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# chap36 - ATOMIC PHYSICS 36 Note For the problems in this...

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36.1 ATOMIC PHYSICS Note: For the problems in this chapter, useful numerical values of Planck’s 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 π = = × ⋅ = 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 π = × ⋅ = × ⋅ = = = × ⋅ = × ⋅ = = × = = × = = × = × h E XERCISES Section 36.1 The Hydrogen Atom 14. 2 34 2 11 0 0 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 πε π π × = = = × × × h while 15. I NTERPRET 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. D EVELOP 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. E VALUATE So 2 1.5 eV 13.6 eV/ , n E n = − = which yields 13.6/1.5 3. n = = A SSESS 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. I NTERPRET This problem is about the allowed values of the magnitude of angular momentum. D EVELOP The magnitude of orbital angular momentum is given by Equation 36.9: ( 1) , 0,1,2, ... L l l l = + = h E VALUATE 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 0 (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 π × = − = = − × = − × × h

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36.2 Chapter 36 6 7 42, × = etc. Evidently, (d) is an erroneous possibility. A SSESS 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. I NTERPRET We’re given the magnitude of the orbital angular momentum of an electron and asked to find its corresponding orbital quantum number. D EVELOP The magnitude of orbital angular momentum is given by Equation 36.9: ( 1) , 0,1,2, ... L l l l = + = h E VALUATE From the above equation, we have ( 1) 30, l l + = or 5. l = A SSESS In this case, the -value l is easily found by inspection. In general, l is a nonnegative integer solution of the quadratic formula. 20. A 6 f state has 6 n = and 3 l = (see the explanation following Example 36.2). Thus, 13.6 eV/36 0.378 eV, E = − = − and 34 34 12(1.054 10 J s) 3.65 10 J s L = × = × (see Equations 36.6 and 35.9). 21. I NTERPRET We’re given the energy and orbital angular momentum of an electron and asked about its state. We need to find both principal and orbital quantum numbers.
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