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Chapter 28

# Chapter 28 - CHAPTER 28 Quick Quizzes 1(b The circumference...

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C HAPTER 28 Quick Quizzes 1. (b). The circumference of the orbit is n times the de Broglie wavelength ( ) 2 rn π λ = , so there are three times as many wavelengths in the 3 n = level as in the level. Also, by combining Equations 28.4, 28.6 and the defining equation for the de Broglie wavelength 1 n = ( ) hmv = , one can show that the wavelength in the 3 n = level is three times as long. 2. The quantum numbers associated with orbital states are n , ± , and m ± . For a specified value of n , the allowed values of ± range from 0 to n – 1. For each value of ± , there are (2 ± + 1) possible values of m ± . (a) If n = 3, then ± = 0, 1, or 2. The number of possible orbital states is then () [ ] () [ ] 20 1 21 1 22 1 1 3 5 9 ++ +=++= . (b) If n = 4, one additional value of ± is allowed ( ± = 3) so the number of possible orbital states is now ( ) [ ] 92 31971 6 = + = . 3. (a) For , there are 5 allowed values of ± , namely ± = 0, 1, 2, 3, and 4. 5 n = (b) Since m ± ranges from – ± to + ± in integer steps, the largest allowed value of ± ( ± = 4 in this case) permits the greatest range of values for m ± . For 5 n = , there are 9 possible values for m ± : -4, -3, -2, –1, 0, +1, +2, +3, and +4. 4. (d). Krypton has a closed configuration consisting of filled shells as well as filled 4 s and 4 p subshells. The filled =1, =2, and =3 nn n 3 n = shell (the next to outer shell in Krypton) has a total of 18 electrons, 2 in the 3 s subshell, 6 in the 3 p subshell and 10 in the 3 d subshell. 383

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CHAPTER 28 Problem Solutions 28.1 The Balmer equation is H 22 11 2 R n λ ⎛⎞ =− ⎜⎟ ⎝⎠ 1 , or 2 2 H 4 4 n Rn = When , 3 n = 7 7- 1 49 6.56 10 m 1.09737 10 m9 4 == × 656 nm = ×− When , 4 n = 7 1 41 6 4.86 m m1 6 4 × 486 nm = When , 5 n = 7 1 42 5 4.34 m m2 5 4 × 434 nm = 28.2 Start with Balmer’s equation, 2 HH 2 1 24 n RR nn 4 = , or 2 2 H 4 4 n = Substituting 1 7 H =1.0973732 10 m R × , we obtain ( ) 72 2 3.645 m 4 n n × 2 2 364.5 nm 4 n n where 3, 4, 5, . . . . n = 28.3 (a) From Coulomb’s law, ( )( ) ( ) 2 92 2 1 9 12 2 2 8.99 10 N m C 1.60 C = 1. 01 0 m e kqq F r ×⋅ × × 8 2. 31 0 N × 384
CHAPTER 28 (b) The electrical potential energy is ( )( )( ) 92 2 1 9 1 9 12 10 8.99 10 N m C 1.60 C1 . 60 C 1. 01 0 m e kqq V r −− ×⋅ × × == × 18 -19 1 eV 2. 31 0 J J ⎛⎞ =− × = ⎜⎟ ⎝⎠ × 14 eV 28.4 (a) From Coulomb’s law, ( )( ) ( ) 2 2 1 9 2 2 15 10 N m C C m e F r × = 2 0 N × × (b) The electrical potential energy is ( )( ) 2 2 1 9 10 N m C C 0 m e V r × × 13 -13 1 MeV J = 1.4 × + 28.5 (a) The electrical force supplies the centripetal acceleration of the electron, so 2 2 2 e ke v m rr = or 2 e v mr = ( )( ) ( )( ) 2 2 1 9 10 N m C C 9.11 kg 1. v × ×× 6 61 ms × (b) No . 6 3 8 5. 0 << 1 3.00 10 m s v c × × × , so the electron is not relativistic. (c) The de Broglie wavelength for the electron is hh pm v λ , or ( ) ( ) 34 6 6.63 Js 4. m × 0.46 nm = (d) Yes. The wavelength and the atom are roughly the same size. 385

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CHAPTER 28 28.6 Assuming a head-on collision, the α -particle comes to rest momentarily at the point of closest approach. From conservation of energy, , or ffi KE PE PE +=+ i ( )( ) ( )( ) 27 9 9 0 ee i fi ke e rr += + With , this gives the distance of closest approach as i r → ∞ ( )( ) ( ) 2 92 2 1 9 2 -13 158 8.99 10 N m C 1.60 10 C 158 5.0 MeV 1.60 J M eV e f i ke r ×⋅ × == × 14 4. 51 0 m = 45 fm 28.7 (a) yields 2 0 n
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Chapter 28 - CHAPTER 28 Quick Quizzes 1(b The circumference...

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