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

Info icon This preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 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=
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 .
Image of page 4
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 = °
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 6
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.
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 8
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 .
Image of page 9

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 10
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern