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| author | nice | 2014-08-27 16:12:51 +0530 |
|---|---|---|
| committer | nice | 2014-08-27 16:12:51 +0530 |
| commit | f873023db6ddb02bba555fb650a4b4c90340f56a (patch) | |
| tree | b866cee9b099fe202c72538b5be2a099d0320a24 /Modern_Physics | |
| parent | 728bf707ac994b2cf05a32d8985d5ea27536cf34 (diff) | |
| download | Python-Textbook-Companions-f873023db6ddb02bba555fb650a4b4c90340f56a.tar.gz Python-Textbook-Companions-f873023db6ddb02bba555fb650a4b4c90340f56a.tar.bz2 Python-Textbook-Companions-f873023db6ddb02bba555fb650a4b4c90340f56a.zip | |
adding book
Diffstat (limited to 'Modern_Physics')
| -rwxr-xr-x | Modern_Physics/Chapter1.ipynb | 404 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter10.ipynb | 214 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter11.ipynb | 141 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter12.ipynb | 180 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter13.ipynb | 501 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter14.ipynb | 349 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter15.ipynb | 63 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter16.ipynb | 62 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter2.ipynb | 418 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter3.ipynb | 496 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter4.ipynb | 447 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter5.ipynb | 295 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter6.ipynb | 268 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter7.ipynb | 230 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter8.ipynb | 222 | ||||
| -rwxr-xr-x | Modern_Physics/Chapter9.ipynb | 148 | ||||
| -rwxr-xr-x | Modern_Physics/README.txt | 10 | ||||
| -rwxr-xr-x | Modern_Physics/screenshots/hubbleslaw.png | bin | 0 -> 45811 bytes | |||
| -rwxr-xr-x | Modern_Physics/screenshots/orbquant.png | bin | 0 -> 50541 bytes | |||
| -rwxr-xr-x | Modern_Physics/screenshots/surtemp.png | bin | 0 -> 56453 bytes |
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diff --git a/Modern_Physics/Chapter1.ipynb b/Modern_Physics/Chapter1.ipynb new file mode 100755 index 00000000..d442fd97 --- /dev/null +++ b/Modern_Physics/Chapter1.ipynb @@ -0,0 +1,404 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:d8ffd8fd29d5d24b3cc8b2545e898e818e9d8ef837edfe4af93ac3d5f1b9fea0" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1: RELATIVITY" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.2, page no. 18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "#Variable declaration\n", + "T1 = 3.0 #proper time(s)\n", + "c = 3 * 10**8 #Speed of light(m/s)\n", + "v = 0.95 * c #speed of observer(m/s)\n", + "\n", + "#calculation\n", + "\n", + "Y = 1/math.sqrt((1-(v/c)**2)) #factor by which it runs slower\n", + "T = Y * T1 #Period of the pendulum (s)\n", + "\n", + "#result\n", + "print \"The Period of the pendulum when measured by the observer is\",round(T,1),\"s\"\n", + "\n", + "\n", + "#Variable declaration\n", + "T1 = 3.0 #proper time(s)\n", + "c = 3 * 10**8 #Speed of light(m/s)\n", + "Vnew = (0.95 + 0.05*0.95) * c #speed of observer(m/s)\n", + "\n", + "#calculation\n", + "\n", + "Y = 1/math.sqrt((1-(Vnew/c)**2))\n", + "T = Y * T1 \n", + "\n", + "#result\n", + "print \"Exercise:The Period of the pendulum when measured by the observer is\",round(T,1),\"s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Period of the pendulum when measured by the observer is 9.6 s\n", + "Exercise:The Period of the pendulum when measured by the observer is 42.5 s\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3, page no. 20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#variable declaration\n", + "\n", + "Lp = 100 #proper length of the spaceship(m)\n", + "c = 3 * 10**8 #speed of light(m/s)\n", + "v = 0.99 * c #speed of the observer(m/s)\n", + "\n", + "#calculation\n", + "\n", + "L = Lp * math.sqrt((1-(v/c)**2)) #Length of the spaceship (m)\n", + "\n", + "#result\n", + "\n", + "print \"The length measured as the spaceship flies by is\",round(L),\"m\"\n", + "\n", + "#variable declaration\n", + "\n", + "Lp = 100 #proper length of the spaceship(m)\n", + "c = 3 * 10**8 #speed of light(m/s)\n", + "v = 0.01 * c #speed of the observer(m/s)\n", + "\n", + "#calculation\n", + "\n", + "L = Lp * math.sqrt((1-(v/c)**2))\n", + "\n", + "#result\n", + "\n", + "print \"Exercise:The length measured as the spaceship flies by is\",round(L,2),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The length measured as the spaceship flies by is 14.0 m\n", + "Exercise:The length measured as the spaceship flies by is 99.99 m\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.4, page no. 20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#variable declaration\n", + "\n", + "Lp = 435 #proper length of the spaceship(m)\n", + "c = 3 * 10**8 #speed of light(m/s)\n", + "v = 0.97 * c #speed of the spaceship(m/s)\n", + "\n", + "#calculation\n", + "\n", + "L = Lp * math.sqrt((1-(v/c)**2)) #altitude of the spaceship (m)\n", + "\n", + "#result\n", + "\n", + "print \"The altitude of the spaceship as measured by an observer in the spaceship is\",round(L),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The altitude of the spaceship as measured by an observer in the spaceship is 106.0 m\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5, page no. 20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#variable declaration\n", + "\n", + "Lp = 50 #horizontal length of the spaceship (m)\n", + "c = 3 * 10**8 #speed of light (m/s)\n", + "v = 0.95 * c #speed of the observer (m/s)\n", + "\n", + "#calculation\n", + "\n", + "L = Lp * math.sqrt((1-(v/c)**2)) #length (m)\n", + "\n", + "#result\n", + "\n", + "print \"The horizontal length measured as seen by the observer\",round(L,1),\"m.\\nThe 25m vertical height is unchanged because it is perpendicular to the direction of relative motion between the observer and the spaceship.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The horizontal length measured as seen by the observer 15.6 m.\n", + "The 25m vertical height is unchanged because it is perpendicular to the direction of relative motion between the observer and the spaceship.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6, page no. 25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#variable declaration\n", + "\n", + "Yobs = 475.0 #observed wavelength (nm)\n", + "Ysource = 394.0 #source wavelength (nm)\n", + "c = 3 * 10**8 #speed of light (m/s)\n", + "\n", + "#calculation\n", + "\n", + "v = c*(((Yobs**2)-(Ysource**2))/((Yobs**2)+(Ysource**2))) #velocity (m/s)\n", + "\n", + "#result\n", + "\n", + "print \"The Hydra is receding from us at\",round(v/10**7,2),\"x 10^7 m/s\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Hydra is receding from us at 5.54 x 10^7 m/s\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.8, page no. 30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "\n", + "c = 3 * 10 **8 #speed of light (m/s)\n", + "v = 0.750 * c #speed of spacecraft A relative to an observer on Earth(m/s)\n", + "ux = -0.850 *c #speed of spacecraft B relative to an observer on Earth(m/s)\n", + "\n", + "#calculation\n", + "\n", + "ux1 = (ux-v)/(1-((ux*v)/(c**2))) #velocity (m/s)\n", + "\n", + "#results\n", + "\n", + "print \"The velocity of B with respect to A is\",round(ux1/c,4),\"c\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The velocity of B with respect to A is -0.9771 c\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.9, page no. 30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "\n", + "c = 3 * 10 **8 #speed of light (m/s)\n", + "v = 0.800 * c #velocity of the motorcycle with respect to the stationary observer (m/s)\n", + "ux1 = 0.700 *c #velocity of the ball in the frame of reference of the motorcyclist (m/s)\n", + "\n", + "#calculation\n", + "\n", + "ux = (ux1+v)/(1+((ux1*v)/(c**2))) #velocity (m/s)\n", + "\n", + "#results\n", + "\n", + "print \"The velocity of the ball relative to the stationary observer is\",round(ux/c,4),\"c\"\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "c = 3 * 10 **8 #speed of light (m/s)\n", + "v = 0.800 * c #velocity of the motorcycle with respect to the stationary observer (m/s)\n", + "ux1 = c #velocity of the beam of light in the frame of reference of the motorcyclist (m/s)\n", + "\n", + "#calculation\n", + "\n", + "ux = (ux1+v)/(1+((ux1*v)/(c**2))) #velocity (m/s)\n", + "\n", + "#results\n", + "\n", + "print \"The speed of the beam of light relative to the stationary observer is\",round(ux/c,4),\"c\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The velocity of the ball relative to the stationary observer is 0.9615 c\n", + "The speed of the beam of light relative to the stationary observer is 1.0 c\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.10, page no. 30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "c = 3 * 10 **8 #speed of light (m/s)\n", + " #observations of stationary police officer\n", + "uxa = 0.75 * c # velocity of pack leader Alpha in x direction (m/s)\n", + "uya = 0 # velocity of pack leader Alpha in y direction (m/s)\n", + "uxb = 0 # velocity of pack leader Beta in x direction (m/s)\n", + "uyb = -0.9 * c # velocity of pack leader Beta in y direction (m/s) \n", + " \n", + "#calculation\n", + "\n", + "ux1 = (uxb-uxa)/(1-((uxb*uxa)/(c**2))) #speed in X axis (m/s)\n", + "uy1 = (math.sqrt(1-((uxa**2)/(c**2)))*(uyb))#/(1-((uxb*uxa)/(c**2))) #speed in Y axis (m/s)\n", + "u1 = math.sqrt(ux1**2 + uy1**2) #total speed (m/s)\n", + "\n", + "#results\n", + "\n", + "print \"The speed of recession of Beta away from Alpha as observed by Alpha is then found to be\",round(u1/c,2),\"c\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The speed of recession of Beta away from Alpha as observed by Alpha is then found to be 0.96 c\n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter10.ipynb b/Modern_Physics/Chapter10.ipynb new file mode 100755 index 00000000..a9cbc233 --- /dev/null +++ b/Modern_Physics/Chapter10.ipynb @@ -0,0 +1,214 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1d13a558dd2cc62e2c7ada350fc7a07d9283efcbc790578d0711fd6c96f50df0" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10: Statistical Physics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.1, page no. 340" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "n1 = 1 #Ground state\n", + "n2 = 2 #First excited state\n", + "n3 = 3 #second excited state\n", + "T = 300 #room temperature(K)\n", + "kb = 8.617 * 10 **-5 #Boltzmann constant(eV/K)\n", + "\n", + "#Calculation\n", + "\n", + "E1 = -13.6 / n1 ** 2\n", + "g1 = 2 * n1 ** 2\n", + "E2 = -13.6 / n2 ** 2\n", + "g2 = 2 * n2 ** 2\n", + "E3 = -13.6 / n3 ** 2\n", + "g3 = 2 * n3 ** 2\n", + "N3 = g3 * math.exp(-E3/(kb*T))\n", + "N2 = g2 * math.exp(-E2/(kb*T))\n", + "N1 = g1 * math.exp(-E1/(kb*T))\n", + "ratio1 = N2 / N1\n", + "ratio2 = N3 / N1\n", + "\n", + "#results\n", + "\n", + "print \"(a) We can see that n2/n1=\",round(ratio1),\"and n3/n1=\",round(ratio2),\"essentially all atoms are in ground state.