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