n",
- "print \"-ve value is not possible. \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ratio of E/V0 = 1.0\n",
- "-ve value is not possible. \n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.3 Page no 156"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=9 #ev\n",
- "v0=5 #ev\n",
- "\n",
- "#Calculation\n",
- "R=((math.sqrt(E)-(math.sqrt(E-v0)))/(math.sqrt(E)+(math.sqrt(E-v0))))**2\n",
- "\n",
- "#Result\n",
- "print\"Reflection ratio is \", R"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Reflection ratio is 0.04\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.4 Page no 156"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=9 #Kinetic energy of a particle in ev\n",
- "v0=10 #ev\n",
- "E1=5 #ev\n",
- "E2=15\n",
- "E3=10 #ev\n",
- "\n",
- "#calculation\n",
- "import math\n",
- "R=((math.sqrt(E2)-(math.sqrt(E2-v0)))/(math.sqrt(E2)+(math.sqrt(E2-v0))))**2\n",
- "T=1-R\n",
- "\n",
- "#Result\n",
- "print\"(a) E1 < vo, therefore R=1, T=0\"\n",
- "print\"(b) reflection coefficient R= \",round(R,3),\"\\n transmission coefficient T= \",round(T,3)\n",
- "print\"(c) E3=v0, therefore R=1 , T=0\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) E1 < vo, therefore R=1, T=0\n",
- "(b) reflection coefficient R= 0.072 \n",
- " transmission coefficient T= 0.928\n",
- "(c) E3=v0, therefore R=1 , T=0\n"
- ]
- }
- ],
- "prompt_number": 52
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.6 Page no 163"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=2 #ev\n",
- "v0=3 #ev\n",
- "m=9*10**-31\n",
- "a=4*10**-10 #m\n",
- "h=1.05*10**-34\n",
- "b=(v0-E)*(1.6*10**-19)\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Ka=((math.sqrt(2*m*(b)))*a)/h\n",
- "x=math.sin(Ka*3.14/180.0)\n",
- "T=(v0**2)/(4.0*E*(v0-E))\n",
- "T1=1/(1+(T*x**2))\n",
- "\n",
- "#Result\n",
- "print\"Transmission coefficient is \", round(T1,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Transmission coefficient is 0.999\n"
- ]
- }
- ],
- "prompt_number": 103
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.7 Page no 164"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=2 #ev\n",
- "v0=3 #ev\n",
- "m=9*10**-31\n",
- "a=1*10**-10 #m\n",
- "h=1.05*10**-34\n",
- "b=(v0-E)*(1.6*10**-19)\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Ka=((math.sqrt(2*m*(b)))*a)/h\n",
- "x=math.sin(Ka*3.14/180.0)\n",
- "T=(v0**2)/(4.0*E*(v0-E))\n",
- "T1=1/(1.0+(T*x))\n",
- "\n",
- "#Result\n",
- "print\"Transmission coefficient is \", round(T1,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Transmission coefficient is 0.99\n"
- ]
- }
- ],
- "prompt_number": 94
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8 Page no 164"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=10*10**6 #ev\n",
- "T=2.0*10**-3\n",
- "m=6.68*10**-27 #kg\n",
- "h=1.054*10**-34 #Js\n",
- "e=1.6*10**-19\n",
- "v0=30.0*10**6 #ev\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "K=(math.sqrt(2*m*(v0-E)*e))/h\n",
- "a=(1/(2.0*K))*(2.303*log10((16/T)*(E/v0)*(1-(E/v0))))\n",
- "\n",
- "#Result\n",
- "print\"The width of the barrier is \", round(a,17),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The width of the barrier is 1.91e-15 m\n"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file |