\"\n", + "\n", + "\n", + "#Variable Declaration\n", + "\n", + "T = 20000 #Temperature(K)\n", + "\n", + "#Calculation\n", + "\n", + "N3 = g3 * math.exp(-E3/(kb*T))\n", + "N2 = g2 * math.exp(-E2/(kb*T))\n", + "N1 = g1 * math.exp(-E1/(kb*T))\n", + "ratio1 = N2 / N1\n", + "ratio2 = N3 / N1\n", + "\n", + "#result\n", + "\n", + "print \"(b) n2/n1=\",round(ratio1,5),\"and n3/n1=\",round(ratio2,5)\n", + "\n", + "\n", + "ratio3 = N3 / N2\n", + "\n", + "print \"(c) S(3->2)/S(2->1)=\",round(ratio3,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) We can see that n2/n1= 0.0 and n3/n1= 0.0 essentially all atoms are in ground state.\n", + "(b) n2/n1= 0.01076 and n3/n1= 0.00809\n", + "(c) S(3->2)/S(2->1)= 0.75\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.2, page no. 345" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable Declaration\n", + "\n", + "m = 3.34 * 10 ** -27 #mass of hydrogen(kg)\n", + "kbT = 3.77 * 10 ** -21 #kb * T (eV)\n", + "N = 6.02 * 10 ** 23 #avogadro's number\n", + "V = 22.4 * 10 ** -3 #Volume of H2 gas (m^3)\n", + "h = 1.055 * 10 ** -34 #Planck's constant (J.s)\n", + "\n", + "#Calculation\n", + "\n", + "r = (N/V)* h ** 3 / (8 * (m * kbT)**1.5)\n", + "\n", + "#result\n", + "\n", + "print \"(a) (N/V)h^3/(8*(mkbT)^3/2)=\",round(r/10**-8,2),\"X10^-8 is much less than 1, we conclude that even hydrogen is described by Maxwell-Boltzmann statistics.\"\n", + "\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "d = 10.5 #density of silver (g/cm^3)\n", + "mw = 107.9 #Molar weight of silver (g/mol)\n", + "me = 9.109*10**-31 #mass of electron(kg)\n", + "kbT = 4.14 * 10 ** -21 #kb*T (J)\n", + "\n", + "#Calculation\n", + "\n", + "Ns = (d/mw)* N * 10 ** 6\n", + "r = (Ns)* h ** 3 / (8 * (me * kbT)**1.5)\n", + "\n", + "#result\n", + "\n", + "print \"(b) (N/V)h^3/(8*(mkbT)^3/2)=\",round(r/8,2),\"is greater than 1, we conclude that the Maxwell-Boltzmann statistics does not hold for electrons of silver.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) (N/V)h^3/(8*(mkbT)^3/2)= 8.83 X10^-8 is much less than 1, we conclude that even hydrogen is described by Maxwell-Boltzmann statistics.\n", + "(b) (N/V)h^3/(8*(mkbT)^3/2)= 4.64 is greater than 1, we conclude that the Maxwell-Boltzmann statistics does not hold for electrons of silver.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.3, page no. 352" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "\n", + "import math\n", + "\n", + "#Variable Declaration\n", + "\n", + "kB = 8.62 * 10 ** -5 #Boltzmann constant(eV/K)\n", + "T1 = 3000 #Cavity walls temperature(K)\n", + "T2 = 3.00 #Cavity walls temperature(K)\n", + "hc = 1.24 * 10 ** -4 #product of planck's constant and speed of light (eV.cm)\n", + "integration = 2.40 #value of integral(z^2/e^z-1,0,+inf)\n", + "\n", + "#Calculation\n", + "\n", + "NbyV_at_3000 = 8* math.pi * (kB * T1/hc)**3 * integration\n", + "NbyV_at_3 = 8* math.pi * (kB * T2/hc)**3 * integration\n", + "\n", + "#result\n", + "\n", + "print \"N/V at 3000K is\",round(NbyV_at_3000/10**11,2),\"X 10^11 photons/cm^3. Likewise N/V at 3.00 K is\",round(NbyV_at_3/10**2,2),\"X 10^2 photons/cm^3\"\n", + "print \"Therefore the photon density decreases by a factor of\",round(NbyV_at_3000/NbyV_at_3/10**9),\" X 10^9 when the temperature drops from 3000K to 3.00K\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "N/V at 3000K is 5.47 X 10^11 photons/cm^3. Likewise N/V at 3.00 K is 5.47 X 10^2 photons/cm^3\n", + "Therefore the photon density decreases by a factor of 1.0 X 10^9 when the temperature drops from 3000K to 3.00K\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter11.ipynb b/Modern_Physics/Chapter11.ipynb new file mode 100755 index 00000000..65f716cb --- /dev/null +++ b/Modern_Physics/Chapter11.ipynb @@ -0,0 +1,141 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e6be7abf8c97801180567f4f6cb83dc9e729f9c81204b1adfe4e1f54ff45dd36" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 11: Molecular Structure" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.1, page no. 380" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#Variable Declaration\n", + "\n", + "f = 1.15 * 10 ** 11 # Frequency(Hz)\n", + "h = 1.055 * 10 ** -34 # Planck's constant(J.s)\n", + " \n", + "#Calculation\n", + "\n", + "w = 2 * math.pi * f\n", + "Icm = h / w\n", + "\n", + "#Results\n", + "\n", + "print \"(a) The moment of inertia is\",round(Icm/10 **-46 , 2),\"X 10^-46 kg.m^2.\"\n", + "\n", + "\n", + "\n", + "#Variable Declaration\n", + "\n", + "Ac = 12.0 #atomic mass of carbon (u)\n", + "Ao = 16.0 #atomic mass of oxygen (u)\n", + "u = 1.66 * 10 ** -27#(kg)\n", + "\n", + "#Calculation\n", + "\n", + "mu = Ac* Ao /(Ac+Ao)\n", + "R0 = math.sqrt(Icm/(mu*u))\n", + "\n", + "#Results\n", + "\n", + "print \"(b) The bond length of the molecule is\",round(R0/10**-9,3),\"nm.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) The moment of inertia is 1.46 X 10^-46 kg.m^2.\n", + "(b) The bond length of the molecule is 0.113 nm.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.2, page no. 383" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable Declaration\n", + "\n", + "f = 6.42 * 10 ** 13 #frequency(Hz)\n", + "mu = 1.14 * 10 ** -26 #(kg)\n", + " \n", + "#Calculation\n", + "\n", + "w = 2 * math.pi * f\n", + "K = mu * w**2\n", + "\n", + "#Results\n", + "\n", + "print \"(a) The force constant for this molecule is \",round(K/10**3,2),\"X 10^3 N/m.\"\n", + "\n", + "\n", + "\n", + "#Variable Declaration\n", + "\n", + "h = 1.055 * 10 ** -34 #Planck's constant (J.s)\n", + "\n", + "#Calculation\n", + "\n", + "A = (h/(mu * w)) ** 0.5\n", + "\n", + "#Results\n", + "\n", + "print \"(b) The Vibration amplitude is\",round(A/10**-9,5),\"nm.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) The force constant for this molecule is 1.85 X 10^3 N/m.\n", + "(b) The Vibration amplitude is 0.00479 nm.\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter12.ipynb b/Modern_Physics/Chapter12.ipynb new file mode 100755 index 00000000..3cc3e98a --- /dev/null +++ b/Modern_Physics/Chapter12.ipynb @@ -0,0 +1,180 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:6dfe8c3f0e395e550057fecf65810f55f95ea60853478f130ac485dd9541147f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12: The Solid State" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.1, page no. 418" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable Declaration\n", + "\n", + "kb = 1.38 * 10 **-23 #Boltzmann constant (J/K)\n", + "T = 300 # room temperature (K)\n", + "me = 9.11 * 10 ** -31 # mass of electron (kg)\n", + "d = 8.96 * 10**6 # density of copper (g/m^3)\n", + "N = 6.023 * 10 **23 #avagodro's number (atoms/mole)\n", + "mw = 63.5 #molar weight (g)\n", + "A = 4 * 10 ** -6 #area (m^2)\n", + "I = 10 # current (A)\n", + "e = 1.6 * 10 ** -19 # charge of electron(C)\n", + "\n", + "#Calculation\n", + "\n", + "Vrms = (3*kb*T/me)**0.5\n", + "n = N * d / mw\n", + "Vd = I /(n*e*A)\n", + "\n", + "#Results\n", + "\n", + "print \"(a) The ratio of drift speed to rms speed is \",round(Vd/Vrms/10**-9,2),\"X 10^-9.\"\n", + "\n", + "\n", + "#Variable Declaration\n", + "\n", + "L = 2.6 * 10 ** -10 #interatomic distance(A') \n", + "\n", + "#Calculation\n", + "\n", + "t = L / Vrms \n", + "\n", + "#Results\n", + "\n", + "print \"(b) The average time between collisions\",round(t/10**-15,1),\"X 10^-15 m/s.\"\n", + "\n", + "\n", + "#Variable Declaration\n", + "\n", + "T = 300 #Room temperature (K)\n", + "\n", + "#Calculation\n", + "\n", + "sigma = n*e**2 * L /(3*kb*T*me)**0.5\n", + "\n", + "#Results\n", + "\n", + "print \"(c) The conductivity is\",round(sigma/10**6,1),\"X 10^6 (ohm.m)^-1.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) The ratio of drift speed to rms speed is 1.57 X 10^-9.\n", + "(b) The average time between collisions 2.2 X 10^-15 m/s.\n", + "(c) The conductivity is 5.3 X 10^6 (ohm.m)^-1.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.2, page no. 429" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable Declaration\n", + "\n", + "V = 7 # voltage(V)\n", + "L = 5 * 10 ** -8 # mean free path (m)\n", + " \n", + "#Calculation\n", + "\n", + "E = V/L\n", + "\n", + "#Results\n", + "\n", + "print \"The electric field required is\",round(E/10**8,1),\"X 10^8 V/m.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electric field required is 1.4 X 10^8 V/m.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.3, page no. 436" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable Declaration\n", + "\n", + "V = 1 #voltage(eV)\n", + "kBT = 0.025 # kB * T (eV)\n", + "\n", + "#Calculation\n", + "\n", + "ratio = (math.exp(V/kBT)-1)/(math.exp(-V/kBT)-1)\n", + "\n", + "#Results\n", + "\n", + "print \"The ratio of forward to reverse current is\",round(ratio/10**17,1),\"X 10^17.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ratio of forward to reverse current is -2.4 X 10^17.