diff options
Diffstat (limited to 'Modern_Physics/Chapter3.ipynb')
| -rwxr-xr-x | Modern_Physics/Chapter3.ipynb | 390 |
1 files changed, 96 insertions, 294 deletions
diff --git a/Modern_Physics/Chapter3.ipynb b/Modern_Physics/Chapter3.ipynb index c4fdd463..72876741 100755 --- a/Modern_Physics/Chapter3.ipynb +++ b/Modern_Physics/Chapter3.ipynb @@ -1,7 +1,6 @@ { "metadata": { - "name": "", - "signature": "sha256:6f3412539d62c2f676626072f86e8478aa55d9f7f8bd139276fa120f78482f67" + "name": "Chapter3" }, "nbformat": 3, "nbformat_minor": 0, @@ -13,7 +12,7 @@ "level": 1, "metadata": {}, "source": [ - "Chapter 3: The Quantum Theory of Light" + "Chapter 3:The particle like properties of electromagnetic radiation" ] }, { @@ -21,28 +20,22 @@ "level": 2, "metadata": {}, "source": [ - "Example 3.1, page no. 69" + "Example 3.1 Page 69" ] }, { "cell_type": "code", "collapsed": false, "input": [ - " \n", + "#initiation of variable\n", + "from math import sin,pi\n", + "w=0.250; theta=26.3;n=1 # n=1 for hydrogen atom and rest all are given values\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", + "#calculation\n", + "d=n*w/(2*sin(theta*pi/180)); # bragg's law\n", "\n", - "#Results\n", - "print '%s %.2f %s' %('the surface temperature of the sun is',T,'K')\n", - "\n" + "#result\n", + "print \"Hence the atomic spacing in nm is\",round(d,3);" ], "language": "python", "metadata": {}, @@ -51,7 +44,7 @@ "output_type": "stream", "stream": "stdout", "text": [ - "the surface temperature of the sun is 5820.79 K\n" + "Hence the atomic spacing in nm is 0.282\n" ] } ], @@ -62,97 +55,23 @@ "level": 2, "metadata": {}, "source": [ - "Example 3.2, page no. 75" + "Example 3.2 Page 73" ] }, { "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", + "#initiation of variable\n", + "from math import pi,sin\n", + "I=120.0;r=0.1*10**-9;Eev=2.3 #I-intensity in W/m^2 r in m & E in electron volt\n", + "A=pi*r**2;K=1.6*10**-19; # A=area and K is conversion factor from ev to joules\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", + "t= Eev*K/(I*A); #time interval\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\"" + "#result\n", + "print \"The value of time interval was found out to be in sec is\",round(t,3);\n" ], "language": "python", "metadata": {}, @@ -161,79 +80,44 @@ "output_type": "stream", "stream": "stdout", "text": [ - "e_total=sigma * T^4 where sigma= 5.67 X 10^-8 W.m^-2.K^-4\n" + "The value of time interval was found out to be in sec is 0.098\n" ] } ], - "prompt_number": 7 + "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 3.4, page no. 83" + "Example 3.3 Page 76" ] }, { "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", + "#initiation of variable\n", + "from math import pi,sin\n", + "w=650.0*10**-9;h=6.63*10**-34;c=3*10**8; #given values and constant taken in comfortable units\n", "\n", "#calculation\n", - "A = math.pi * 10 ** -16\n", - "t= O / (I * A)\n", + "E=h*c/w; \n", + "E1=E/(1.6*10**-19);\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", + "print \"The Energy of the electron in J \",E,\"which is equivalent to in eV is \", round(E1,3);\n", + "print \"The momentum of electron is p=E/c i.e is \", round(E1,3);\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", + "#part b\n", + "E2=2.40; #given energy of photon.\n", "\n", "#calculation\n", - "vmax = math.sqrt( 2* e* Vs /Me)\n", - "Kmax = e *Vs\n", + "w2=h*c*10**9/(E2*1.6*10**9); #converting the energy in to eV and nm \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')" + "print \"The wavelength of the photon in m is\",round(w2*10**28,0)" ], "language": "python", "metadata": {}, @@ -242,76 +126,49 @@ "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" + "The Energy of the electron in J 3.06e-19 which is equivalent to in eV is 1.912\n", + "The momentum of electron is p=E/c i.e is 1.912\n", + "The wavelength of the photon in m is 518.0\n" ] } ], - "prompt_number": 13 + "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 3.6, page no. 85" + "Example 3.4 Page 78" ] }, { "cell_type": "code", "collapsed": false, "input": [ + "#initiation of variable\n", + "hc=1240.0; phi=4.52 #both the values are in eV\n", "\n", - "#variable declaration\n", - "I0 = 1 * 10 ** -6 #intensity of light falling W/cm^2\n", - "\n", - "#calculation\n", - "I = .03 * .04 * I0\n", + "#calcualtion\n", + "w1=hc/phi; \n", "\n", "#result\n", - "print '%s %s %s' %('The actual intensity available is',I,'W/cm^2')\n", + "print \"The cutoff wavelength of the tungsten metal in nm is \",round(w1,3);\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", + "#part b\n", + "w2=198.0; #given value of wavelength \n", "\n", "#calculation\n", - "Ne = I *lamda / (h * c)\n", + "Kmax=(hc/w2)-phi;\n", "\n", "#result\n", - "print '%s %s %s' %('number of electrons is',round(Ne/10**9,1),'x 10^9')\n", - "\n", + "print 'The max value of kinetic energy in eV is',round(Kmax,3);\n", "\n", - "#variable declaration\n", - "e = 1.6 * 10 **-19 #electron charge(c)\n", - "\n", - "#calculation\n", - "i = e * Ne\n", + "#part c\n", + "Vs=Kmax;\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')" + "print \"The numerical value of the max kinetic energy is same as stopping potential in volts.Hence in V is\",round(Vs,3);" ], "language": "python", "metadata": {}, @@ -320,44 +177,47 @@ "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" + "The cutoff wavelength of the tungsten metal in fnm is 274.336\n", + "The max value of kinetic energy in eV is 1.743\n", + "The numerical value of the max kinetic energy is same as stopping potential in volts.Hence in V is 1.743\n" ] } ], - "prompt_number": 16 + "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 3.7, page no. 93" + "Example 3.5 Page 80" ] }, { "cell_type": "code", "collapsed": false, "input": [ + "#initiation of variable\n", + "T1=293.0; Kw=2.898*10**-3;\n", "\n", + "#calculation\n", + "w1=Kw/T1;\n", "\n", - "import math\n", + "#result\n", + "print \"The wavelength at which emits maximum radiation in um. is\",round(w1*10**6,3);\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", + "#part b\n", + "w2=650.0*10**-9; \n", + "T2=Kw/w2;\n", "\n", - "#calculation\n", - "dl= h *(1 - math.cos(math.pi * theta /180)) /(Me * c)\n", + "#result\n", + "print 'The temperature of the object must be raised to in K. is',round(T2,3);\n", + "\n", + "#part c\n", + "x=(T2/T1)**4; \n", "\n", "#result\n", - "print '%s %s %s' %('the wavelength off the scattered x-ray at this angle is',dl+lamda,'m')" + "print \"Thus the thermal radiation at higher temperature in times the room (lower) tempertaure. is\",round(x,3);" ], "language": "python", "metadata": {}, @@ -366,115 +226,54 @@ "output_type": "stream", "stream": "stdout", "text": [ - "the wavelength off the scattered x-ray at this angle is 2.00711312103e-10 m\n" + "The wavelength at which emits maximum radiation in um. is 9.891\n", + "The temperature of the object must be raised to in K. is 4458.462\n", + "Thus the thermal radiation at higher temperature in times the room (lower) tempertaure. is 53612.939\n" ] } ], - "prompt_number": 18 + "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 3.8, page no. 93" + "Example 3.6 Page 82" ] }, { "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", + "#initiation of variable\n", + "#part a\n", + "from math import cos, sin, pi,atan\n", + "w1=0.24;wc=0.00243;theta=60.0; #given values w=wavelength(lambeda)\n", "\n", "#calculation\n", - "f= dl/ lamdaX\n", + "w2=w1+(wc*(1-cos(theta*pi/180))); \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", + "print \"The wavelength of x-rays after scattering in nm is\",round(w2,5);\n", "\n", - "#variable declaration\n", - "lamdaGr = 5461\n", - "\n", - "#calculation\n", - "f= dl/ lamdaGr\n", + "#part b;\n", + "hc=1240;\n", + "E2=hc/w2;E1=hc/w1; \n", "\n", "#result\n", - "print \"the fractional change in wavelength of green rays is\",round(f/10**-6,3),\"x 10^-6\"\n", + "print \"The energy of scattered x-rays in eV is\",round(E2,3);\n", "\n", + "#part c\n", + "K= E1-E2; #The kinetic energy is the difference in the energy before and after the collision;\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", + "print \"The kinetic energy of the x-rays in eV is\",round(K,3);\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", + "#part d\n", + "phi2=atan(E2*sin(theta*pi/180)/(E1-E2*cos(theta*pi/180)))\n", "\n", "#result\n", - "print '%s %s %s'%(\"the shift in wavelength\",round(lamda * fraction *10**9,4),'nm')" + "print \"The direction of the scattered eletron in degrees is\",round(phi2*180/pi,3);" ], "language": "python", "metadata": {}, @@ -483,11 +282,14 @@ "output_type": "stream", "stream": "stdout", "text": [ - "the shift in wavelength 0.0695 nm\n" + "The wavelength of x-rays after scattering in nm is 0.24121\n", + "The energy of scattered x-rays in eV is 5140.642\n", + "The kinetic energy of the x-rays in eV is 26.025\n", + "The direction of the scattered eletron in degrees is 59.749\n" ] } ], - "prompt_number": 22 + "prompt_number": 9 } ], "metadata": {} |