Magnetic Field Lines for a Current Loop

µr

nAA ˆ I I ==rrµ

A circulating charge q

TqI =, where T is period of motion.

Magnetic field like bar magnet.

A current loop =

Current I flowing in circle in x-y plane.Magnetic dipole moment

TAm2

Tr m2

Tr 2rmprL q

2

qq ===×=ππrrr

L2m

qˆATqˆA I

q

rr=== nnµ

Orbital angular momentum

Magnetic dipole moment

Dipole moment vector is normal to orbit,with magnitude proportional to angular momentum.

Holds for any set of charges with same ratio q/mq in orbital motion.

electron For electrons, q=−e and L2m

e

e

rr−=µ

Magnetic dipole moment vector is anti-parallel to angular momentum vector.

Bound electron in the external B field

sincedtLdBr

rrr=×= µτ

Torque results in precession of angular momentum vector.

Larmor precession frequency B2m

e

e

=Lω

( )B and ˆLd µ⊥L

grrr mr ×=τ

Just like gyroscope :spinning in gravity field

the rate at which the axle rotates about the vertical axis

pd Mghdt Iφω

ω= =

( ) φθ dsinLLd =r

( ) dtdt

= θτ sinBL

2me

e

Potential energy of system

( )Brr

•−=−=−= µθτ dddWdUChange in orientation of µ relative to B produces change in potential energy,

Defining orientation potential θµµ BcosB −=•−=rrU

0.0 0.5 1.0

-1.0-0.8-0.6-0.4-0.20.00.20.40.60.81.0

U (µ

B)

θ (π)B // rrµ B //

rrµ−For electron B //

rrL− B //

rrL

Quantum considerationElectron bound to Hydrogen atom

L and Lz are quantized on atomic scale

Magnetic dipole moment

( )12meL

2me

ee

+== llhrrµ

l

h m2meL

2me

ez

e

==zµ

magnitude

z-component

Electron has probability distribution, not classical orbit.Quantization of L and Lz means that µ and µz are also quantized.

ll hhrrrr mU L

ez

ee

m2meBBL

2meBL

2meB ωµ ===•−=•−=

llh BmEmEE Bnn µω +=+= LTotal energy −l, …, 0, …, l

Degeneracy partially broken: total energy depends on n and ml.

Energy diagram for Z=1

l=2l=0 l=1E B=0 B=0B≠0 B≠0

-13.6eV

-3.4eV

1s

2s 2p

n=1

n=2 n=2

n=2, l=1, ml=1n=2, l=1, ml=0n=2, l=1, ml=−1

B≠0

BE Bµω ==∆ Lh

L0,0,112,1,

0,0,12,1,0

L0,0,12,1,1

ωωωωω

ωωω

−=

=

+=

→−

→

→

o

o

ooω

-1.5eV-0.85eV

3s4s

3p4p

3d4d

A triplet spectral lines when B≠0.

Normal Zeeman Effect 1896 Lorentz Zeeman

1902

1853~1928

1902

1865~1943

First observation of spectral linesplitting due to magnetic field.

Requires Quantum Mechanics (1926)to explain.

ml=1ml=0ml=−1

Magnetic field present

Lo ωω hh +Lo ωω hh −

n=2, l=1No magnetic field

oωhml=0n=1, l=0

ml210−1−2

−110 l=1

l=2

2,3ω

n=3

n=1

n=2

Selection rule

1,0m1

±=∆=∆

l

l

l=00

1,2ω 0,2ω

Bm2eB

eL

hh == Bµω

A triplet of equally spaced spectral lines when B≠0 is expected.Normal Zeeman effect

Selection ruleEnergy spacing =(5.79×10-5 [eV/T])×B[T]

For B=1Tesla, rad/s1078.8 and eV1079.5 10L

5L ×=×= − ωωh

Mysteries :Other splitting patterns such as four, six or even moreunequally spaced spectral lines when B≠0 are observed.Anomalous Zeeman effect

Existence of electron spin

Electron spin Stern

1943

1888~19691889~1979

( ) [ ]BrUrrrr

•−−∇=−∇= µF

zdzdBzFz zF µ==

r

Translational force in z-direction is proportional to z-component of magnetic dipole moment.

Quantum prediction

zdzdBgmz

dzdB

Bµµ l

r−== zF

Gerlach

Direct observation of energy level splitting in an imhomogeneous B field.

Assuming B depends only on zBx, By : constant ; Bz=B(z) only

Bx, By : constantBz=B(z) only

Stern and Gerlach, Z. Physik 9, 349 (1922).

?