\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter13.ipynb b/Modern_Physics/Chapter13.ipynb new file mode 100755 index 00000000..e35d94ac --- /dev/null +++ b/Modern_Physics/Chapter13.ipynb @@ -0,0 +1,501 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:8b9d9c8500ecb20deeeb6f065d1408d228b300a0b95f2800ee645e0184760d45" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 13: Nuclear Structure " + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.1, page no. 466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Variable Declaration\n", + "\n", + "NA = 6.02 * 10 ** 23 # Avogadro's number\n", + "m = 0.012 # one mole of carbon (kg)\n", + "mC = 12 # mass of one atom(1u)\n", + " \n", + "#Calculation\n", + "\n", + "ma = m/NA\n", + "u = ma / mC\n", + "\n", + "#Results\n", + "\n", + "print \"The atomic mass unit is \",round(u/10**-27,2),\"X 10^-27 kg.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The atomic mass unit is 1.66 X 10^-27 kg.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.2, page no. 468" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#Variable Declaration\n", + "\n", + "m = 1.67 * 10 ** -27 # mass of neutron(kg)\n", + "r0 = 1.2 * 10 ** -15 # radius os the nucleus(m)\n", + "\n", + "#Calculation \n", + "\n", + "pn = 3* m /(4 * math.pi * r0**3)\n", + "\n", + "#Results\n", + "\n", + "print \"The nuclear density is\",round(pn/10**17,1),\"X 10^17 kg/m^3.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The nuclear density is 2.3 X 10^17 kg/m^3.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.3, page no. 473" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "\n", + "#Variable Declaration\n", + "\n", + "MH = 1.007825 #mass of hydrogen (u)\n", + "mn = 1.008665 #mass of neutron (u)\n", + "M2 = 2.014102 #mass of deuteron (u)\n", + "\n", + "#Calculation\n", + "\n", + "Eb = (MH + mn - M2) * 931.494\n", + "\n", + "#results\n", + "\n", + "print \"The binding energy of deuteron is\",round(Eb,3),\"MeV.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The binding energy of deuteron is 2.224 MeV.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.4, page no. 482" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "\n", + "Thalf = 5730 #Half-life (yr)\n", + "N = 1000 #No of carbon nuclei\n", + "T = 22920 #(yr)\n", + "\n", + "#calculation\n", + "n=N\n", + "t = 0\n", + "while(t!=T):\n", + " n = n/2\n", + " t = t + Thalf\n", + "\n", + "#result\n", + "\n", + "print \"The number of Carbon nuclei left after 22,920 yr is\",n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of Carbon nuclei left after 22,920 yr is 62\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.5, page no. 483" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "Thalf = 1.6 * 10 ** 3 #Half-life (yr)\n", + "s = 3.16 * 10 ** 7 #number of seconds in a year (s/yr)\n", + "\n", + "#Calculation\n", + "\n", + "lamda = 0.693 / (Thalf * s)\n", + "\n", + "#result\n", + "\n", + "print \"(a) The decay constant is\",round(lamda/10**-11,1),\"X 10^-11 s^-1.\"\n", + "\n", + "\n", + "#Variable Declaration\n", + "\n", + "N0 = 3.0 * 10 ** 16 #number of radioactive nuclei at t=0\n", + "Ci = 3.7 * 10 **10 \n", + "\n", + "#Calculation\n", + "\n", + "R0 = lamda * N0\n", + "\n", + "#results\n", + "\n", + "print \"(b) Its activity is\",round(R0/Ci/10**-6,1),\"X 10^-6 Ci.\"\n", + "\n", + "\n", + "#Variable Declaration\n", + "\n", + "T = 2.0 * 10 ** 3 #(yr)\n", + "\n", + "#Calculation\n", + "\n", + "R = R0 * math.exp(-lamda*(T * s))\n", + "\n", + "#results\n", + "\n", + "print \"(c) The decay rate after 2.0 X 10^3 yr is\",round(R/10**5,1),\"X 10^5 decays/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) The decay constant is 1.4 X 10^-11 s^-1.\n", + "(b) Its activity is 11.1 X 10^-6 Ci.\n", + "(c) The decay rate after 2.0 X 10^3 yr is 1.7 X 10^5 decays/s\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.6, page no. 483" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable Declaration\n", + "\n", + "A = 6.02 * 10 ** 23 #avogadro's number\n", + "m = 3.50 * 10 ** -6 # mass of carbon(g)\n", + "ma = 11.0 #atomic mass of carbon (g)\n", + "\n", + "#Calculation\n", + "\n", + "N = m * A / ma\n", + "\n", + "#Result\n", + "\n", + "print \"(a) The number of nuclei samples at t=0 is\",round(N/10**17,2),\"X 10^17 nuclei.\"\n", + "\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "Thalf = 20.4 * 60 #half-life (s)\n", + "T = 8.00 #(h)\n", + "\n", + "#Calculation\n", + "\n", + "lamda = 0.693 / Thalf\n", + "R0 = lamda * N\n", + "R = R0 * math.exp(-lamda* T*60*60)\n", + "\n", + "#result\n", + "\n", + "print \"(b) The activity initially is\",round(R0/10**14,2),\"X 10^14 decay/s and after t=8.0h is\",round(R/10**6,2),\"X 10^6 decay/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) The number of nuclei samples at t=0 is 1.92 X 10^17 nuclei.\n", + "(b) The activity initially is 1.08 X 10^14 decay/s and after t=8.0h is 8.99 X 10^6 decay/s\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.7, page no. 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable Declaration\n", + "\n", + "Thalf = 8.04 # half-life (days)\n", + "R0 = 5.0 #Activity at t=0 (mCi)\n", + "R = 4.2 #Activity (mCi)\n", + "\n", + "#Calculation\n", + "\n", + "t = - (Thalf/0.693)* math.log(R/R0)\n", + "\n", + "#Result\n", + "\n", + "print \"The time elapsed is\",round(t,2),\"days.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The time elapsed is 2.02 days.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.8, page no. 486" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "\n", + "#Variable Declaration\n", + "\n", + "Mx = 226.025406 #atomic mass of 226Ra(u)\n", + "My = 222.017574 #atomic mass of 222Rn(u)\n", + "Ma = 4.002603 #atomic mass of 4He (u)\n", + "\n", + "#Calculation\n", + "\n", + "Q = (Mx - My - Ma) * 931.494\n", + "\n", + "#Result\n", + "\n", + "print \"Q value is\",round(Q,2),\"MeV.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q value is 4.87 MeV.\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.9, page no. 487" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "#Variable declaration\n", + "\n", + "r0 = 7.25 * 10**-5 #Bohr radius for alpha particle(A')\n", + "Z = 86 #daughter nucleus radon's atomic number\n", + "A = 222 #radon's mass number\n", + "ke2 = 14.40 #boltzmann constant X (charge of electron)^2 (eV.A')\n", + "E = 5 #disintegration energy (MeV)\n", + "\n", + "#Calculation\n", + "\n", + "E0 = ke2 / (2.0*r0) * 10**-6 #Energy unit analogous to Rydberg (MeV)\n", + "R = (1.2 * 10**-5) *(A)**(1.0/3.0) #radius of Radon nucleus(A')\n", + "Te = math.exp(round(-4*math.pi*Z*math.sqrt(E0/E)+8*math.sqrt(Z*R/r0)))\n", + "\n", + "#result\n", + "print \"The probability of escape of alpha particle is\",round(Te/10**-34,2),\"X 10^-34\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The probability of escape of alpha particle is 1.33 X 10^-34\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.11, page no. 490" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable Declaration\n", + "\n", + "Thalf = 5730 # half-life (yr)\n", + "s = 3.16 * 10 ** 7 # (s/yr)\n", + "A = 6.02 * 10 ** 23 # Avogadro's number\n", + "mw = 12.0 # molar weight of carbon(g)\n", + "m = 25.0 # mass of carbon(g) \n", + "r = 1.3 * 10 ** -12 # ratio of 14C to 12C\n", + "R = 250 # activity observed (decays/min)\n", + "\n", + "#Calculation\n", + "\n", + "lamda = 0.693 /(Thalf * s)\n", + "N1 = A * m/mw\n", + "N0 = r * N1\n", + "R0 = N0 * lamda\n", + "t = -(1/lamda)*math.log(R/(R0*60))\n", + " \n", + "#Result\n", + "\n", + "print \"The tree has been dead for\",round(round(t/10**11)*10**11/s/10**3,1),\"X 10^3 yr.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The tree has been dead for 3.2 X 10^3 yr.\n" + ] + } + ], + "prompt_number": 25 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter14.ipynb b/Modern_Physics/Chapter14.ipynb new file mode 100755 index 00000000..4fff498c --- /dev/null +++ b/Modern_Physics/Chapter14.ipynb @@ -0,0 +1,349 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:bb0d2851c786dd28ae3ad8cbd1e6a7fc5db7cc1384cc2031f2bb6380ed7515e2" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 14: Nuclear Physics Apllications" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.1, page no. 505" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Variable declaration\n", + "\n", + "MLi = 7.016003 #atomic mass of Lithium\n", + "MH = 1.007825 #atomic mass of Hydrogen\n", + "MHe = 4.002603 #atomic mass of Helium\n", + "c2 = 931.50 #Square of speed of light (MeV/u)\n", + "\n", + "#Calculation\n", + "\n", + "Q = (MLi + MH - 2*MHe) * c2\n", + "\n", + "#Results\n", + "\n", + "print \"(a) The Q value is\",round(Q,1),\"MeV.\"\n", + "\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "Kincident = 0.6 #kinetic energy of incident particle (MeV)\n", + "\n", + "#Calculation\n", + "\n", + "Kproducts = Q + Kincident\n", + "\n", + "#Results\n", + "\n", + "print \"(b) The kinetic energy of the products is\",round(Kproducts,1),\"MeV.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) The Q value is 17.3 MeV.\n", + "(b) The kinetic energy of the products is 17.9 MeV.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.2, page no. 509" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Variable declaration\n", + "\n", + "A = 27 #Atomic number of Aluminum\n", + "d = 2.7 #density of aluminum\n", + "Av = 6.02 * 10 ** 23 #Avogadro number nuclei/mol\n", + "sigma = 2.0 * 10 **-31 #capture cross section (m^2)\n", + "x = 0.3 * 10 ** -3 #thickness of the foil(m)\n", + "R0 = 5.0 * 10 ** 12 #rate of incident particles(neutrons/cm^2.s)\n", + "\n", + "#Calculation\n", + "\n", + "n = Av * d / A * 10**6\n", + "R = R0 * sigma * x * n\n", + "\n", + "#Result\n", + "\n", + "print \"The number of neutrons captured by 1.0cm^2/s is \",round(R/10**7,1),\"X 10^7 neutrons/cm^2.s\"\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of neutrons captured by 1.0cm^2/s is 1.8 X 10^7 neutrons/cm^2.s\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.3, page no. 512" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "\n", + "A = 235 #atomic number of uranium\n", + "m = 10**3 #mass of 235U\n", + "Av = 6.02*10**23#avogadro number (nuclei/mol)\n", + "Q = 208 #disintegration energy per event (MeV)\n", + "\n", + "#Calculation\n", + "\n", + "N = Av * m / A\n", + "E = N * Q\n", + "\n", + "#Result\n", + "\n", + "print \"The disintegration energy is\",round(E/10**26,2),\"X 10^26 MeV.