magresspec - Magnetic Resonance Spectroscopy In our...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Magnetic Resonance Spectroscopy In our discussion of spectroscopy, we have shown that absorption of E.M. radiation occurs on resonance: When the frequency of applied E.M. field matches the energy splitting between two quantum states. Magnetic resonance differs from these other methods in the sense that we need to immerse the same in a magnetic field in order to see the levels that we probe with an external (rf or μ wave) field. (Two fields: Static magnetic and E.M.) We will be probing the energy levels associated with the spin angular momentum of nuclei and electrons: NMR--nuclear magnetic resonance and ESR/EPR--electron spin resonance. Angular momentum: In our treatment of rotational energy levels, we said that the energy levels depended on the rotational angular momentum, L , which was quantized: L 2 = = 2 J J ( + 1 ) J = 0,1,2 rot . quant . number Degeneracy of J was ( m J = 0, , ± J ) ( 2 J + 1 ) 2 We related L to the energy levels E rot = L 2 ( + ) BJ J 1 2I Actually, all angular momentum is quantized. If a particle can spin, it has A.M. and quantized E levels. In particular, we also have to be concerned with the spin of individual nuclei and electrons. 5.33 Lecture Notes: Magnetic Resonance Spectroscopy Page 1
Image of page 1

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

View Full Document Right Arrow Icon
You already know that electrons have… ORBITAL angular momentum 2 2 M = = A A ( + 1 ) A = 0,1,2 orbital angular momentum quantum number degeneracy of orbitals: 2 A + 1 from… m A = − A 1 , + A magnetic quantum number m A represents the quantization of the components of M : z A = 1 Projection of M onto ˆ z axis: M = = 2 M + 1 = M Z = m A = (How we choose ˆ z doesn’t matter until 0 = we apply a magnetic field.) 1 = Now, the angular momentum that we are concerned with is: Electron Spin Angular Momentum S 2 = = 2 s s + 1 s: electron spin quantum number = ½ 1 ( ) for each unpaired e 2 S = m = m : ± 1 ( S, S 1, , + S + ) z s s 2 one unpaired e Two paired electrons: s = 0. Two unpaired electrons (triplet): s = 1. 5.33 Lecture Notes: Magnetic Resonance Spectroscopy Page 2
Image of page 2
Nuclear spin angular momentum I 2 = = 2 I I ( + 1 ) I: nuclear spin quantum number I z = m I = m I : I , I + 1, , I What is I ? ¾ Each proton/neutron has a spin quantum number of 1/2. ¾ Spin of many nucleons add to give I . ¾ Pairing dictated by a shell model of nucleus. (analogous -not identical- to electron spin pairing) ¾ Protons and neutrons add separately. ¾ Spins pair up. Paired spins I = 0 . Some basic rules: 1. For even number of protons plus even number of neutrons: = 12 C, 16 O I 0 . 2. For mixed even/odd number of nucleons, spin is half-integer. ( 1 / 2 > I > 9 / 2 ) For one unpaired nucleon I = 1 2 m I = ± 1 2 degeneracy 2 I + 1 = 2 1 H 13 C 15 N , , So the proton and electron are similar—both spin 1 2 particles . We’ll talk about these two particles more specifically… 3. For odd/odd number of nucleons, I is integer > 0.
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
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