Ag atom in ground state

l=0, ml=0outermost electrons

Expectation No separation

Expectations for l=1 :

Three separate lines

Experimental results

B

Not zero, but two linesNo B field With B field

Two lines were observed.

Total magnetic moment is not zero.Orbital angular momentum cannot be the source of quantized magnetic moment. l = 0, 1

Same result for Hydrogen atom (1927) : two lines were observedPhipps and Taylor

B-field off: No splitting

B-field on: Two peaks!

Gerlach's postcard, dated 8 February 1922, to Niels Bohr. It shows a photograph of the beam splitting, with the message, in translation: “Attached [is] the experimental proof of directional quantization. We congratulate [you] on the confirmation of your theory.”

Goudsmit Uhlenbeck1925, Goudsmit and Uhlenbeck

proposed that electron carries intrinsic angular momentum. spin

Experimental result requires

212 =+s21

=s 1900~19881902~1978

spin quantum number

New angular momentum operator S

[ ] eezS Ψ=Ψ hsm

[ ] ( ) ee ssS Ψ+=Ψ 22 1 h

eΨ±= h21

eΨ

= 2

23

21

h

Both cannot be changed in any way. Intrinsic property

Electron SpinThe new kind of angular momentum is called the electron The new kind of angular momentum is called the electron SPINSPIN. . Why call it spin?Why call it spin?

If the electron were spinning on its axis, it would have angular momentum and a magnetic moment regardless of its spatial motion.

However, this “spinning” ball picture is not realistic, because it would require that the tiny electron be spinning so fast that parts would travel faster than c !So we cannot picture the spin in any simple way … the electron’s spin is simply another degree-of-freedom available to electron

B≠0+µsBB=0

24

24e

9.2848 10 J/Tesla -- electron magnetic moment

= e /2m 9.2741 10 J/Tesla (the "Bohr Magneton" related to orbital motion)s

B

µ

µ

−

−

= ×

≈ = ×h

Note: All particles have spin (e.g., protons, neutrons, quarks, photons)

B∆E = 2µs|B|−µsB

Image of Spinning Electron

Convenient to picture electron spin as the result of spinning charge distribution.

Spin is a quantum property.

Electron is a point-like object with no internal coordinates.

Magnetic dipole moment

S2m

eg

e

err

−=sµ

ge : electron gyromagnetic ratio =2.0023 by measurementAgree with prediction from Quantum Electrodynamics

Sr

Only two allowed orientations of spin vector

Recent Breakthrough Recent Breakthrough ––Detection of a Detection of a singlesingle electron spin!electron spin!

(Nature July 15, 2004) (Nature July 15, 2004) ---- IBM scientists IBM scientists achieved a breakthrough in achieved a breakthrough in nanoscalenanoscalemagnetic resonance imaging (MRI) by magnetic resonance imaging (MRI) by directly detecting the faint magnetic directly detecting the faint magnetic signal from a signal from a singlesingle electron buried electron buried inside a solid sample. inside a solid sample.

Nature 430, 329 (2004).

Next step Next step –– detection of single nuclear spin (660x smaller).detection of single nuclear spin (660x smaller).

So, we need FOUR quantum numbers to specify the electronic state of a hydrogen atom

n, l , ml, ms (where ms = −1/2 and +1/2)Complete wavefunction : product of spatial wave function ψ

and spin wave function χ±.

[ ] ±± ±= χχ h21 zS [ ] ±± = χχ 22

43 hS

Spin wave function χ± : eigenfunction of [Sz] and [S2]

( ) ( ) ±=± χφθψ ,),r( mne YrR ll

r

statesn=1,2,3,….l=0,1,2,…, n-1ml=0,±1,±2,…, ±lms=±1/2

EigenvaluesEn=−13.6(z/n)2eV

( )hll 1+=LhlmLz =hsmSz =

Wavefunction

Degeneracy in the absence of magnetic field :Each state l has degenerate states.2×(2l+1) ( )∑ +

−

=

1

0122

n

l

lEach state n has degenerate states.2n2

In the strong magnetic fields,

( )zeze

SgL2meBBB +=•−•−=

rrrrsLU µµ

S and Lrr

precess independently around Br

( )Bmgm2me

ee

s+= l

h

Taking spin contribution into account

For an electron (ge=2)

( )B1m2me

e

+=↑ l

hU

( )B1m2me

e

−=↓ l

hU

spin up ms=1/2

spin down ms=−1/2ml=0,±1,±2,…, ±l

e2mBe2 h

×=−≡∆ ↓↑ UUUFor a given ml, Lωh2=

Contribution to energy shifts.

