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Unformatted text preview: 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 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, × = × = × = 36.1 36 2 34 2 18 1 2 31 11 2 ( . J s/ ) 6 626 10 2 . J . eV 2 180 10 13 61 ( . kg)( . m) 2 2 9 109 10 5 292 10 E ma π × ⋅ =  = =  × =  × × h 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, thevalue l is easily found by inspection. In general, l is a nonnegative integer solution of the quadratic formula....
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 Spring '09
 Prof.Kim
 Physics, Atom, Angular Momentum, orbital angular momentum

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