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Mc2 " Mc2 , E = K + Mc2 = ! Mc2 for Particle of Charge Q and Mass M Moving in B Field : P = ! Mu = Qbr E2 = ( Pc)2 + (Mc2 )2

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Department of Physics Modern Physics (2D) University of California Prof. V. Sharma San Diego Quiz # 3 (Jan 28, 2005) Some Relevant Formulae, Constants and Identities p = ! mu, K = ! mc2 " mc2 , E = K + mc2 = ! mc2 For particle of charge q and mass m moving in B field : p = ! mu = qBR E2 = ( pc)2 + (mc2 )2 In Photoelectric Effect E= hf = Kmax + # 23 Avagadro's Number NA = 6.022 $ 10 particles/mole Planck's Constant h=6.626 $ 10-34J.s=4.136 $ 10-15eV.s 1 eV = 1.602 $ 10-19 J -14 2 Electron rest mass = 8.2 $ 10 J = 0.511 MeV/c Proton rest mass = 1.673$ 10-27Kg = 938.3MeV/c2 Speed of Light in vaccum c = 2.998 $ 108m/s Electron charge = 1.602 $ 10-19 C Atomic mass unit u = 1.6605 $ 10-27kg = 931.49 MeV/c2 Please write your scratch work in pencil and write your answer in indelible ink in your Blue book. Please write your code number clearly on each page. Please plug in numbers only at the very end of your calculations. Department of Physics Modern Physics (2D) University of California Prof. V. Sharma San Diego Quiz # 3 (Jan 28, 2005) Problem 1: Weapons of Mass Destruction [10 pts] (a) How much energy is released in the explosion of a fission bomb containing 3.0kg of fissionable material? Assume that 0.10% of the rest mass is converted to released energy? (b) What mass of TNT would have to explode to provide the same energy release? Assume that each mole of TNT liberates 3.4MJ of energy on exploding. The molecular weight of TNT is 0.227kg/mol. (c) For the same mass of explosive, how much more effective are nuclear explosions than TNT? This is to say, compare the fractions of the rest mass that are converted to energy in each case. Problem 2: Designing A Photocell [10 pts] Light of wavelength 200nm falls on an aluminum surface. In metallic aluminum 4.2 eV are required to remove the electron. What is the kinetic energy of (a) the fastest (b) the slowest photoelectrons? (c) What is the stopping potential (including sign) with respect to the photocathode for this wavelength? (d) If the intensity of incident light is 2.0W/m2, what is the average rate per unit area at which photons strike the aluminum surface? (e) What is the cutoff wavelength for aluminum? quiz3solutions.nb 1 Physics 2D, Winter 2005 Quiz 3 Solutions 1. Weapons of mass destruction? (a) Only 0.003 kg of the fission bomb is converted to energy. Thus we have 2 8 2 14 E = meffective c = 0.003 kg 3 µ 10 meters second º 2.7 µ 10 Joules (b) A good way to do this is using the concept of unit conversion. H L H ê L 1 mole mass in kg mTNT = energy released µ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅenergy perÅÅÅÅÅÅÅÅmoleÅÅÅÅÅÅÅÅÅÅ µ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ1 moleÅÅÅÅÅÅÅ 14 1 mole 0.227 kg = 2.7 µ 10 Joules µ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ3.4µ106ÅÅÅÅÅÅÅÅ JoulesÅÅÅÅÅÅÅÅ µ ÅÅÅÅÅÅÅÅ1 moleÅÅÅÅÅÅÅÅÅÅÅ 7 ºH1.8 µ 10 kg L I M I M (c) For fission, we areH told that the fractionL H of the rest LmassI that isM converted to energy is f1 = 0.001 For TNT, we must calculate this, given the mass that we use up, 2.2 µ 106 kg, and the mass equivalent of the energy released, 2.7 µ 1014 Joules. We already know the mass equivalent, since in part (a), we found this energy from precisely the mass equivalent, 0.003 kg. We compute 0.003 kg -10 f2 = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ1.8µ107ÅÅÅÅ kgÅÅÅÅÅ º 1.7 µ 10 Finally, the ratio of these fractions is (I prefer to divide the bigger by the smaller, but either way is fine, so long as you know what the result means). 