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diff --git a/Modern_Physics/Chapter4.ipynb b/Modern_Physics/Chapter4.ipynb index 505cc127..4ae3fd7c 100755 --- a/Modern_Physics/Chapter4.ipynb +++ b/Modern_Physics/Chapter4.ipynb @@ -1,7 +1,6 @@ { "metadata": { - "name": "", - "signature": "sha256:fcebe7a1b857b1b190100018ff76ede788e3c249d5181d4bebb1457ad5175559" + "name": "Chapter4" }, "nbformat": 3, "nbformat_minor": 0, @@ -13,7 +12,7 @@ "level": 1, "metadata": {}, "source": [ - "Chapter 4: The Particle Nature of Matter" + "Chapter 4:The Wave Like Properties of Particles" ] }, { @@ -21,31 +20,36 @@ "level": 2, "metadata": {}, "source": [ - "Example 4.1, page no. 109" + "Example 4.1 Page 101" ] }, { "cell_type": "code", "collapsed": false, "input": [ + "#initiation of variable\n", + "from math import sqrt\n", + "h=6.6*10**-34; # h(planck's constant)= 6.6*10^-34 \n", + "m1= 10.0**3;v1=100.0; # for automobile\n", + "\n", + "#calculation\n", + "w1= h/(m1*v1); # ['w'-wavelength in metre'm'-mass in Kg 'v'-velocity in metres/sec.] of the particles \n", + "m2=10.0*(10**-3);v2= 500; # for bullet\n", + "w2=h/(m2*v2);\n", + "m3=(10.0**-9)*(10.0**-3); v3=1.0*10**-2;\n", + "w3=h/(m3*v3);\n", + "m4=9.1*10**-31;k=1*1.6*10**-19; # k- kinetic energy of the electron & using 1ev = 1.6*10^-19 joule\n", + "p=sqrt(2.0*m4*k); # p=momentum of electron ;from K=1/2*m*v^2\n", + "w4=h/p;\n", + "hc=1240;pc=100 # In the extreme relativistc realm, K=E=pc; Given pc=100MeV,hc=1240MeV \n", + "w5= hc/pc;\n", "\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.\"" + "#result\n", + "print \"Wavelength of the automobile in m is\",w1;\n", + "print \"Wavelength of the bullet in m is \",w2 ;\n", + "print\"Wavelength of the smoke particle in m is\",w3 ;\n", + "print \"Wavelength of the electron(1ev) in nm is\",round(w4*10**9,3) ;\n", + "print \"Wavelength of the electron (100Mev) in fm is\",w5;" ], "language": "python", "metadata": {}, @@ -54,45 +58,40 @@ "output_type": "stream", "stream": "stdout", "text": [ - "The mass of Barium and Chlorine obtained is 25.6 g 13.2 g respectively.\n" + "Wavelength of the automobile in m is 6.6e-39\n", + "Wavelength of the bullet in m is 1.32e-34\n", + "Wavelength of the smoke particle in m is 6.6e-20\n", + "Wavelength of the electron(1ev) in nm is 1.223\n", + "Wavelength of the electron (100Mev) in fm is 12\n" ] } ], - "prompt_number": 2 + "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 4.2, page no. 113" + "Example 4.2 Page 113" ] }, { "cell_type": "code", "collapsed": false, "input": [ + "#initiation of variable\n", + "from math import pi\n", + "# w=wavelength; consider k=2*(pi/w); \n", + "# differentiate k w.r.t w and replace del(k)/del(w) = 1 for equation.4.3\n", + "# which gives del(w)= w^2 /(2*pi*del(x)), hence \n", + "w=20; delx=200; # delx=200cm and w=20cm\n", "\n", + "#calculation\n", + "delw=(w**2)/(delx*2*pi);\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.\"" + "#result\n", + "print \"Hence uncertainity in length in cm is\",round(delw,3);" ], "language": "python", "metadata": {}, @@ -101,8 +100,7 @@ "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" + "Hence uncertainity in length is in cm 0.318\n" ] } ], @@ -113,68 +111,25 @@ "level": 2, "metadata": {}, "source": [ - "Example 4.3, page no. 117" + "Example 4.3 Page 114" ] }, { "cell_type": "code", "collapsed": false, "input": [ + "#initiation of variable\n", + "from math import pi\n", + "delt=1.0; #consider time interval of 1 sec\n", + "delw=1/delt; # since