ml=1ml=0ml=−1

ml=±1, ms=m1/2ml=0, ms=1/2

ml=−1, ms=−1/2ml=0, ms=−1/2

Magnetic field present

n=1,ml=0

Lo ωω hh +

ml=1, ms=1/2

ml=0, ms=1/2

ml=0, ms=−1/2

Lo ωω hh −

The spin-orbit interactionIn the absence of magnetic field, internal field generated by electron motion (proportional to orbital angular momentum) will interact with spin dipole moment.

e−

BLrr

→Sr

Zp

Nucleus circulates around electron

S2m

eg

e

err

−=sµ

Frame of electronS with parallel is Lwhen rr

Frame of nucleus

e−

Lr

Sr

ZpB-field due to nuclear motionOrbital dipole moment is anti-parallelwith spin dipole moment.

Brr

•−= sU µhigher energy state

>0

e−

Sr

BLrr

→

ZpS

2meg

e

err

−=sµ

Frame of electronS with parallel-anti is Lwhen rr

Frame of nucleus

Nucleus circulates around electron

e− ZpB-field due to nuclear motionOrbital dipole moment is parallel with spin dipole moment.

Brr

•−= sU µlower energy state

<0

( )SLUo

LS

rr•= 322

e

e2

rc4m1

2g

4Zeπε

2p

1s

2p3/22p1/2

Lr Sr

sµr

Sr

sµr

.separately conserved is Snor LNeither rr

SLJrrr

+=

( )hr

1jjJ +=

hjmJz =

Square of total angular momentum ( ) SL2SLSLJ 2222rrrrrrr

•++=+=

2SLJSL

222rrr

rr −−=•

( ) ( )

−+−+=•∝

4311jj

2SL

2

llhrr

LSUEnergy shifts due to spin-orbit interaction

magnitudeDefining total angular momentumz-component

Back to 2p state ml=1, ms=±1/2

z

ml=1

ms=1/2

Lr

Jr

Sr

z

23j = mj=3/2,1/2,−1/2,−3/2

Projections :mj=3/2

zml=1

Lr

Jr

Sr

z

mj=1/2,−1/221j =

Projections :

mj=1/2ms=−1/2

How to determine j and mj ?

21mmmm ±=+= ll sjSince Jz=Lz+Sz

Range for j obtained from extremes of aligned and anti-aligned

S and Lrr

nsorientatio z maximum with S and LBoth rr

21j += lAligned

mj=l+1/2, l+1/2-1, … , − (l+1/2)Possible projections :

21j −= lnsorientatio z maximum with S and LBoth

rrAnti-aligned

mj=l-1/2, l+1/2-1, … , − (l-1/2)Possible projections :

( ) ( )

−+−+=

4311jj lloLS UUSpin-orbital energy shift

2p3/2

2p1/2

2s1/2

j=3/2,l=1,s=1/2

j=1/2,l=1,s=1/2

j=1/2,l=0,s=1/2

( ) ( )

−+−

++=

431111

23

23E2pE 23/2 oU

( ) ( )

−+−

++=

431111

21

21E2pE 21/2 oU

( ) ( )

−+−

++=

431101

21

21E2sE 21/2 oU

For a hydrogen atom, Uo=1.5×10-5eV

=1

=−2

=0

mjB=0 B≠0, but small

(2)2s,2p

2p3/2

2p1/2 j=1/2,

3/21/2−1/2−3/2

−1/21/2

j=3/2,l=1(4)

2s1/2ms=±1/2 j=1/2,l=0(8)

−1/21/2

l=1(2)

Spectroscopic notation

j 12s L + symbol

l=0 s “sharp”

l=1 p “principle”

“diffuse”l=2 d

l=3 f “fundamental”

l=4 g

l=5 h

l=6 i

Pauli

19451900~1958

Pauli Exclusion Principle

From spectra of complex atoms, Wolfgang Pauli (1925) deduced a new rule:

“In a given atom, no two electrons* can be in the same quantum state, i.e. they cannot have the same set of quantum numbers n, l , ml, ms”-- i.e., every “atomic orbital with n, l , ml” can hold 2 electrons: (↑↓)

“Pauli Exclusion Principle”

Therefore, electrons do not pile up in the lowest energy state, i.e, the (1,0,0) orbital.

They are distributed among the higher energy levels according to the Pauli Principle.Particles that obey the Pauli Principle are called “fermions”

**Note: More generally, no two identical fermions (any particle with spin of ħ/2, 3ħ/2, etc.) can be in the same quantum state.state.

