Chapter40 - 1. (a) Using Table 40-1, we find = [ m ]max =...

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1. (a) Using Table 40-1, we find A = [ m A ] max = 4. (b) The smallest possible value of n is n = A max +1 A + 1 = 5. (c) As usual, m s 1 2 , so two possible values.
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2. (a) For 3 = A , the greatest value of m A is 3 m = A . (b) Two states ( m s 1 2 ) are available for 3 m = A . (c) Since there are 7 possible values for m A : +3, +2, +1, 0, – 1, – 2, – 3, and two possible values for s m , the total number of state available in the subshell 3 = A is 14.
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3. (a) For a given value of the principal quantum number n , the orbital quantum number A ranges from 0 to n 1. For n = 3, there are three possible values: 0, 1, and 2. (b) For a given value of A , the magnetic quantum number m A ranges from A to + A . For A = 1, there are three possible values: – 1, 0, and +1.
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4. For a given quantum number A there are (2 A + 1) different values of m A . For each given m A the electron can also have two different spin orientations. Thus, the total number of electron states for a given A is given by N A = 2(2 A + 1). (a) Now A = 3, so N A = 2(2 × 3 + 1) = 14. (b) In this case, A = 1, which means N A = 2(2 × 1 + 1) = 6. (c) Here A = 1, so N A = 2(2 × 1 + 1) = 6. (d) Now A = 0, so N A = 2(2 × 0 + 1) = 2.
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5. (a) We use Eq. 40-2: () ( ) ( ) 34 34 1 3 3 1 1.055 10 J s 3.65 10 J s. L −− =+ × = × AA = (b) We use Eq. 40-7: z Lm = A = . For the maximum value of L z set m A = A . Thus [ ] ( ) 34 34 max 3 1.055 10 J s 3.16 10 J s. z L == × = × A=
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6. For a given quantum number n there are n possible values of A , ranging from 0 to n – 1. For each A the number of possible electron states is N A = 2(2 A + 1). Thus, the total number of possible electron states for a given n is () 11 2 00 22 1 2 . nn n ll NN n −− == + = ∑∑ A A (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, nN n =→ = = 2 8 2 c h .
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7. The magnitude L of the orbital angular momentum L G is given by Eq. 40-2: (1 ) L =+ AA = . On the other hand, the components z L are z Lm = A = , where ,... m =− + A A . Thus, the semi-classical angle is cos / z LL θ = . The angle is the smallest when m = A , or 1 cos cos ) ) θθ ⎛⎞ =⇒ = ⎜⎟ ++ ⎝⎠ A= A = With 5 = A , we have 1 cos (5/ 30) 24.1 . == °
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8. For a given quantum number n there are n possible values of A , ranging from 0 to 1 n . For each A the number of possible electron states is N A = 2(2 A + 1). Thus the total number of possible electron states for a given n is () 11 2 00 22 1 2 . nn n NN n −− == + = ∑∑ A AA A Thus, in this problem, the total number of electron states is N n = 2 n 2 = 2(5) 2 = 50.
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For a given value of A , the greatest that m A can be is A , so the smallest that LL xy 22 + can be is =AA A =A +− = 1 2 b g . The smallest possible magnitude of m A is zero, so the largest + can be is + 1 b g . Thus, =A ≤+ + 1 bg . 9. Since LLLLLL xyz xy z 222222 =++ += , . Replacing L 2 with AA = + 1 2 and L z with m A = , we obtain m 2 1 + A
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10. (a) For n = 3 there are 3 possible values of A : 0, 1, and 2. (b) We interpret this as asking for the number of distinct values for m A (this ignores the multiplicity of any particular value). For each A there are 2 A + 1 possible values of m A .
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This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter40 - 1. (a) Using Table 40-1, we find = [ m ]max =...

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