delw*delt =1 from equation 4.4\n", + "delf=0.01 #calculated accuracy is 0.01Hz\n", "\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", + "#calculation\n", + "delwc =2*pi*delf # delwc-claimed accuracy from w=2*pi*f\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\"" + "#result\n", + "print \"The minimum uncertainity calculated is 1rad/sec. The claimed accuracy in rad/sec is \\n\",round(delwc,3);\n", + "print \"thus there is a reason to doubt the claim\"\n" ], "language": "python", "metadata": {}, @@ -183,46 +138,39 @@ "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" + "The minimum uncertainity calculated is 1rad/sec. The claimed accuracy is in rad/sec\n", + "0.063\n", + "thus there is a reason to doubt the claim\n" ] } ], - "prompt_number": 16 + "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 4.4, page no. 121" + "Example 4.4 Page 115" ] }, { "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", + "#initiation of variable\n", + "from math import pi\n", + "m=9.11*10**-31;v=3.6*10**6; #'m','v' - mass an velocity of the electron in SI units\n", + "h=1.05*10**-34; #planck's constant in SI\n", + "p=m*v; #momentum\n", + "delp=p*0.01; #due to 1% precision in p\n", + "delx = h/delp; #uncertainity in position\n", "\n", "#result\n", + "print \"Uncertainty in position in nm is\",round(delx*10**9,2);\n", "\n", - "print \"The percentage change in the velocity of alpha particle is\",change,\"%\"" + "#partb\n", + "print \"Since the motion is strictly along X-direction, its velocity in Y direction is absolutely zero.\\n So uncertainity in velocity along y is zero=> uncertainity in position along y is infinite. \\nSo nothing can be said about its position/motion along \"" ], "language": "python", "metadata": {}, @@ -231,39 +179,42 @@ "output_type": "stream", "stream": "stdout", "text": [ - "The percentage change in the velocity of alpha particle is -40.0 %\n" + "Uncertainity in position in nm is 3.2\n", + "Since the motion is strictly along X-direction, its velocity in Y direction is absolutely zero.\n", + " So uncertainity in velocity along y is zero=> uncertainity in position along y is infinite. \n", + "So nothing can be said about its position/motion along \n" ] } ], - "prompt_number": 18 + "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 4.5, page no. 124" + "Example 4.5 Page 116" ] }, { "cell_type": "code", "collapsed": false, "input": [ + "#initiation of variable\n", + "from math import pi\n", + "m=0.145;v=42.5; #'m','v' - mass an velocity of the electron in SI units\n", + "h=1.05*10**-34; #planck's constant in SI\n", + "p=m*v; #momentum\n", + "delp=p*0.01;#due to 1% precision in p\n", "\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", + "#calculation\n", + "delx = h/delp#uncertainty in position\n", "\n", "#result\n", + "print \"Uncertainity in position is %.1e\" %delx;\n", "\n", - "print \"The radius of the aluminum nucleus is\",round(dmin/10**-15,1),\"X 10^-15 m.\"" + "#part b\n", + "print \"Motion along y is unpredictable as long as the velocity along y is exactly known(as zero).\";" ], "language": "python", "metadata": {}, @@ -272,54 +223,40 @@ "output_type": "stream", "stream": "stdout", "text": [ - "The radius of the aluminum nucleus is 4.9 X 10^-15 m.\n" + "Uncertainity in position is 1.7e-33\n", + "Motion along y is unpredictable as long as the veloity along y is exactly known(as zero).