Two particles is an infinite quantum well of length L

Review : single particle solution 2

222

n 2mLnE hπ

=( )

= x

Lnsin

L2xn

πψ with

( )

= 111 x

L4sin

L2x πψ

( )

= 222 x

L3sin

L2x πψ

particle 1 is at n1=4 stateparticle 2 is at n2=3 state

Assuming

What is the solution for the two particle system ?

( ) =21 x,xψ ( )

= 111 x

L4sin

L2 x πψ ( )

222 xL

3sinL2 x πψ?

Right or False

distinguishable or indistinguishable

≠

2L,

4L

4L,

2L PPe.g.Case (I) distinguishable

( ) =21 x,xψ ( )

= 111 x

L4sin

L2 x πψ ( )22 xψ

2xL

3sinL2 π

( ) ( )

== 2

21

22

22121 x

L3πsinx

L4πsin

L4x,xx,x ψPProbability density

x1=L/2

x2=L/2 0x,2L

2 =

ψ

0x,2L

2 =

P

−=

11 xL4πsin

L2

2L,xψ

=

12

21 xL4πsin

L4

2L,xP

Case (II) indistinguishable ( ) ( ) 212

221 x,xx,x ψψ =

Probability density must be unchanged if particles are switched.

( ) ( ) ( ) ( )[ ] xxxx2

112212211 ψψψψ +

( ) =21 x,xψ( ) ( ) ( ) ( )[ ] xxxx

21

12212211 ψψψψ −

+

= 2121 x

L4πsinx

L3πsinx

L3πsinx

L4πsin

L2

21( )21 x,xSψ

−

= 2121 x

L4πsinx

L3πsinx

L3πsinx

L4πsin

L2

21( )21 x,xAψ

2

22

2

222

2

222

2mL52

2mL4

2mL3E hhh πππ

=+=Eigenvalue

( ) ( )1221 x,xx,x SS ψψ =

( ) ( )1221 x,xx,x AA ψψ −=

( ) ( )1221 x,xx,x SS PP =

( ) ( )1221 x,xx,x AA PP =However,

Any crucial difference between PS(x1, x2) and PA(x1, x2) ?

The particles are on average farther apart in an antisymmetric spatial statethan they are in a symmetric one.

Exchange Symmetry

The simplest case : Two identical particle systemindistinguishable

Suppose two particles satisfy 3D Schrödinger equations with coordinates r1 and r2, and no interaction between them,

( ) ( ) ( ) ( ) ( ) ( )1a11aa1a11a21

2

rrrErrUrm2

rrrrrrh ψψψψ H==+∇−

( ) ( ) ( ) ( ) ( ) ( )2b22bb2b22b22

2

rrrErrUrm2

rrrrrrh ψψψψ H==+∇−

( ) ( )[ ] ( ) ( )212121 r,rEr,rrr rrrrrrψψ =+HH

( ) ( ) ( )2b1a21 rrr,rrrrr

ψψψ =Particle a has energy Ea and particle b has energy Eb.

w/. E=Ea+Eb

Distinguishable solution( ) ( ) ( )1b2a12 rrr,r

rrrrψψψ =r1 ⇔ r2Interchange particles

Particle a has energy Eb and particle b has energy Ea.

Try combination of the two wave functions

( ) ( ) ( ) ( ) ( )[ ]1b2a2b1a21 rrrr2

1r,rrrrrrr

ψψψψψ +=

( ) ( ) ( ) ( ) ( )[ ]1b2a2b1a21 rrrr2

1r,rrrrrrr

ψψψψψ −=

( ) ( )[ ] ( ) ( ) ( )[ ] ( ) ( ) ( ) ( )[ ]1b2a2b1a212121 rrrr2

1rrr,rrrrrrrrrrrrr

ψψψψψ ++=+ HHHH

( ) ( ) ( ) ( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )[ ]1b2a22b1a2

1b2a12b1a1

rrrrrr2

1

rrrrrr2

1

rrrrrr

rrrrrr

ψψψψ

ψψψψ

HH

HH

++

+=

( ) ( ) ( ) ( )[ ]

( ) ( ) ( ) ( )[ ]1b2aa2b1ab

1b2ab2b1aa

rrErrE2

1

rrErrE2

1

rrrr

rrrr

ψψψψ

ψψψψ

++

+=

also solutions of Schrödinger Eq.w/. E=Ea+Eb

( ) ( ) ( ) ( ) ( )[ ]1b2a2b1aba rrrr2

1EE rrrr

ψψψψ ++= ( ) ( )21ba r,rEE rr

ψ+=

Interchange particles r1 ⇔ r2

( ) ( ) ( ) ( ) ( )[ ]1b2a2b1a21 rrrr2

1r,rrrrrrr

ψψψψψ +=

( ) ( ) ( ) ( ) ( )[ ]2b1a1b2a12 rrrr2

1r,rrrrrrr

ψψψψψ +=

symmetric

SIdentical !