6 f1 f2 º 6 µ 10 º 6 million 1 So, fission is 6 million times more effective than TNT (or TNT is ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ6 millionthÅÅÅÅÅÅÅÅ as effective as fission). ê 2. Designing a Photocell (a) The energy of the incoming photon, and therefore the initial kinetic energy of the electron, is Eg = h c l = 1240 eV nm 200 nm = 6.2 eV We are told the work function is f = 4.2 eV. Once we take 4.2 eV from the electron we are left with ê H L ê Kmax = 2.0 eV (b) The work function is the minimum amount of energy taken from the electron as it escapes from the metal. Most electrons lose more energy as they escape, either because they are more tightly bound to atoms on the surface, or because they are coming from atoms below the surface (and escaping from below the surface takes energy). The point is that some electrons don't have any kinetic energy at all by the time they escape, so they just barely leave the surface, and then don't go anywhere. Kmin = 0. (c) The stopping potential is the potential that takes all of the Kmax away, i.e., q Vs = Kmax Vs = 2.0 eV -e = -2.0 Volts That was a little bit slick, but you can verify it like H L ê H L 1.6µ10-19 Joules -19 Vs = 2.0 eV µ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ1ÅÅÅÅÅÅÅÅeV ÅÅÅÅÅÅÅÅÅÅÅÅÅ -1.6 µ 10 Coulombs = -2.0 Volts Now, qualitatively, the stopping potential is applied to the receiver, and it stops electrons; therefore, it must be negative - to Hrepel theL electronsI tryingM toë H reach the receiver. L (d) This can be done just converting units (using the fact that each photon has an energy of 6.2 eV). ÅÅÅÅÅÅÅÅWattsÅÅÅÅÅÅÅÅ µ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ1 Joule ÅÅÅÅÅÅÅÅsecondÅÅÅÅÅÅÅ µ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ1ÅÅÅÅÅÅÅÅeV ÅÅÅÅÅÅÅÅÅÅÅÅÅ µ ÅÅÅÅÅÅÅÅ1 photonÅÅÅÅÅÅÅÅÅÅÅ = µ 18 2 2.0 meter2 1 Watt 1.6µ10-19 Joules 6.2 eV 2 10 photons second meter ê (e) The cutoff wavelenth is the wavelenght for which Kmax = 0, i.e., ê ê lcut = h c f = 1240 eV nm 4.2 eV º 300 nm ê ê (a) The energy of the incoming photon, and therefore the initial kinetic energy of the electron, is Eg = h c l = 1240 eV nm 200 nm = 6.2 eV We are told the work function is f = 4.2 eV. Once we take 4.2 eV from the electron we are left with ê H L ê Kmax = 2.0 eV quiz3solutions.nb(b) The work function is the minimum amount of energy taken from the electron as it escapes from the metal. Most 2 electrons lose more energy as they escape, either because they are more tightly bound to atoms on the surface, or because they are coming from atoms below the surface (and escaping from below the surface takes energy). The point is that some electrons don't have any kinetic energy at all by the time they escape, so they just barely leave the surface, and then don't go anywhere. Kmin = 0. (c) The stopping potential is the potential that takes all of the Kmax away, i.e., q Vs = Kmax Vs = 2.0 eV -e = -2.0 Volts That was a little bit slick, but you can verify it like H L ê H L 1.6µ10-19 Joules -19 Vs = 2.0 eV µ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ1ÅÅÅÅÅÅÅÅeV ÅÅÅÅÅÅÅÅÅÅÅÅÅ -1.6 µ 10 Coulombs = -2.0 Volts Now, qualitatively, the stopping potential is applied to the receiver, and it stops electrons; therefore, it must be negative - to Hrepel theL electronsI tryingM toë H reach the receiver. L (d) This can be done just converting units (using the fact that each photon has an energy of 6.2 eV). ÅÅÅÅÅÅÅÅWattsÅÅÅÅÅÅÅÅ µ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ1 Joule ÅÅÅÅÅÅÅÅsecondÅÅÅÅÅÅÅ µ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ1ÅÅÅÅÅÅÅÅeV ÅÅÅÅÅÅÅÅÅÅÅÅÅ µ ÅÅÅÅÅÅÅÅ1 photonÅÅÅÅÅÅÅÅÅÅÅ = µ 18 2 2.0 meter2 1 Watt 1.6µ10-19 Joules 6.2 eV 2 10 photons second meter ê (e) The cutoff wavelenth is the wavelenght for which Kmax = 0, i.e., ê ê lcut = h c f = 1240 eV nm 4.2 eV º 300 nm ê ê.
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