\n" ] } ], - "prompt_number": 20 + "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 4.7, page no. 135" + "Example 4.7 Page 119" ] }, { "cell_type": "code", "collapsed": false, "input": [ + "#initiation of variable\n", + "from math import sqrt\n", + "mc2=2.15*10**-4; #mc2 is the mass of the electron, concidered in Mev for the simplicity in calculations\n", + "hc=197.0 # The value of h*c in Mev.fm for simplicity\n", + "delx= 10.0 # Given uncertainty in position=diameter of nucleus= 10 fm\n", + "\n", + "#calculation\n", + "delp= hc/delx ; #Uncertainty in momentum per unit 'c' i.e (Mev/c) delp= h/delx =(h*c)/(c*delx);hc=197 Mev.fm 1Mev=1.6*10^-13 Joules')\n", + "p=delp; # Equating delp to p as a consequence of equation 4.10\n", + "K1=p**2+mc2**2 # The following 3 steps are the steps invlolved in calculating K.E= sqrt((p*c)^2 + (mc^2)^2)- m*c^2\n", + "K1=sqrt(K1)\n", + "K1= K1-(mc2);\n", "\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.\"" + "#result\n", + "print \"Kinetic energy was found out to be in Mev is\", round(K1,3)" ], "language": "python", "metadata": {}, @@ -328,59 +265,39 @@ "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" + "Kinetic energy was found out to be in Mev 19.7\n" ] } ], - "prompt_number": 22 + "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 4.8, page no. 136" + "Example 4.8 Page 120" ] }, { "cell_type": "code", "collapsed": false, "input": [ + "#initiation of variable\n", + "h=6.58*10**-16; # plack's constant\n", + "delt1=26.0*10**-9;E1=140.0*10**6 #given values of lifetime and rest energy of charged pi meson\n", + "delt2=8.3*10**-17;E2=135.0*10**6; #given values of lifetime and rest energy of uncharged pi meson\n", + "delt3=4.4*10**-24;E3=765*10**6; #given values of lifetime and rest energy of rho meson\n", "\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", + "#calculation\n", + "delE1=h/delt1; k1=delE1/E1; # k is the measure of uncertainity\n", + "delE2=h/delt2; k2=delE2/E2;\n", + "delE3=h/delt3; k3=delE3/E3;\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.\"" + "#result\n", + "print \"Uncertainty in energy of charged pi meson is %.1e\" %k1;\n", + "print \"Uncertainty in energy of uncharged pi meson is %.1e\" %k2;\n", + "print \"Uncertainty in energy of rho meson is \",round(k3,2);" ], "language": "python", "metadata": {}, @@ -389,43 +306,37 @@ "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" + "Uncertainity in energy of charged pi meson is 1.8e-16\n", + "Uncertainity in energy of uncharged pi meson is 5.9e-08\n", + "Uncertainity in energy of rho meson is 0.2\n" ] } ], - "prompt_number": 24 + "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 4.9, page no. 137" + "Example 4.9 Page 121" ] }, { "cell_type": "code", "collapsed": false, "input": [ + "#initiation of variable\n", + "h=1.05*10**-34; #value of planck's constant in J.sec\n", + "delx= 1.0; # uncertainty in positon= dimension of the ball\n", + "delp=h/delx; # uncertainty in momentum \n", + "m=0.1; #mass of the ball in kg\n", "\n", + "#calculation\n", + "delv=delp/m; # uncertainty in velocity\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.\"" + "#result\n", + "print \"The value of minimum velocity was found out to be in m/sec\",delv;" ], "language": "python", "metadata": {}, @@ -434,11 +345,11 @@ "output_type": "stream", "stream": "stdout", "text": [ - "The temperature is 78886.0 K and by Boltzmann distribution the temperature is 51390.0 K.\n" + "The value of minimum velocity was found out to be in m/sec 1.05e-33\n" ] } ], - "prompt_number": 26 + "prompt_number": 17 } ], "metadata": {} |