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The disintegration energy is 5.33 X 10^26 MeV.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.4, page no. 513" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Variable declaration\n", + "\n", + "A = 235 #atomic mass of Uranium\n", + "Av = 6.02 * 10**23 # Avagadro's number\n", + "Q = 208 # disintegration energy per event (MeV)\n", + "\n", + "#Calculation\n", + "\n", + "N = Av / A *10**3 #No of nuclei in 1kg of 235U\n", + "E = N*Q #disintegration energy (MeV)\n", + "\n", + "#result\n", + "\n", + "print \"The disintegration energy is\",round(E/10**26,2),\" X 10^26 MeV.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The disintegration energy is 5.33 X 10^26 MeV.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.5, page no. 513" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "\n", + "rBa = 6.2 * 10 **-15 #nuclear radius of Ba(m)\n", + "rKr = 5.4 * 10**-15 #nuclear radius of Kr(m)\n", + "Z1 = 56 #atomic number of Ba\n", + "Z2 = 36 #atomic number of Kr\n", + "k = 1.440 * 10 **-9 #Coulomb constant (eV.nm)\n", + "\n", + "#Calculation\n", + "\n", + "r = rBa + rKr\n", + "U = k * Z1 * Z2 /( round(r/10**-15) * 10**-15)\n", + "\n", + "#Result\n", + "\n", + "print \"The potential energy of the two nuclei is\",round(U/10**6),\"MeV.\" " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The potential energy of the two nuclei is 242.0 MeV.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.6, page no. 519" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#variable declaration\n", + "\n", + "e = 1.6 * 10 ** -19 #electron charge (C)\n", + "r = 1.0 * 10**-14 #separation (m)\n", + "k = 8.99 * 10**9 #Coulomb constant(N.m^2/C^2)\n", + "\n", + "#Calculation\n", + "\n", + "U = k * e**2/ r\n", + "\n", + "#result\n", + "\n", + "print \"(a)The height of the potential barrier is\",round(U/e/10**6,2),\"MeV.\"\n", + "\n", + "\n", + "#Variable Declaration\n", + "\n", + "kB = 1.38 * 10**-23 #Boltzmann's constant (J/K)\n", + "\n", + "#Calculation\n", + "\n", + "T = (U/2.0)*(2.0/3.0)/(kB+.07*10**-23)\n", + "\n", + "#Result\n", + "\n", + "print\"(b)The effective temperature is\",round(T/10**8,1),\"X 10^8 K.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The height of the potential barrier is 0.14 MeV.\n", + "(b)The effective temperature is 5.3 X 10^8 K.\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.7, page no. 530" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#variable declaration\n", + "\n", + "mu = 55 #mu for x-rays \n", + "\n", + "#Calculation\n", + "\n", + "x = math.log(2)/mu\n", + "\n", + "#Result\n", + "\n", + "print \"The half value thickness for lead is\",round(x/10**-2,2),\"X 10^-2 cm.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The half value thickness for lead is 1.26 X 10^-2 cm.\n" + ] + } + ], + "prompt_number": 24 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter15.ipynb b/Modern_Physics/Chapter15.ipynb new file mode 100755 index 00000000..d83185b4 --- /dev/null +++ b/Modern_Physics/Chapter15.ipynb @@ -0,0 +1,63 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fbb8ae4f61a2f656bdccdaea83dd7c5e0f4029b53a892411eec2330f543018a0" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 15: Elementary Particles" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.5, page no. 570" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#variable declaration\n", + "\n", + "mp = 938.3 #mass of proton (MeV/c^2)\n", + "mpi = 135 #mass of pion (MeV/c^2)\n", + "\n", + "#Calculation\n", + "\n", + "Kth = ((2*mp + mpi)**2 - (2*mp)**2)/(2*mp)\n", + "\n", + "#result\n", + "\n", + "print \"The kinetic energy of protons\",round(Kth),\"MeV.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The kinetic energy of protons 280.0 MeV.\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter16.ipynb b/Modern_Physics/Chapter16.ipynb new file mode 100755 index 00000000..f8fb36e8 --- /dev/null +++ b/Modern_Physics/Chapter16.ipynb @@ -0,0 +1,62 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f4f136f0af18cf143b33cec13dfb6a06f3807644fa35b75af1bd5439584b255b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 16: Cosmology" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.1, page no. 15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "\n", + "c = 3.0 * 10 ** 5 #speed of light (km/s)\n", + "H0 = 20 * 10**-6 #(km/s/lightyear)\n", + "\n", + "#Calculation\n", + "\n", + "Rmax = c/(4*H0)\n", + "\n", + "#Result\n", + "\n", + "print \"The maximum distance at which the Hubble's law applies without relativistic correction is\",round(Rmax/10**9),\"X 10^9 lightyears.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum distance at which the Hubble's law applies without relativistic correction is 4.0 X 10^9 lightyears.\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter2.ipynb b/Modern_Physics/Chapter2.ipynb new file mode 100755 index 00000000..b0fca27c --- /dev/null +++ b/Modern_Physics/Chapter2.ipynb @@ -0,0 +1,418 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:eb498fc54ec9607231510853c13ca78b6bdacd256b3a6b8f272ed771d86d795c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2: Relativity II" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1, page no. 44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "m = 9.11 * 10 **-31 #mass of electron(kg)\n", + "c = 3.0 * 10**8 #speed of light(m/s)\n", + "u = 0.750 * c #speed of electron(m/s)\n", + "\n", + "#calculation\n", + "\n", + "p = m*u/(math.sqrt(1-(u**2/c**2))) \n", + "momentum = m * u\n", + "\n", + "#result\n", + "\n", + "print \"The correct relativistic momentum is\",round(p/10**-22,2),\"X 10^-22 kg m/s\"\n", + "print \"The incorrect classical expression results in momentum equal to\",round(momentum/10**-22,2),\"X 10^-22 kg m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The correct relativistic momentum is 3.1 X 10^-22 kg m/s\n", + "The incorrect classical expression results in momentum equal to 2.05 X 10^-22 kg m/s\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3, page no. 47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "me = 9.11 * 10 **-31 #mass of electron(kg)\n", + "c = 3* 10** 8 #speed of light(m/s)\n", + "u = 0.850 * c #speed of electron (m/s)\n", + "e = 1.6 *10 **-19 #charge of electron(C)\n", + "\n", + "#Calculation\n", + "\n", + "E = me*c**2/(e*math.sqrt(1-(u**2/c**2)))\n", + "K = E - (me*c**2/e)\n", + "\n", + "#results\n", + "\n", + "print \"The Total energy is\",round(E/10**6,2),\"Mev and the kinetic energy is \",round(K/10**6,3),\"Mev.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Total energy is 0.97 Mev and the kinetic energy is 0.46 Mev.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.4, page no. 47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "\n", + "#Variable Declaration\n", + "\n", + "mp = 1.67 * 10 ** -27 #mass of proton (kg)\n", + "c = 3.0 * 10 ** 8 #speed of light(m/s)\n", + "e = 1.602 * 10 ** -19 #charge of electron (C)\n", + "\n", + "#Calculation\n", + "\n", + "restenergy = mp * c**2/ e \n", + "\n", + "#result\n", + "\n", + "print \"The rest energy of proton is \",round(restenergy/10**6),\"Mev\"\n", + "\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "E = 3.0 * restenergy\n", + "\n", + "#Calculation\n", + "\n", + "u = (math.sqrt(1-(1/(E/restenergy)**2))*c)\n", + "\n", + "#result\n", + "\n", + "print \"The speed of proton is\",round(u/10**8,2),\"X 10^8 m/s. \"\n", + "\n", + "\n", + "#calculation\n", + "\n", + "K = E - restenergy\n", + "\n", + "#result\n", + "\n", + "print \"The kinetic energy of proton is\",round(K/10**6),\"Mev\"\n", + "\n", + "\n", + "#calculation\n", + "\n", + "p=math.sqrt(round(E)**2 - round(restenergy)**2)/c\n", + "\n", + "#result\n", + "\n", + "print \"The proton's momentum is \",round(p * c/10**6),\"Mev/c\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rest energy of proton is 938.0 Mev\n", + "The speed of proton is 2.83 X 10^8 m/s. \n", + "The kinetic energy of proton is 1876.0 Mev\n", + "The proton's momentum is 2654.0 Mev/c\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5, page no. 49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#Variable Declaration\n", + "\n", + "u = 450.0 #speed of the balls(m/s)\n", + "c = 3.0 * 10 ** 8 #speed of light(m/s)\n", + "m = 5.0 #mass of the ball (kg)\n", + "\n", + "#calculation\n", + "\n", + "dM = 2* m *(1+(u**2/(2*c**2))-1) #because u^2/c^2 << 1\n", + "\n", + "#result\n", + "\n", + "print \"The mass increment is\",round(dM/10**-11,1),\"X10^-11 kg\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The mass increment is 1.1 X10^-11 kg\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7, page no. 50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Variable declaration\n", + "\n", + "u = 931.5 #atomic mass unit (MeV/c^2)\n", + "Mu = 236.045563 #mass of uranuim (u)\n", + "MRb = 89.914811 #mass of Rb (u)\n", + "MCs = 142.927220 #mass of cs(u)\n", + "mn = 1.008665 #mass of neutron (u)\n", + "\n", + "#Calculation\n", + "\n", + "dm = Mu - (MRb + MCs + 3 * mn)\n", + "\n", + "#result\n", + "\n", + "print \"The reaction products have \",dm,\"u less than the initial uranium mass\"\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "c = 3.0 * 10 ** 8 #speed of light (m/s)\n", + "\n", + "#calculation\n", + "\n", + "dm = dm * u\n", + "Q = dm\n", + "\n", + "#result\n", + "\n", + "print \"the energy given off per fission event is\",-round(Q,1),\"MeV\"\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "A = 6.02 * 10 ** 23 #Avagadro number\n", + "N = A * 1000/ Mu #number of nuclei\n", + "efficiency = 0.4\n", + "kWh = 4.435 * 10 **-20 #conversion (kWh/MeV)\n", + "\n", + "#calculation\n", + "\n", + "E = efficiency * N * Q *kWh\n", + "\n", + "#result\n", + "\n", + "print \"The total energy produced is \" ,round(E/10 **6,2),\"X 10^6 kWh\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reaction products have 0.177537 u less than the initial uranium mass\n", + "the energy given off per fission event is -165.4 MeV\n", + "The total energy produced is 7.48 X 10^6 kWh\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.8, page no. 