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hw25 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 25...

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PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 25. 41.17, 41.22, 41.46, 41.48; Q41.07, Q41.10 41.17. Since the Fermi energy is known, the number of electrons per volume can be found with the help of equation 41-9, on page 1150. Then, the number of atoms per volume can be found from the density and molar mass given, as in problem 1). Finally, the number of electrons per volume can be divided by the number of atoms per volume to give the number of electrons per atom. E h m n n h mE n F F 3 -3 , so J s) kg)(11.6 eV)( J eV) 0.121 m = = = × × × = × - - - 0121 1 0121 1 6 63 10 911 10 16 10 179 10 2 2 3 3 3 2 34 31 19 3 2 29 . . ( . ( . . / . / / / Then, atoms volume = 2.70 g/cm /mol atoms/mol atoms/cm atoms/m 3 3 3 27.0 g 6 02 10 6 02 10 6 02 10 23 22 28 . . . × = × = × Finally, electrons atom = electrons/m atoms/m 3 3 179 10 6 02 10 29 28 . × × . Electrons/atom = 2.98 Since aluminum has three electrons in its outer shell, this is a very reasonable result. 41.22. a) Since zinc is divalent, the number of electrons per volume equals two times the number of zinc atoms per volume. The method of calculating this is similar to that
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