ch40 - 1515 Chapter 40 1. The magnitude L of the orbital...

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Unformatted text preview: 1515 Chapter 40 1. The magnitude L of the orbital angular momentum L is given by Eq. 40-2: ( 1) L . On the other hand, the components z L are z L m , where ,... m . Thus, the semi-classical angle is cos / z L L . The angle is the smallest when m , or 1 cos cos ( 1) ( 1) . With 5 , we have 1 cos (5/ 30) 24.1 . 2. 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 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(6) 2 = 72. 3. (a) We use Eq. 40-2: 34 34 1 3 3 1 1.055 10 J s 3.65 10 J s. L (b) We use Eq. 40-7: z L m . 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 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 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. CHAPTER 40 1516 (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 c h . 5. (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. 6. 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, = 2, which means N = 2(2 2 + 1) = 10. (c) Here = 1, so N = 2(2 1 + 1) = 6. (d) Now = 0, so N = 2(2 0 + 1) = 2. 7. (a) Using Table 40-1, we find = [ m ] max = 4. (b) The smallest possible value of n is n = max +1 + 1 = 5....
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ch40 - 1515 Chapter 40 1. The magnitude L of the orbital...

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