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Unformatted text preview: 1549 Chapter 40 1. (a) Using Table 401, we find l = [ m l ] max = 4. (b) The smallest possible value of n is n = l max +1 ≥ l + 1 = 5. (c) As usual, m s = ± 1 2 , so two possible values. 2. (a) For 3 = l , the greatest value of m l is 3 m = l . (b) Two states ( m s = ± 1 2 ) are available for 3 m = l . (c) Since there are 7 possible values for m l : +3, +2, +1, 0, – 1, – 2, – 3, and two possible values for s m , the total number of state available in the subshell 3 = l is 14. 3. (a) For a given value of the principal quantum number n , the orbital quantum number l ranges from 0 to n – 1. For n = 3, there are three possible values: 0, 1, and 2. (b) For a given value of l , the magnetic quantum number m l ranges from  l to + l . For l = 1, there are three possible values: – 1, 0, and +1. 4. For a given quantum number l there are (2 l + 1) different values of m l . For each given m l the electron can also have two different spin orientations. Thus, the total number of electron states for a given l is given by N l = 2(2 l + 1). (a) Now l = 3, so N l = 2(2 × 3 + 1) = 14. (b) In this case, l = 1, which means N l = 2(2 × 1 + 1) = 6. (c) Here l = 1, so N l = 2(2 × 1 + 1) = 6. (d) Now l = 0, so N l = 2(2 × 0 + 1) = 2. 5. (a) We use Eq. 402: ( 29 ( 29 ( 29 34 34 1 3 3 1 1.055 10 J s 3.65 10 J s. L = + = + × ⋅ = × ⋅ l l h (b) We use Eq. 407: z L m = l h . For the maximum value of L z set m l = l . Thus CHAPTER 40 1550 [ ] ( 29 34 34 max 3 1.055 10 J s 3.16 10 J s. z L = = × ⋅ = × ⋅ lh 6. For a given quantum number n there are n possible values of l , ranging from 0 to n – 1. For each l the number of possible electron states is N l = 2(2 l + 1). Thus, the total number of possible electron states for a given n is ( 29 1 1 2 2 2 1 2 . n n n l l N N n = = = = + = ∑ ∑ l l (a) In this case n = 4, which implies N n = 2(4 2 ) = 32. (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 . 7. The magnitude L of the orbital angular momentum L r is given by Eq. 402: ( 1) L = + l l h . On the other hand, the components z L are z L m = l h , where ,... m =  + l l& l . Thus, the semiclassical angle is cos / z L L θ = . The angle is the smallest when m = l , or 1 cos cos ( 1) ( 1) θ θ = ⇒ = + + lh l l l h l l With 5 = l , we have 1 cos (5/ 30) 24.1 . θ = = ° 8. For a given quantum number n there are n possible values of l , ranging from 0 to 1 n . For each l the number of possible electron states is N l = 2(2 l + 1). Thus the total number of possible electron states for a given n is ( 29 1 1 2 2 2 1 2 . n n n N N n = = = = + = ∑ ∑ l l l l Thus, in this problem, the total number of electron states is N n = 2 n 2 = 2(5) 2 = 50....
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 Summer '10
 Prof.Yang
 Physics, Photon, Atomic orbital, Principal quantum number

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