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Physical Chemistry (II)
Lecture 22
CHEM 3172-80
Lecturer: Hanning Chen, Ph.D.04/17/2017
Spin-Magnetic Field Coupling
Quiz 21
10 minutes
Please stop writing when the timer stops !
Origin of Electric and Magnetic FieldsHow to generate an electric field using an electron ?
−
How to generate a magnetic field using an electron ?
−
−Coulomb’s law
(1758)
E = 14πε0
qr2
Charles de Coulomb
electron rotation
µ = IS
SI B = µ0
4π3r µ i r( )
r5− µr3
⎛⎝⎜
⎞⎠⎟
Ampere’s law (1825)
Andre M. Ampere
Spinssome magnetic moments do NOT stem from the rotations of electric charges
Spin “looks” like an “intrinsic” rotation
Sun
Earth
Spin: self-rotation
Angular momentum: collective rotation
!µtotal =
!µL +
!µS
!µL : orbital rotation
!µS : spin rotation
Spin quantum number: S
Bosons: integer spin S = 0, 1, 2,....
Fermions: half-integer spin S = 1/2, 3/2, 5/2,....
electron is the most chemically important Fermion with S = 1/2
Spin magnetic quantum number: ms
−S,−S +1,...,S −1,S{ }degeneracy: 2S +1
Total Angular Momentum
z
l(l+1)!
ml!
l: magnitude of the angular moment ml: orientation of the angular moment
ml! +ms!
s: magnitude of the spin moment ms: orientation of the spin moment
s(s +
1)! ms!
l,ml , s,ms( )
ml : magnetic quantum number
ms : spin magnetic quantum number
µzz
BBz
magnetic coupling energy
Em = −!µ i!B
a magnetic dipole tends to be stabilized (destabilized) by a parallel (anti-parallel) magnetic field
Em = −!µzBz
parallel dipole
anti-parallel dipole +!µzBz
−!µzBz
ΔEm = 2µzBz µz must be quantized !!!
Energy-level Splitting in a Magnetic Field
Quantization of the Magnetic Dipole Moment
!µL : orbital rotation
!µL = γ
!L
!L : orbital angular momentumγ : magnetogyric ratio
γ = q2m
chargemass
for electron, γ e = −1.760 ×1011 rad/(s i T)
Lz : projection on axis z Lz = ml!
µz = mlγ e! = mlµB µB : Bohr magneton: 9.274 ×10−24JT-1
µB : fundamental unit of electron magnetic dipole moment
Energy gap between px ml = −1( ) and py ml = +1( )orbitals under 100 Tesla magnetic field
ΔEm = (+1)µBBz − (−1)µBBz = 2µBBz = 2 × 9.274 ×10−24 ×100 = 1.12 ×10−2 eV
(the world’s most powerful magnet)
(comparable with kT at room temperature)low temperature environment is preferred.
Separation of Two Electron Spin States
!µS : spin rotation
!µS = geγ e
!S
!S : spin angular momentumγ e : magnetogyric ratio of electron
ge : g-factor 2.002319.... inseparable space and time (relativistic theory)
ΔEm = geµBB
↑ ms = + 12
↓ ms = − 12
under 100 Tesla magnetic field
ΔEm = 1.12 ×10−2 eV
basis of electron paramagnetic resonance (EPR)radical molecules (unpaired electrons)
Larmor frequency
vL =γ eB2π
("electron rotation frequency")
= hvL
Nuclear Magnetic Moment
nucleus 1H 13C 15N 14N 19F 31P
I 1/2 1/2 1/2 1 1/2 1/2natural
abundance 99.98% 1.11% 0.37% 99.64% 100% 100%
g-factor 5.586 1.405 -0.566 0.404 5.255 2.268
γ I GHzT−1( ) 0.267 0.067 −0.027 0.252 0.1080.0193
nuclear orbital magnetic dipole moment is usually negligible, its spin component may NOT be !
12C and 16O have I=0 and hence no nuclear magnetic dipole momentno energy-level splitting under a magnetic field
natural abundance: 12C : 98.89% natural abundance: 16O : 99.76%
Proton Magnetic Resonance Spectroscopyapparently proton is the most suitable isotope for magnetic resonance spectroscopy (NMR)
ΔEp = gIµI B
↑ mI = + 12
↓ mI = − 12
I : nuclear spin quantum number
I p =12
µI : nuclear magneton
µP =
e!2mP
= 5.051×10−27 JT −1
under 100 Tesla magnetic field ΔEp = gIµI B = 1.92 ×10−5 eV
ΔEe = 1.12 ×10−2 eV ≫ ΔEPfor an unpaired electron,
an electron is much more sensitive to an external magnetic field than a proton
why do not we detect electrons ? Do not we have more electrons than nuclei?
