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# CH-40 - Chapter 40 1(a For a given value of the principal...

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Chapter 40 1. (a) For a given value of the principal quantum number n , the orbital quantum number ranges from 0 to n 1. For n = 3, there are three possible values: 0, 1, and 2. (b) For a given value of , the magnetic quantum number m ranges from   to  . For  1 , there are three possible values: – 1, 0, and +1. 2. For a given quantum number there are (2 + 1) different values of m . For each given m the electron can also have two different spin orientations. Thus, the total number of electron states for a given is given by N = 2(2 + 1). (a) Now = 3, so N = 2(2 3 + 1) = 14. (b) In this case, = 1, which means N = 2(2 1 + 1) = 6. (c) Here = 1, so N = 2(2 1 + 1) = 6. (d) Now = 0, so N = 2(2 0 + 1) = 2. 3. (a) We use Eq. 40-2: ( ) ( ) ( ) 34 34 1 3 3 1 1.055 10 J s 3.65 10 J s. L �� ��   h (b) We use Eq. 40-7: z L m h . For the maximum value of L z set m = . Thus [ ] ( ) 34 34 max 3 1.055 10 J s 3.16 10 J s. z L �� �� h 4. For a given quantum number n there are n possible values of , ranging from 0 to n – 1. For each the number of possible electron states is N = 2(2 + 1) . Thus, the total number of possible electron states for a given n is ( ) 1 1 2 0 0 2 2 1 2 . n n n l l N N n (a) In this case n = 4, which implies N n = 2(4 2 ) = 32. 1585

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CHAPTER 40 (b) Now n = 1, so N n = 2(1 2 ) = 2. (c) Here n = 3, and we obtain N n = 2(3 2 ) = 18. (d) Finally, n N n 2 2 2 8 2 ch . 5. The magnitude L of the orbital angular momentum L r is given by Eq. 40-2: ( 1) L   h . On the other hand, the components z L are z L m h , where ,... m  , . Thus, the semi-classical angle is cos / z L L q . The angle is the smallest when m  , or 1 cos cos ( 1) ( 1) q q h     h With 5 , we have 24.1 . q 6. (a) For 3 , the greatest value of m is 3 m . (b) Two states ( m s  1 2 ) are available for 3 m . (c) Since there are 7 possible values for m : +3, +2, +1, 0, – 1, – 2, – 3, and two possible values for s m , the total number of state available in the subshell 3 is 14. 7. (a) Using Table 40-1, we find = [ m ] max = 4. (b) The smallest possible value of n is n = max +1  + 1 = 5. (c) As usual, m s  1 2 , so two possible values. 8. For a given quantum number n there are n possible values of , ranging from 0 to 1 n . For each the number of possible electron states is N = 2(2 + 1). Thus the total number of possible electron states for a given n is ( ) 1 1 2 0 0 2 2 1 2 . n n n N N n Thus, in this problem, the total number of electron states is N n = 2 n 2 = 2(5) 2 = 50. 9. (a) For  3 , the magnitude of the orbital angular momentum is ( ) 1 L   h ( ) 3 3 1 12 h h . So the multiple is 12 3.46. 50
(b) The magnitude of the orbital dipole moment is orb   1 12 b g B B . So the multiple is 12 3.46. (c) The largest possible value of m is 3 m

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