PHYS
ch40

# ch40 - 1(a For a given value of the principal quantum...

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1. (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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2. 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.
3. (a) We use Eq. 40-2: ( ) ( ) ( ) 34 34 1 3 3 1 1.055 10 J s 3.65 10 J s. L = + = + × = × A A = (b) We use Eq. 40-7: z L m = 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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4. 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 ( ) 1 1 2 0 0 2 2 1 2 . n n n l l N N 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, n N n = = = 2 2 2 8 2 c h .
5. The magnitude L of the orbital angular momentum L G is given by Eq. 40-2: ( 1) L = + A A = . On the other hand, the components z L are z L m = A = , where ,... m = − + A A A . Thus, the semi-classical angle is cos / z L L θ = . The angle is the smallest when m = A , or 1 cos cos ( 1) ( 1) θ θ § · = ¡ = ¨ ¸ ¨ ¸ + + © ¹ A= A A A = A A With 5 = A , we have 24.1 . θ = °

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6. (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.
7. (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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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 ( ) 1 1 2 0 0 2 2 1 2 . n n n N N n = = = = + = ¦ ¦ A A A A Thus, in this problem, the total number of electron states is N n = 2 n 2 = 2(5) 2 = 50.
9. (a) For A = 3 , the magnitude of the orbital angular momentum is ( ) 1 L = + = A A = ( ) 3 3 1 12 + = = = . So the multiple is 12 3.46. (b) The magnitude of the orbital dipole moment is µ µ µ orb = + = A A 1 12 b g B B . So the multiple is 12 3.46. (c) The largest possible value of m A is 3 m = = A A . (d) We use L m z = A = to calculate the z component of the orbital angular momentum. The multiple is 3 m = A . (e) We use µ µ z B m = − A to calculate the z component of the orbital magnetic dipole moment. The multiple is 3 m = − A .

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