Experimental Setup of NMR
the sample is subject to two incident electromagnetic wavesstrong magnetic field: ~10 Tesla resonant radio: ~100 MHz
ΔEp = gIµI B
hv = ΔEp
v = vLLarmor
frequency
vL =γ I B2π
Intensities in NMR Spectra
ΔEp = gIµI B
α
β
Nα
Nβ
I sti−abs∝B 12N
α
I sti−emi∝B 2
1Nβ
B12 = B21Nα ?Nβ
Population difference:Nβ
Nα
= e−gIµI BkbT
INMR ∝ Nα − Nβ ∝ (1− e−gIµI BkbT ) ≈1− 1− gIµI B
kbT⎛⎝⎜
⎞⎠⎟= gIµI B
kbT
(Boltzmann law)
NMR intensity:
two possible ways to improve the NMR resolution:stronger magnetic field
lower temperature800 MHz77K
radio wave
Chemical Shift
− E0E '
a charged particle in a dielectric medium
Eloc = E0 − E ' = (1−α )E0E ' =αE0 α : electrical shielding constant
µI B0B '
a magnetic dipole in an electron cloud
Bloc = B0 − B ' = (1−σ )B0B ' =σ B0 σ : magnetic shielding constant
vL =γ I Bloc2π
= 1−σ( )γ I B02π
reduced Larmor (resonant) frequency
Definition of chemical shift: δ = vL − vL0
vL0 ×106 vL
0 : standard resonant frequency
1H : Si(CH4 )4Reference systems: 13C : Si(CH4 )415 P : H3PO4
δ = 0
Range of Typical Chemical Shifts for 1H
δ =1−σ( )− (1−σ 0 )
(1−σ 0 )×106 = σ 0 −σ
1−σ 0
×106 ≈ σ 0 −σ( )×106The relation between chemical shift and magnetic shielding constant:
The systems with large chemical shifts are well deshielded (small shielding constants).deshielding shielding
Range of Typical Chemical Shifts for 13C
deshielding shielding
the Origin of Shielding Constants
σ total =σ local +σ neighbour +σ solvent
Local effect: σ local =σ diamagnetic +σ paramagnetic
Bz
Bz'
the precession of magnetic dipole moment
Lamb formula:
σ d =e2µ012πme
1rNe
> 0
the contribution from a given shielding electron
the electron density and
wavefunction compactness
Bz
the density of unpaired electrons
σ p < 0
Bz'
Neighboring Group Contributions
H
XBzθ
µz
Bz'
Bz' =
µz 1− 3cos2θ( )
r3If group X is perfectly isotropic
Bz' = dθ
0
2π
∫µz 1− 3cos
2θ( )r3
= 0
If group X is anisotropic
θ = 0 µz = χ!Bzθ = π
2µz = χ⊥Bz
σ neighbour =
χ! − χ⊥( ) 1− 3cos2Θ( )r3
H
X
symmetry axis of X
Θ
Θ : constant for a rigid molecule
positive ornegative
σ neighbour
Solvent Contributions
Benzene is one of the most widely used organic solvents
Solute
deshielded region δ ↑
shielded region
δ ↓
sometimes, the formation of hydrogen bonds may lead to a change of δ too
Review of Homework 6Review of Homework 2113.C1 Light-induced degradation of molecules, also called photobleaching, is a serious problemin fluorescence microscopy. A molecule of a fluorescent dye commonly used to label biopolymerscan withstand about 106 excitations by photons before light-induced reactions destroy its π system, andthe molecule no longer fluoresces. For how long will a single dye molecule fluoresce while beingexcited by 1.0mW of 488 nm radiation from a continuous-wave argon ion laser? You may assume thatthe dye has an absorption spectrum that peaks at 488 nm and that every photon delivered by the laser is absorbed by the molecule.
The energy of each photon: E = hcλ
= 4 ×10−19 J
The rate of photon emission: 1.0mW4 ×10−19 J
= 1.0 ×10−3Js−1
4 ×10−19 J= 2.5 ×1015 s−1
t = 106
2.5 ×1015 s−1= 4 ×10−10 s = 0.4ns
Homework 22
Reading assignment: Chapters 14.1, 14.2, 14.3, 14.4 and 14.5
Homework assignment: Exercises 14.1 Problems 14.1
Homework assignments must be turned in by 5:00 PM, April 18th, Tuesday
to my mailbox in the Department Main Office located at Room 4000, Science and Engineering Hall
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