CHAPTER 45 - 9 - 49. Wefindtheequivalentmassofthephotonfrom...

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    E e = 3 5 E F 1 2 N = 3 5 1 2 N h 2 8 m e 3 N π 2 V 2/ 3 .     E e = 3 5 1 2 N h 2 8 m e 9 N 8 π 2 R 3 2/ 3 = 3 Nh 2 320 m e R 2 9 N π 2 2/ 3 . 49. We find the equivalent mass of the photon from m  =  E / c 2 . To escape from a mass  M  and radius  R , the energy of the photon must be greater than the gravitational potential  energy at the surface: mc 2  =  GMm / R , so the Schwarzschild radius is         R  =  GM / c 2 . 50. If there are  N  nucleons, approximately half of them will be protons.  Thus there will be  ! N  electrons.  From Eqs.  41–12 and 41–13, the average energy of an electron for a system where the energy levels are filled to the Fermi  energy is 3 E F /5.  Thus the total energy of the electrons is For a spherical volume, we get The Fermi energy for the nucleons will have a similar expression; however, because  m n  »  m e   , their Fermi energy 
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This note was uploaded on 09/13/2010 for the course PHYSICS 7 taught by Professor ? during the Spring '08 term at University of California, Berkeley.

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CHAPTER 45 - 9 - 49. Wefindtheequivalentmassofthephotonfrom...

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