{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}