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Unformatted text preview: 36.1 ATOMIC PHYSICS Note: For the problems in this chapter, useful numerical values of Plancks constant, in SI and atomic units, are: 34 15 6.626 10 J s 4.136 10 eV s 1240 eV nm/c, h = = = and 34 /2 1.055 10 J s h = = = h 16 6.582 10 eV s 197.3 MeV fm/ . c = Useful constants and combinations, in SI and atomic units, are: 34 15 34 16 8 2 27 2 23 5 6.626 10 J s 4.136 10 eV s 1240 eV nm/ /2 1.055 10 J s 6.582 10 eV s 197.3 MeV fm/ 2.998 10 m/s, 1.440 eV nm, 1 1.661 10 kg 931.5 MeV/ 1.381 10 J/K 8.617 10 eV/K B h c h c c ke u c k = = = = = = = = = = = = = h EXERCISES Section 36.1 The Hydrogen Atom 14. 2 34 2 11 2 31 19 2 4 4 (8.854 pF/m)(6.626 10 J S/2 ) 5.293 10 m (9.109 10 kg)(1.602 10 C) a me = = = h while 15. INTERPRET Our system consists of a group of hydrogen atoms in the same excited state characterized by the quantum number n . The 1.5 eV minimum energy is the ionization energy. DEVELOP The hydrogen energy levels are given by Equation 36.6: 1 2 2 13.6 eV n E E n n = = where n is the principal quantum number. The ionization energy for this state is the difference between the zero of energy and the energy of the state. EVALUATE So 2 1.5 eV 13.6 eV/ , n E n = = which yields 13.6/1.5 3. n = = ASSESS The result makes sense since principal quantum must be a positive integer. 16. The quantum number l can take integer values from 0 to 1, n so its maximum value is 6. From Equation 36.9, 6 7 42 . L = = h h 17. INTERPRET This problem is about the allowed values of the magnitude of angular momentum. DEVELOP The magnitude of orbital angular momentum is given by Equation 36.9: ( 1) , 0,1,2, ... L l l l = + = h EVALUATE Successive values of ( 1), l l + starting with 3, l = are: 3 4 12, 4 5 20, 5 6 30, = = = 36 2 34 2 18 1 2 31 11 2 (6.626 10 J s/2 ) 2.180 10 J 13.61 eV 2 2(9.109 10 kg)(5.292 10 m) E ma = = = = h 36.2 Chapter 36 6 7 42, = etc. Evidently, (d) is an erroneous possibility. ASSESS Since orbital angular momentum is quantized, only certain discrete values are allowed. 18. From Equation 36.9, 2 2 ( 1) ( / ) (2.585/1.054) 6.01 6. l l L + = = = h Therefore 2. l = Since 1, l n the smallest value of n is 3; thus the minimum energy is 2 13.6 eV/(3) 1.51 eV = (from Equation 36.6). 19. INTERPRET Were given the magnitude of the orbital angular momentum of an electron and asked to find its corresponding orbital quantum number. DEVELOP The magnitude of orbital angular momentum is given by Equation 36.9: ( 1) , 0,1,2, ... L l l l = + = h EVALUATE From the above equation, we have ( 1) 30, l l + = or 5. l = ASSESS In this case, the value l is easily found by inspection. In general, l is a nonnegative integer solution of the quadratic formula....
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.
 Fall '10
 Pheong

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