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "\n", + "#Variable Declaration\n", + "\n", + "e = 1.6 * 10 **-19 #charge of electron(C)\n", + "BE = 3 #binding energy of water(eV)\n", + "c = 3.0 * 10**8 #speed of light (m/s)\n", + "\n", + "#Calculation\n", + "\n", + "dm = BE * e / c**2\n", + "\n", + "#result\n", + "\n", + "print \"The mass difference is\",round(dm/10 **-36,1),\"X 10^-36 kg\"\n", + "\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "MH2O = 3.0 * 10 **-26 #mass of water molecule (kg)\n", + "\n", + "#Calculation\n", + "\n", + "fractional_loss= dm / MH2O\n", + "\n", + "#result\n", + "\n", + "print \"The fractional loss of mass per gram of water formed is\",round(fractional_loss/10 ** -10,1),\"X 10^-10 \"\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "dm = 1.8 * 10 ** -13 #change in mass when 1 gram of water is formed (kg)\n", + "\n", + "#calculation\n", + "\n", + "E = dm * c**2\n", + "\n", + "#result\n", + "\n", + "print \"The energy released when 1 gram of H2O is formed is\",round(E/10**3),\"kJ\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The mass difference is 5.3 X 10^-36 kg\n", + "The fractional loss of mass per gram of water formed is 1.8 X 10^-10 \n", + "The energy released when 1 gram of H2O is formed is 16.0 kJ\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.9, page no. 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "muCsq = 106 #energy of muon (Mev)\n", + "Ku = 4.6 #kinetic energy of muon (Mev)\n", + "\n", + "#calculation\n", + "\n", + "mpiCsq = math.sqrt(muCsq**2 + Ku**2 + 2*Ku *muCsq)+math.sqrt(Ku**2 + 2*Ku*muCsq)\n", + "\n", + "#result\n", + "\n", + "print \"The mass of the pion is \",round(mpiCsq),\"MeV/c^2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The mass of the pion is 142.0 MeV/c^2\n" + ] + } + ], + "prompt_number": 17 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter3.ipynb b/Modern_Physics/Chapter3.ipynb new file mode 100755 index 00000000..c4fdd463 --- /dev/null +++ b/Modern_Physics/Chapter3.ipynb @@ -0,0 +1,496 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:6f3412539d62c2f676626072f86e8478aa55d9f7f8bd139276fa120f78482f67" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3: The Quantum Theory of Light" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1, page no. 69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "\n", + "#Variable declaration\n", + "Rs=7.0 * (10 ** 8) \t#sun's radius (m)\n", + "R = 1.5 *(10 ** 11)\t#earth to sun distance (m)\n", + "a = 1 #since sun is considered as a blackbody \n", + "k = 5.6 * (10 ** (-8)) #Stefan-Boltzmann constant ( W.m ^-2 .K^-4)\n", + "eTotalR = 1400\t#power per unit area (W/m^2)\n", + "\n", + "#Calculations\n", + "T = ((eTotalR * R * R) / (k * Rs * Rs) ) ** .25\n", + "\n", + "#Results\n", + "print '%s %.2f %s' %('the surface temperature of the sun is',T,'K')\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the surface temperature of the sun is 5820.79 K\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2, page no. 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "h = 6.63 * 10 ** -34 #planck's constant(Js)\n", + "c = 3 * 10 ** 8 #speed of light(m/s)\n", + "lgreen = 540 * 10 ** -9#wavelength of green light(m)\n", + "lred = 700 *10**-9 #wavelength of red light(m)\n", + "e = 1.602 * 10 ** -19 #charge of an electron(C)\n", + "\n", + "#calculation\n", + "dEg = h*c /(lgreen* e)\n", + "dEr = h*c/(lred * e)\n", + "\n", + "#results\n", + "print '%s %s %s %s %s' %('the minimum energy change for ',lgreen * 10 ** 9,'nm is ',round(dEg,2),'eV')\n", + "print '%s %s %s %s %s' %('the minimum energy change for ',lred * 10 ** 9,'nm is ',round(dEr,2),'eV')\n", + "\n", + "\n", + "#Variable declaration\n", + "l=1 #length of the pendulum(m)\n", + "m = 0.1 # mass of the pendulum(kg)\n", + "g = 9.8 #acceleration due to gravity(m.s^-2)\n", + "h = 6.63 *10 **-34 #planck's constant(J.s)\n", + "theta = 10 # displaced angle\n", + "\n", + "#Calculations\n", + "E = m * g * l *(1 - math.cos(math.pi * theta /180))\n", + "f = math.sqrt(g /l) /(2* math.pi)\n", + "Edelta = h *f\n", + "\n", + "#results\n", + "print '%s %s %s' %('the pendulum frequency is',round(f,2),'Hz')\n", + "print '%s %s %s' %('the total energy of the pendulum is',round(E,3),'J')\n", + "print '%s %s %s' %('therefore an energy change of one quantum corresponds to',round(Edelta/10**-34,2),'x 10^-34 J')\n", + "print '%s %s %s' %('Therefore the fractional change in energy ^E/E is ',round(Edelta/E/10**-32,2),'x 10^-32 ')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the minimum energy change for 540.0 nm is 2.3 eV\n", + "the minimum energy change for 700.0 nm is 1.77 eV\n", + "the pendulum frequency is 0.5 Hz\n", + "the total energy of the pendulum is 0.015 J\n", + "therefore an energy change of one quantum corresponds to 3.3 x 10^-34 J\n", + "Therefore the fractional change in energy ^E/E is 2.22 x 10^-32 \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3, page no. 80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Variable declaration\n", + "\n", + "pi = 3.141 \n", + "k = 1.381 * 10 **-23 #Boltzmann constant (J/K)\n", + "c = 2.998 * 10 ** 8 #Speed of light (m/s)\n", + "h = 6.626 * 10 ** -34 #Planck's constant (J.s)\n", + "\n", + "#Calculation\n", + "\n", + "sigma = 2 * pi**5 * k**4 / (15 * c**2 * h**3)\n", + "\n", + "#Result\n", + "\n", + "print \"e_total=sigma * T^4 where sigma=\",round(sigma/10**-8,2),\"X 10^-8 W.m^-2.K^-4\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "e_total=sigma * T^4 where sigma= 5.67 X 10^-8 W.m^-2.K^-4\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4, page no. 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#variable declaration\n", + "e= 1.68 * 10 **-19 #electron charge(C)\n", + "O = 2.28 * e #work function of sodium\n", + "I = 10 ** -10 #power per unit area(W/cm^2)\n", + "\n", + "#calculation\n", + "A = math.pi * 10 ** -16\n", + "t= O / (I * A)\n", + "\n", + "#result\n", + "print '%s %s %s' %('the time lag is given by',round(t/(60*60*24)),'days which is approximated to 130 days in the text book')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the time lag is given by 141.0 days which is approximated to 130 days in the text book\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5, page no. 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#variable declaration\n", + "Vs = 4.3 #Stopping voltage(V)\n", + "e = 1.6 * 10 **-19 #electron charge(C)\n", + "Me = 9.1 *10**-31 #mass of electron(kg)\n", + "\n", + "#calculation\n", + "vmax = math.sqrt( 2* e* Vs /Me)\n", + "Kmax = e *Vs\n", + "\n", + "#result\n", + "print '%s %s %s' %('the Kmax of these electrons are', Kmax ,'J')\n", + "print '%s %s %s' %('vmax of these electrons are',round(vmax/10**6,2),' x 10^6 m/s')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the Kmax of these electrons are 6.88e-19 J\n", + "vmax of these electrons are 1.23 x 10^6 m/s\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6, page no. 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#variable declaration\n", + "I0 = 1 * 10 ** -6 #intensity of light falling W/cm^2\n", + "\n", + "#calculation\n", + "I = .03 * .04 * I0\n", + "\n", + "#result\n", + "print '%s %s %s' %('The actual intensity available is',I,'W/cm^2')\n", + "\n", + "#variable declaration\n", + "lamda = 250 *10 ** -9 #wavelength of violet light(m)\n", + "c= 3*10**8 #speed of light(m/s)\n", + "h = 6.63 *10 **-34 #planck's constant(J.s)\n", + "\n", + "#calculation\n", + "Ne = I *lamda / (h * c)\n", + "\n", + "#result\n", + "print '%s %s %s' %('number of electrons is',round(Ne/10**9,1),'x 10^9')\n", + "\n", + "\n", + "#variable declaration\n", + "e = 1.6 * 10 **-19 #electron charge(c)\n", + "\n", + "#calculation\n", + "i = e * Ne\n", + "\n", + "#result\n", + "print '%s %s %s' %('current in the phototube is ',round(i/10**-10,1),'x 10^-10 A')\n", + "\n", + "\n", + "#variable declaration\n", + "f0 = 1.1 *10**15 #cutoff frequency (Hz)\n", + "\n", + "#calculation\n", + "O = h *f0 / e \n", + "\n", + "#result\n", + "print '%s %s %s' %('the work function is ',round(O,1),'eV')\n", + "\n", + "\n", + "\n", + "#variable declaration\n", + "lamda = 250 * 10 ** -9 #wavelength(m)\n", + "\n", + "#calculation\n", + "Vs = ((h*c )/(lamda * e )) - O\n", + "\n", + "#result\n", + "print '%s %s %s' %('stopping voltage for iron is ',round(Vs,2),'V')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The actual intensity available is 1.2e-09 W/cm^2\n", + "number of electrons is 1.5 x 10^9\n", + "current in the phototube is 2.4 x 10^-10 A\n", + "the work function is 4.6 eV\n", + "stopping voltage for iron is 0.41 V\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7, page no. 93" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#variable declaration\n", + "lamda = .2 * 10 ** -9 #wavelength(m)\n", + "theta = 45 #observed angle(degrees)\n", + "h = 6.63 * 10 ** -34 #planck's constant(J.s)\n", + "Me = 9.1 * 10 ** -31 #electron mass(kg)\n", + "c = 3* 10 ** 8 #speed of light(m/s)\n", + "\n", + "#calculation\n", + "dl= h *(1 - math.cos(math.pi * theta /180)) /(Me * c)\n", + "\n", + "#result\n", + "print '%s %s %s' %('the wavelength off the scattered x-ray at this angle is',dl+lamda,'m')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the wavelength off the scattered x-ray at this angle is 2.00711312103e-10 m\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8, page no. 93" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#variable declaration\n", + "lamdaG = 0.0106 #wavelenght(m)\n", + "\n", + "#calculation\n", + "dl = 0.0243 * (1-math.cos(math.pi/2))\n", + "f= dl/ lamdaG\n", + "\n", + "#result\n", + "print \"the compton shift is \",dl,\"A'\"\n", + "print \"the fractional change in wavelength of gamma rays is\",round(f,4)\n", + "\n", + "#(2)X-rays from molybdenum, lamda = 0.712 x 10 ^-10 m,\n", + "\n", + "#variable declaration\n", + "lamdaX = 0.712 #wavelenght(m)\n", + "\n", + "#calculation\n", + "f= dl/ lamdaX\n", + "\n", + "#result\n", + "print \"the fractional change in wavelength of X rays is\",round(f,4)\n", + "\n", + "#(3)green light from a mercury lamp, lamda = 5461 *10 ^ -10 \n", + "\n", + "#variable declaration\n", + "lamdaGr = 5461\n", + "\n", + "#calculation\n", + "f= dl/ lamdaGr\n", + "\n", + "#result\n", + "print \"the fractional change in wavelength of green rays is\",round(f/10**-6,3),\"x 10^-6\"\n", + "\n", + "\n", + "#variable declaration\n", + "h = 6.63 * 10 ** -34 #planck's constant(J.s)\n", + "c = 3* 10 ** 8 #speed of light(m/s)\n", + "lamda = 0.712 * 10 **-10\n", + "e = 1.6 * 10 **-19 #electron charge(c)\n", + "\n", + "#calculation\n", + "E = h*c/(lamda * e)\n", + "\n", + "#result\n", + "print \"the energy of incident x-ray is\",round(E,2),\"ev. It is large when compared to binding energy of 4eV\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the compton shift is 0.0243 A'\n", + "the fractional change in wavelength of gamma rays is 2.2925\n", + "the fractional change in wavelength of X rays is 0.0341\n", + "the fractional change in wavelength of green rays is 4.45 x 10^-6\n", + "the energy of incident x-ray is 17459.62 ev. It is large when compared to binding energy of 4eV\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9, page no. 96" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#variable declaration\n", + "M = 1.99 * 10 ** 30 #mass of sun(kg)\n", + "lamda = 300 *10**-9 #wavelength(m)\n", + "Rs = 6.37 * 10 ** 6 #radius of earth(m)\n", + "G = 6.67 * 10 ** -11 #gravitational constant(N.m^2.kg^-2)\n", + "c = 3 * 10 ** 8 #speed of light(m/s)\n", + "\n", + "#calculation\n", + "fraction = G * M / (Rs * c *c)\n", + "\n", + "#result\n", + "print '%s %s %s'%(\"the shift in wavelength\",round(lamda * fraction *10**9,4),'nm')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the shift in wavelength 0.0695 nm\n" + ] + } + ], + "prompt_number": 22 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter4.ipynb b/Modern_Physics/Chapter4.ipynb new file mode 100755 index 00000000..505cc127 --- /dev/null +++ b/Modern_Physics/Chapter4.ipynb @@ -0,0 +1,447 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fcebe7a1b857b1b190100018ff76ede788e3c249d5181d4bebb1457ad5175559" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4: The Particle Nature of Matter" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1, page no. 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "\n", + "I = 10.0 #current (A)\n", + "t = 3600 #time (s)\n", + "molar_mass_Ba = 137 #molar mass of Barium (g)\n", + "valence_Ba = 2 #valence of Barium\n", + "molar_mass_Cl = 35.5 #molar mass of chlorine(g)\n", + "valence_Cl = 1 #valence of chlorine\n", + "\n", + "#Calculation\n", + "\n", + "mBa = I * t * molar_mass_Ba /(96500 * valence_Ba)\n", + "mCl = I * t * molar_mass_Cl /(96500 * valence_Cl)\n", + "\n", + "#Results\n", + "\n", + "print \"The mass of Barium and Chlorine obtained is \",round(mBa,1),\"g\",round(mCl,1),\"g respectively.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The mass of Barium and Chlorine obtained is 25.6 g 13.2 g respectively.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2, page no. 113" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "V = 200 #voltage (V)\n", + "theta = 0.20 #deflection (rad)\n", + "l = 0.05 #length (m)\n", + "d = 0.015 #spacing (m)\n", + "e_by_me = 1.76 * 10 **11 #e/me (C/kg)\n", + "B_earth = 0.5 * 10**-4 #earth's magnetic field (T)\n", + "\n", + "#Calculation\n", + "\n", + "B = math.sqrt(V * theta /(l * d * e_by_me))\n", + "\n", + "#results\n", + "\n", + "print \"The magnetic field required to produce the deflection is\",round(B/10**-4,1),\"X 10^-4 T\"\n", + "print \" The earth's magnetic field is \",B_earth,\"T so we require\",round(B/B_earth),\"times stronger field.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The magnetic field required to produce the deflection is 5.5 X 10^-4 T\n", + " The earth's magnetic field is 5e-05 T so we require 11.0 times stronger field.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3, page no. 117" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "dy = 0.006 #distance of rise or fall (m)\n", + "dt = 21.0 #average time of fall (s)\n", + "n = 1.83 * 10 **-5 #viscosity of air (kg/ms)\n", + "p = 858 #oil density (kg/m^3)\n", + "g = 9.81 #acceleration due to gravity (m/s^2)\n", + "\n", + "#Calculation\n", + "\n", + "v = dy/dt\n", + "a = math.sqrt(9*n*v/(2*p*g))\n", + "V = 4*math.pi*a**3/3\n", + "m = p * V\n", + "\n", + "#results\n", + "\n", + "print \"The radius,volume and mass of the drop are\",round(a/10**-6,2),\"X10^-6 m,\",round(V/10**-17,2),\"X10^-17 m^3,\",round(m/10**-14,2),\"X10^-14 kg\"\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "Vt = 4550 #potential difference (V)\n", + "d = 0.0160 #plate separation (m)\n", + "#successive rise times(s)\n", + "dt1 = 46.0\n", + "dt2 = 15.5\n", + "dt3 = 28.1\n", + "dt4 = 12.9\n", + "dt5 = 45.3\n", + "dt6 = 20.0\n", + "\n", + "#Calculation\n", + "\n", + "E = Vt/d #Electric field\n", + "v1 = dy / dt1\n", + "v2 = dy / dt2\n", + "v3 = dy / dt3\n", + "v4 = dy / dt4\n", + "v5 = dy / dt5\n", + "v6 = dy / dt6\n", + "q1 = (m*g/E)*((v+v1)/v)\n", + "q2 = (m*g/E)*((v+v2)/v)\n", + "q3 = (m*g/E)*((v+v3)/v)\n", + "q4 = (m*g/E)*((v+v4)/v)\n", + "q5 = (m*g/E)*((v+v5)/v)\n", + "q6 = (m*g/E)*((v+v6)/v)\n", + "\n", + "#Results\n", + "\n", + "print \"The successive charges on the drop is as follows \\n q1 =\",round(q1/10**-19,2),\"X 10^-19 C \\n q2 = \",round(q2/10**-19,2),\"X 10^-19 C \\n q3 = \",round(q3/10**-19,2),\"X 10^-19 C \\n q4 = \",round(q4/10**-19,2),\"X 10^-19 C \\n q5 = \",round(q5/10**-19,2),\"X 10^-19 C \\n q6 = \",round(q6/10**-19,2),\"X 10^-19 C\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The radius,volume and mass of the drop are 1.67 X10^-6 m, 1.96 X10^-17 m^3, 1.68 X10^-14 kg\n", + "The successive charges on the drop is as follows \n", + " q1 = 8.44 X 10^-19 C \n", + " q2 = 13.65 X 10^-19 C \n", + " q3 = 10.12 X 10^-19 C \n", + " q4 = 15.23 X 10^-19 C \n", + " q5 = 8.48 X 10^-19 C \n", + " q6 = 11.88 X 10^-19 C\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4, page no. 121" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Variable declaration\n", + "\n", + "ma_by_mp = 4.0 #as alpha has 2 protons and 2 neutrons\n", + "\n", + "#Calculation\n", + "\n", + "vp = (2*ma_by_mp/(ma_by_mp+1))\n", + "va1 = (ma_by_mp -1)/(ma_by_mp + 1)\n", + "change = (va1 - 1)*100\n", + "\n", + "#result\n", + "\n", + "print \"The percentage change in the velocity of alpha particle is\",change,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The percentage change in the velocity of alpha particle is -40.0 %\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5, page no. 124" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "\n", + "e = 1.6 * 10 ** -19 #Charge of electron (C)\n", + "Z = 13\n", + "k = 8.99 * 10 ** 9 #(Nm^2/C^2)\n", + "Ka = 7.7 * 10 ** 6 * e #potential energy at closest approach(J)\n", + "\n", + "#Calculation\n", + "\n", + "dmin = k * 2 * Z * e ** 2/Ka\n", + "\n", + "#result\n", + "\n", + "print \"The radius of the aluminum nucleus is\",round(dmin/10**-15,1),\"X 10^-15 m.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The radius of the aluminum nucleus is 4.9 X 10^-15 m.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7, page no. 135" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "\n", + "R = 1.097 * 10**7 #Rydberg constant(m^-1) \n", + "nf = 1.0 #energy state\n", + "ni = 2.0 #energy state\n", + "c = 3 * 10 ** 8 #speed of light (m/s)\n", + "h = 4.136 * 10 ** -15 #Planck's constant (eV.s)\n", + "\n", + "#Calculation\n", + "\n", + "lamda = (R * ((1/nf**2) - (1/ni**2))) ** -1\n", + "f = c / lamda\n", + "E = h * f\n", + "\n", + "#Results\n", + "\n", + "print \"The wavelength of the emitted photon is\",round(lamda/10**-9,2),\"nm and frequency is\",round(f/10**15,2),\"X 10^15 Hz and energy is\",round(E,1),\"eV\"\n", + "\n", + "\n", + "#Variable declarartion\n", + "mc2 = 938.8 * 10 ** 6 #m*c^2 of hydrogen atom(eV)\n", + "#Calculation\n", + "\n", + "p = E / c\n", + "K = 0.5 * E**2 /mc2\n", + "\n", + "#results\n", + "\n", + "print \"The momentum of the recoiling hydrogen atom is \",round(p*c,1),\"eV/c and the energy is\",round(K/10**-8,2),\"X 10^-8 eV.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The wavelength of the emitted photon is 121.54 nm and frequency is 2.47 X 10^15 Hz and energy is 10.2 eV\n", + "The momentum of the recoiling hydrogen atom is 10.2 eV/c and the energy is 5.55 X 10^-8 eV.\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8, page no. 136" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "\n", + "R = 1.097 * 10**7 #Rydberg constant(m^-1) \n", + "nf = 2.0 #energy state\n", + "ni = 3.0 #energy state\n", + "c = 3 * 10 ** 8 #speed of light (m/s)\n", + "h = 4.136 * 10 ** -15 #Planck's constant (eV.s)\n", + "\n", + "#Calculation\n", + "\n", + "lamda = (R * ((1/nf**2) - (1/ni**2))) ** -1\n", + "f = c / lamda\n", + "E = h * f\n", + "\n", + "#Results\n", + "\n", + "print \"The longest-wavelength photon in the Balmer series is\",round(lamda/10**-9,2),\"nm and frequency is\",round(f/10**15,2),\"X 10^15 Hz and energy is\",round(E,2),\"eV\"\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "R = 1.097 * 10**7 #Rydberg constant(m^-1) \n", + "nf = 2.0 #energy state\n", + "c = 3 * 10 ** 8 #speed of light (m/s)\n", + "h = 4.136 * 10 ** -15 #Planck's constant (eV.s)\n", + "\n", + "#Calculation\n", + "\n", + "lamda = (R * (1/nf**2)) ** -1\n", + "\n", + "#Results\n", + "\n", + "print \"The shortest-wavelength photon emitted in the Balmer series is\",round(lamda/10**-9,2),\"nm.