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# Elementary Modern Physics PHY 215

WFU

GPA 3.91

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This 3 page Class Notes was uploaded by Miss Lucienne Hamill on Wednesday October 28, 2015. The Class Notes belongs to PHY 215 at Wake Forest University taught by Staff in Fall. Since its upload, it has received 24 views. For similar materials see /class/230736/phy-215-wake-forest-university in Physics 2 at Wake Forest University.

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Date Created: 10/28/15

Chapter 4 Q1 An emission spectrum for hydrogen can be obtained by analyzing light from hydrogen gas that has been heated to very high temperatures the heating populates many excited states of hydrogen An absorption spectrum can be obtained by passing light from a broadband incandescent source through hydrogen gas If the absorption spectrum is obtained at room temperature when all atoms are in the ground state the absorption spectrum will 1 be identical to the emission spectrum 2 contain some but not all of the lines appearing in the emission spectrum 3 contain all the lines seen in the emission spectrum plus additional lines 4 look nothing like the emission spectrum Pl Tipler 415 Calculate the longest wavelengths in the Lyman series Hf l in nm and indicate their position on a horizontal linear scale Indicate the series limit shortest wavelength on this scale Are any of these in the visible spectrum P2 Tipler 424a Calculate the energies of the three lowest states of positronium Questions 24 68 from Tipler Q2 If an electron moves in an orbit of greater radius does its total energy increase or decrease Does its Kinetic Energy increase or decrease Q3 What is the energy of the shortest wavelength photon that can me emitted from hydrogen Q4 How would you characterize the motion and location of an electron with E0 and n gt oo in Figure 416 F 3 Tipler 429 What is the approximate radius of the n l orbit of gold Z 79 Compare this with the radius of the gold nucleus 7 about 71 fm EX 4 Tipler 437 If one could ll the FranckHertz tube with positronium what cathode grid voltage would be needed to reach the 2quotd current decrease More Derivation of Compton s Equation Let 1 and M be the wavelengths of the incident and scattered x rays respectively as shown in Figure 321 The corresponding momenta are plicic 1 and 7 E2 7 11 P2 C 2 using f G Since Compton used the KX line of molybdenum 00711 nm see Figure 318b the energy of the incident x ray 174 keV is much greater than the binding energy of the valence electrons in the carbon scattering block about 11 eV therefore the carbon electron can be considered to be free Conservation of momentum gives P1P2Pe 52 hi2 m2 p2 Fig 321 The scattering of x rays can be treated as a collision of a photon of initial momen tum hlx1 and a free electron Using conservation of momentum and energy the momentum of the scattered photon m2 can be related to the initial momentum the electron mass and the scattering angle The resulting Compton equation for the change in the wavelength of the x ray is Equation 340 Continued 16 More p 1 12 7 2p1p2 2 27 341 p1pz 21211220080 where pa is the momentum of the electron after the collision and 6 is the scattering angle for the photon measured as shown in Figure 321 The energy of the electron before the collision is simply its rest energy E0 me2 see Chapter 2 After the col lision the energy of the electron is E3 pgcz Conservation of energy gives Pic E0 PZC E3 PECZWZ Transposing the term pzc and squaring we obtain E3 C2071 1722 2CE0p1 P2 E3 PECZ 342 p2 pl 12 2121122 2E p1 7 W C If we eliminate pg from Equations 341 and 342 we obtain E 7 0P1C P2 plpza 7 COS 0 Multiplying each term by help1 sz0 and using hp we obtain Compton s equation he he A27A1El icose l 7cos0 0 h M 7 1 a l 7 cos 6 340

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