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The longest-wavelength photon in the Balmer series is 656.34 nm and frequency is 0.46 X 10^15 Hz and energy is 1.89 eV\n", + "The shortest-wavelength photon emitted in the Balmer series is 364.63 nm.\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9, page no. 137" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "E = 10.2 #average thermal energy per atom(eV)\n", + "kB =8.62 * 10**-5 #Boltzmann constant (eV/K)\n", + "N2_by_N1 = 0.10 #N2/N1\n", + "\n", + "#Calculation\n", + "\n", + "T1 = E /((3.0/2.0)* kB)\n", + "T2 = -E/(kB * math.log(N2_by_N1))\n", + "\n", + "#Results\n", + "\n", + "print \"The temperature is \",round(T1),\"K and by Boltzmann distribution the temperature is\",round(T2),\"K.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The temperature is 78886.0 K and by Boltzmann distribution the temperature is 51390.0 K.\n" + ] + } + ], + "prompt_number": 26 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter5.ipynb b/Modern_Physics/Chapter5.ipynb new file mode 100755 index 00000000..175eed34 --- /dev/null +++ b/Modern_Physics/Chapter5.ipynb @@ -0,0 +1,295 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e257dccb197e3cab0c059eb9e1d236e5359e2a825fc6be941cab77026f236087" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Chapter 5: Matter Waves" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1, page no. 154" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Variable declaration\n", + "\n", + "h = 6.63 * 10 ** -34 #Planck's constant (J.s)\n", + "m = 0.14 #mass of the baseball (kg)\n", + "v = 27.0 #speed of the baseball (m/s)\n", + "\n", + "#Calculation\n", + "\n", + "lamda = h / (m * v)\n", + "\n", + "#Result\n", + "\n", + "print \"The de Broglie wavelength of the baseball is\",round(lamda/10**-34,2),\"X 10^-34 m.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The de Broglie wavelength of the baseball is 1.75 X 10^-34 m.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2, page no. 154" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "h = 6.63 * 10 ** -34 #Planck's constant (J.s)\n", + "me = 9.11 * 10 ** -31 #mass of electron (kg)\n", + "q = 1.6 * 10 ** -19 #charge of electron (C)\n", + "V = 50 #potential difference (V)\n", + "\n", + "#Calculation\n", + "\n", + "lamda = h / math.sqrt(2*me*q*V)\n", + "\n", + "#Result\n", + "\n", + "print \"The de Broglie wavelength of electron is\",round(lamda/10 ** -10,1),\"X 10^-10 m.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The de Broglie wavelength of electron is 1.7 X 10^-10 m.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3, page no. 158" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Variable declaration\n", + "\n", + "h = 6.63 * 10 ** -34 #Planck's constant (J.s)\n", + "lamda = 1.0 * 10 ** -10 #de Broglie wavelength of the neutron (m)\n", + "mn = 1.66 * 10**-27 #mass of neutrons (kg)\n", + "e = 1.602 * 10 **-19 #charge of electron(C)\n", + "\n", + "#Calculation\n", + "\n", + "p = h / lamda\n", + "K = p**2/(2*mn)\n", + "\n", + "#result\n", + "\n", + "print \"The kinetic energy is\",round(K/e,4),\"eV.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The kinetic energy is 0.0826 eV.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.8, page no. 177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable Declaration\n", + "\n", + "h = 1.05 * 10 ** -34 #(J.s)\n", + "dx = 15 #length of the room (m)\n", + "m = 0.1 #mass of the ball (kg)\n", + "vx = 2.0 #velocity of the ball (m/s)\n", + "\n", + "#Calculation\n", + "\n", + "dpx = h /( 2* dx)\n", + "dvx = dpx /m\n", + "uncertainity = dvx/vx\n", + "\n", + "#Result\n", + "\n", + "print \"The relative uncertainty is\",round(uncertainity/10**-35,1),\"X 10^-35 which is not measurable.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The relative uncertainty is 1.8 X 10^-35 which is not measurable.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.9, page no. 178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "h = 6.58 * 10 ** -16 #(eV.s)\n", + "dx = 1.0 * 10 ** -14 / 2.0 # dx is half the length of confinement (m)\n", + "c = 3.00 * 10 ** 8 #speed of light (m/s)\n", + "me = 9.11 * 10 ** -31 # mass of electron (kg)\n", + "e = 1.6 * 10 ** -19 #charge of electron (C)\n", + "\n", + "#Calculation\n", + "\n", + "dpx = h * c / (2 * dx)\n", + "E = math.sqrt(dpx**2 + (me * c**2/e)**2)\n", + "K = E - (me * c**2/e)\n", + "\n", + "#result\n", + "\n", + "print \"The kinetic energy of an intranuclear electron is\",round(K/10**6,2),\"MeV.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The kinetic energy of an intranuclear electron is 19.23 MeV.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.10, page no. 178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "dt = 1.0 * 10 ** -8 #lifetime (s)\n", + "\n", + "#calculation\n", + "\n", + "df = 1/(4*math.pi*dt)\n", + "\n", + "#result\n", + "\n", + "print \"The frequency of the light emitted is\",round(df/10**6,1),\"X 10^6 Hz.\"\n", + "\n", + "#Variable declaration\n", + "\n", + "c = 3.0 * 10 ** 8 #speed of light (m/s)\n", + "lamda = 500 * 10 ** -9 #wavelength (m)\n", + "\n", + "#Calculation\n", + "\n", + "f = c/ lamda\n", + "df_by_f0 = df / f\n", + "\n", + "#result\n", + "\n", + "print \"The fractional broadening is\",round(df_by_f0/10**-8,1),\"X 10^-8.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of the light emitted is 8.0 X 10^6 Hz.\n", + "The fractional broadening is 1.3 X 10^-8.\n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter6.ipynb b/Modern_Physics/Chapter6.ipynb new file mode 100755 index 00000000..76cd1700 --- /dev/null +++ b/Modern_Physics/Chapter6.ipynb @@ -0,0 +1,268 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:846e8e3b3770f7cb30a2e91a53718bf5de841338951843c54481c2acfda5e63d" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6: Quantum Mechanics in One Dimension" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4, page no. 197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "me = 9.11 * 10 ** -31 #mass of electron (kg)\n", + "h = 1.055 * 10**-34 #h/2*pi (J.s)\n", + "dx0 = 1.0 * 10**-10 #initial location of electron(m)\n", + "m = 1.0 * 10**-3 #mass of marble (kg)\n", + "dx0m = 10**-4 #initial location of marble (m)\n", + "\n", + "#Calculation\n", + "\n", + "te = math.sqrt(99)* (2* me / h) * dx0**2\n", + "tm = math.sqrt(99)* (2* m / h) * dx0m**2\n", + "\n", + "#result\n", + "\n", + "print \"The time elapsed for electron is\",round(te/10**-15,1),\"X 10^-15 s and that of marble is \",round(tm/10**24,1),\"X 10^24 s.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The time elapsed for electron is 1.7 X 10^-15 s and that of marble is 1.9 X 10^24 s.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5, page no. 202" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "m = 1.0 * 10 **-6 #mass (kg)\n", + "h = 6.626 * 10 **-34 #Planck's constant(J.s)\n", + "n = 1.0\n", + "L = 1.0 * 10**-2 #separation(m)\n", + "\n", + "#Calculation\n", + "\n", + "E1 = n**2 * h**2 /(8*m*L**2)\n", + "v1 = math.sqrt(2*E1/m)\n", + "\n", + "#result\n", + "\n", + "print \"(a) The minimum speed of the particle is\",round(v1/10**-26,2),\"X 10^-26 m/s.\"\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "v = 3.00 * 10**-2 #speed of the particle (m/s)\n", + "\n", + "#Calculation\n", + "\n", + "E = m* v**2 /2\n", + "n = math.sqrt(8*m*L**2*E)/h\n", + "\n", + "#results\n", + "\n", + "print \"(b) We get n = \",round(n/10**23,2),\"X 10^23.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) The minimum speed of the particle is 3.31 X 10^-26 m/s.\n", + "(b) We get n = 9.06 X 10^23.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6, page no. 203" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "L = 0.2 #length of the box (nm)\n", + "me = 511 * 10 ** 3 #mass of electron (eV/c^2)\n", + "hc = 197.3 #(eV.nm)\n", + "\n", + "#Calculation\n", + "\n", + "E1 = math.pi ** 2 * hc**2 /(2* me * L**2)\n", + "E2 = 2**2 * E1\n", + "dE = E2-E1\n", + "lamda = hc*2*math.pi / dE\n", + "\n", + "#result\n", + "\n", + "print \"The energy required is\",round(dE,1),\"eV and the wavelength of the photon that could cause this transition is\",round(lamda),\"nm.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy required is 28.2 eV and the wavelength of the photon that could cause this transition is 44.0 nm.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8, page no. 211" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "h = 197.3 #(eV.nm/c)\n", + "m = 511 * 10**3 #mass of electron (eV/c**2)\n", + "U = 100 #(eV)\n", + "L = 0.200 #width(nm)\n", + "\n", + "#Calculation\n", + "\n", + "d = h /math.sqrt(2*m*U)\n", + "E = math.pi**2 * h**2 /(2*m*(L+2*d)**2)\n", + "new_U = U - E\n", + "d = h/math.sqrt(2*m*new_U)\n", + "E = math.pi**2 * h**2 /(2*m*(L+2*d)**2)\n", + "\n", + "#result\n", + "\n", + "print \"The ground-state energy for the electron is\",round(E,3),\"eV.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ground-state energy for the electron is 6.506 eV.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.13, page no. 217" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "m = 0.0100 #mass of the spring (kg)\n", + "K = 0.100 #force constant of spring (N/m)\n", + "Kh = 510.5 #force constant of hydrogen (N/m)\n", + "h = 6.582 * 10**-16#Planck's constant (eV.s)\n", + "mu = 8.37 * 10**-28#mass of hydrogen molecule(kg)\n", + "\n", + "#calculation\n", + "\n", + "w = math.sqrt(K / m)\n", + "dE = h * w\n", + "wh =math.sqrt(Kh / mu)\n", + "dEh = h * wh\n", + "\n", + "#results\n", + "\n", + "print \"The quantum level spacing in the spring case is\",round(dE/10**-15,2),\"X 10^-15 eV, while in case of hydrogen molecule it is\",round(dEh,3),\"eV which is easily measurable.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The quantum level spacing in the spring case is 2.08 X 10^-15 eV, while in case of hydrogen molecule it is 0.514 eV which is easily measurable.\n" + ] + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter7.ipynb b/Modern_Physics/Chapter7.ipynb new file mode 100755 index 00000000..5f1ccec9 --- /dev/null +++ b/Modern_Physics/Chapter7.ipynb @@ -0,0 +1,230 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:75c249ea2e8c0a6e5f6f1e7aa12f02f35c3e7e62df28f6611e18b53d1b7e6dcd" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7: Tunneling Phenomena" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.1, page no. 235" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "h = 1.973 * 10**3 #planck's constant (eV.A'/c)\n", + "me = 511 * 10**3 #mass of electron (eV/c^2)\n", + "U = 10.0\n", + "E = 7.0\n", + "L = 50.00 #thickness of layer (A')\n", + "\n", + "#Calculation\n", + "\n", + "a = math.sqrt(2*me*(U-E))/h\n", + "T=(1.0+(1.0/4.0)*(U**2/(E*(U-E)))*(math.sinh(a*L))**2)**-1\n", + "\n", + "#Result\n", + "\n", + "print \"The transmission coefficient for L=\",L,\"A' is\",round(T/10**-38,3),\"X 10^-38\"\n", + "\n", + "#(b)if the layer thickness is 1.00nm.\n", + "\n", + "#Variable Declaration\n", + "\n", + "L = 10 #thickness of layer (A')\n", + "\n", + "#Calculation\n", + "\n", + "T=(1.0+(1.0/4.0)*(U**2/(E*(U-E)))*(math.sinh(a*L))**2)**-1\n", + "\n", + "#Result\n", + "\n", + "print \"The transmission coefficient for L=\",L,\"A' is\",round(T/10**-7,3),\"X 10^-7\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The transmission coefficient for L= 50.0 A' is 0.963 X 10^-38\n", + "The transmission coefficient for L= 10 A' is 0.657 X 10^-7\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.2, page no. 236" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "\n", + "e = 1.6 * 10 ** -19 #charge of electron (C)\n", + "I = 1.00 * 10 ** -3 #electron current(A)\n", + "T = 0.657 *10**-7 #Transmission coefficient\n", + "\n", + "#Calculation\n", + "\n", + "Ne = I / e\n", + "Nadj = Ne * T\n", + "Iadj = Nadj * e\n", + "\n", + "#Result\n", + "\n", + "print \"The number of electrons per second continuing on the adjacent wire is\",round(Nadj/10**8,2),\"X 10^8 and the transmitted current is\",round(Iadj/10**-12,1),\"pA.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of electrons per second continuing on the adjacent wire is 4.11 X 10^8 and the transmitted current is 65.7 pA.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5, page no. 241" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable Declaration\n", + "\n", + "e = 1.6 * 10 **-19 #charge of electron(C)\n", + "f = 1.0*10**30 #collision frequency (s^-1.cm^-2)\n", + "Ec = 5.5 * 10 ** 10 \n", + "V = 10 * 10 ** 3 #potential difference(V)\n", + "d = 0.010 * 10**-3 #plate separation(m)\n", + "\n", + "#Calculation\n", + "\n", + "E = V /d\n", + "Te = math.exp(-Ec/E)\n", + "rate = f * Te\n", + "I = e * rate\n", + "\n", + "#result\n", + "\n", + "print \"The tunneling current is\",round(I/10**-12,2),\"pA.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The tunneling current is 0.21 pA.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6, page no. 244" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "Zth = 90 #atomic number of thorium\n", + "Zdth = 88 #atomic number of thorium's daughter nucleus\n", + "E = 4.05 #energy of ejected alphas(MeV)\n", + "Zpo = 84 #atomic number of polonium\n", + "Zdpo = 82 #atomic number of polonium's daughter nucleus\n", + "Epo = 8.95 #energy of ejected alphas(MeV)\n", + "R = 9.00 #nucleus size(fm)\n", + "r0 = 7.25 #Bohr radius of alpha(fm)\n", + "E0 = 0.0993 #(MeV)\n", + "f = 10 ** 21 #collision frequency(Hz)\n", + "\n", + "\n", + "#Calculation\n", + "\n", + "Te = math.exp(-4*math.pi*Zdth*math.sqrt((E0/E))+ 8 * math.sqrt(Zdth*R/r0))\n", + "rate = f * Te\n", + "t = math.log(2)/rate\n", + "Tep = math.exp(-4*math.pi*Zdpo*math.sqrt((E0/Epo))+ 8 * math.sqrt(Zdpo*R/r0))\n", + "ratep = f * Tep\n", + "tp = math.log(2)/ratep\n", + "\n", + "\n", + "#Result\n", + "\n", + "print \"The half life of thorium is\",round(t/10**17,1),\"X 10^17 s and that of polonium is\",round(tp/10**-10,1),\"X 10^-10 s.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The half life of thorium is 5.4 X 10^17 s and that of polonium is 8.4 X 10^-10 s.\n" + ] + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter8.ipynb b/Modern_Physics/Chapter8.ipynb new file mode 100755 index 00000000..a12d49fc --- /dev/null +++ b/Modern_Physics/Chapter8.ipynb @@ -0,0 +1,222 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:8315bc6aae714998ec5db8a7cf2faf25a033ca960fbcf8e976cdfc44e47b09f3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8: Quantum Mechanics in Three Dimensions" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4, page no. 270" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "R = 1.0 #radius(m)\n", + "T = 1.0 #period of revolution(s)\n", + "m = 1.0 #mass of stone(kg)\n", + "h = 1.055 * 10**-34 #planks constant (kg.m^2/s)\n", + "\n", + "#Calculation\n", + "\n", + "v = 2*math.pi*R/T\n", + "L = m * v * R\n", + "l = L / h\n", + "\n", + "#Results\n", + "\n", + "print \"The orbital quantum number l is\",round(l/10**34,2),\"X 10^34.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The orbital quantum number l is 5.96 X 10^34.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.6, page no. 272" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "l = 3 #orbital quantum number\n", + "\n", + "#Calculation and results\n", + "\n", + "L = math.sqrt(l*(l+1))\n", + "print \"The allowed values of Lz are\"\n", + "for i in range(-l,l+1):\n", + " print i,\"h\"\n", + "print \"The allowed values of theta are\"\n", + "for i in range(-l,l+1):\n", + " theta = round(math.acos(i/L)*180/math.pi,1)\n", + " if theta > 90:\n", + " print theta-180\n", + " else:\n", + " print theta" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The allowed values of Lz are\n", + "-3 h\n", + "-2 h\n", + "-1 h\n", + "0 h\n", + "1 h\n", + "2 h\n", + "3 h\n", + "The allowed values of theta are\n", + "-30.0\n", + "-54.7\n", + "-73.2\n", + "90.0\n", + "73.2\n", + "54.7\n", + "30.0\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.7, page no. 281" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Variable declaration\n", + "\n", + "n = 2\n", + "Z = 1\n", + "\n", + "#Calculation and results\n", + " \n", + "print \"The states of hydrogen atom are\"\n", + "for i in range(n):\n", + " for j in range(-i,i+1):\n", + " print \"n = \",n,\"l = \",i,\"ml = \",j\n", + "\n", + "E2 = -13.6 * Z**2/n**2\n", + "print \"All states have same energy of \",E2,\"eV.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The states of hydrogen atom are\n", + "n = 2 l = 0 ml = 0\n", + "n = 2 l = 1 ml = -1\n", + "n = 2 l = 1 ml = 0\n", + "n = 2 l = 1 ml = 1\n", + "All states have same energy of -3.4 eV.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.8, page no. 284" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import scipy\n", + "from scipy import integrate\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "z2 = lambda z: z**2 * math.e ** -z\n", + "\n", + "#Calculation\n", + "inf = float('inf')\n", + "integ,err = scipy.integrate.quad(z2,2.0,inf)\n", + "P = integ * 0.5\n", + "\n", + "#result\n", + "\n", + "print \"The probability is\",round(P,3)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The probability is 0.677\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/Chapter9.ipynb b/Modern_Physics/Chapter9.ipynb new file mode 100755 index 00000000..52c8dc08 --- /dev/null +++ b/Modern_Physics/Chapter9.ipynb @@ -0,0 +1,148 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a22a4cd44988289c63c9da64c71766c244fd20324a2c5a80b53d48938212c4cc" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9: Atomic Structure" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1, page no. 300" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Variable declaration\n", + "\n", + "Ub = 9.27 * 10 ** -24 #(J/T)\n", + "B = 1.00 #magnetic field strength (T)\n", + "h = 6.58 * 10 ** -16 #Planck's constant (eV.s)\n", + "e = 1.6 * 10 ** -19 #electron charge (C)\n", + "\n", + "#Calculation\n", + "\n", + "hwl = Ub * B / e\n", + "wl = hwl / h\n", + "\n", + "#results\n", + "\n", + "print \"The magnetic energy is\",round(hwl/10**-5,2),\"x 10^-5 eV and the Larmor frequency is\",round(wl/10**10,2),\"x 10^10 rad/s.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The magnetic energy is 5.79 x 10^-5 eV and the Larmor frequency is 8.81 x 10^10 rad/s.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4, page no. 311" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "\n", + "l2 = 589.592 #sodium doublet wavelength (nm)\n", + "l1 = 588.995 #sodium doublet wavelength (nm)\n", + "hc = 1240 #planck's constant X speed of light (eV.nm)\n", + "\n", + "#Calculation\n", + "\n", + "dE = hc * (l2-l1)/(l2*l1)\n", + "\n", + "#Results\n", + "\n", + "print \"dE =\",round(dE/10**-3,2),\"X 10^-3 eV represents the spin orbit splitting of the initial levels.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "dE = 2.13 X 10^-3 eV represents the spin orbit splitting of the initial levels.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.6, page no. 317" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "E = -5.14 #Energy of the 3s electron in sodium (eV)\n", + "Eh = 13.6 #energy of the 3s electron in hydrogen (eV)\n", + "\n", + "\n", + "#Calculation\n", + "\n", + "Zeff = 3 * math.sqrt(-E/Eh)\n", + "\n", + "#Results\n", + "\n", + "print \"Zeff for the 3s electron in sodium is\",round(Zeff,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Zeff for the 3s electron in sodium is 1.84\n" + ] + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Modern_Physics/README.txt b/Modern_Physics/README.txt new file mode 100755 index 00000000..a4569c81 --- /dev/null +++ b/Modern_Physics/README.txt @@ -0,0 +1,10 @@ +Contributed By: Puneetha Ramachandra +Course: be +College/Institute/Organization: National Institute of Technology +Department/Designation: Electronics & CommunicationsEngg +Book Title: Modern Physics +Author: Serway, Moses & Moyer +Publisher: Thomson Learning, Inc. +Year of publication: 2005 +Isbn: 0-534-40624-6 +Edition: 3rd
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