( ) ( ) ( ) ( ) ( )[ ]1b2a2b1a21 rrrr2

1r,rrrrrrr

ψψψψψ −=

( ) ( ) ( ) ( ) ( )[ ]2b1a1b2a12 rrrr2

1r,rrrrrrr

ψψψψψ −= ( )21 r,r rrψ−=

anti-symmetric

A

Both satisfy the Exchange Symmetry Principles

( ) ( ) 212

221 r,rr,r rrrr

ψψ = No observable difference

For both the symmetric and antisymmetric total eigenfunctions, the probability density function are not changed

by an exchange of the particle labels.

A two electron system : such as the Helium atom

( ) ( ) ( ) ( )1aa1a1

1a21

e

2

rErr4ee2r

m2rrrh ψψ

πεψ =

−+∇−

o

2e+

( )1a rrψ

( )2b rrψ

( ) ( ) ( )1aa1a rEr1 rrψψ =h

( ) ( ) ( )2bb2b rEr2 rrψψ =h

( ) ( ) ( ) ( ) ( )[ ]1b2a2b1a21S rrrr2

1r,r rrrrrrψψψψψ +=

( ) ( ) ( ) ( ) ( )[ ]1b2a2b1a21A rrrr2

1r,r rrrrrrψψψψψ −=

( ) ( )χψ rr spacerr

=Ψspin

ΨA for FermionsΨS for Bosons

Pauli exclusion principle

( ) ( ) ( ) ( )2bb2b2

2b22

2

rErr4e2er

m2rrrh ψψ

πεψ =

−+∇−

o

Total wave function

The symmetry character of various particles

1symmetricDeuteron1symmetricPhoton0symmetricπ meson0symmetricHe atom (G)0symmetricα particle

1/2AntisymmetricMuon1/2AntisymmetricNeutron1/2AntisymmetricProton1/2AntisymmetricPositron1/2AntisymmetricElectron

Spin (s)Spin (s)symmetrysymmetryparticleparticle

Bosonzero or integer spin

Fermionhalf-integer spin

Ex. Ground state of the Helium atom

Construct explicitly the two electron ground state wave functionfor the Helium atom in the independently particle approximation.

Each Helium electron sees only the doubly charged Helium nucleus.

Hydrogen like atom with the lowest energy, Z=2

( ) oa2r2/3

o100 a

21r −

= e

πψ w/. energy E1=−22(13.6eV)

Total eigenfunction= (space eigenfunction) × (spin eigenfunction)

( ) ( ) 2/111001a rr ±= χψψrr ( )1100 rr±ψ

and Ea= Eb=−54.4eV( ) ( ) 2/121002b rr ±= χψψ

rr ( )2100 rr±ψ

Experimentally, −24.6eV

( ) ( ) ( ) ( ) ( )[ ]110021002100110021S rrrr2

1r,r rrrrrrψψψψψ +=

( ) ( ) ( ) ( ) ( )[ ]110021002100110021A rrrr2

1r,r rrrrrrψψψψψ −=

Space eigenfunction

E=−108.8eV

Spin eigenfunction

( ) ( )[ ]21 ,2

12

1 ,21

21

+−−−+=AχSinglet state

s=0, ms=0

( ) ( )[ ]

+−+−+= 21 ,2

12

1 ,21

21

Sχ

( )21 ,2

1 ++

( )21 ,2

1 −−

ms=0

ms=1

ms=−1

Triplet state

s=1, ms=0, ±1

The ground state : only one choice 1S2, spin singlet E=−108.8eV

( ) ( ) ( ) ( ) ( ) ( )[ ]110021002100110021s21A rrrr2

1r,rr,rrrrrrrrr

−+−+ −== ψψψψχψ AΨ

The first excited state : 1S2S and E=−13.6eV×[(4/1)+(4/4)]=−68eV

E (eV)

-70

dataexpectation S n=1, l=0; n=2, l=1-50

-110

-90

singlet

n=1; n=1

n=1; n=2

n=1, l=0; n=1, l=0

n=1, l=0; n=2, l=0ST

T

Electron interactions and Screening effects

( ) ( ) ( )21r,r 21 hhH +=rr

?( )( )

21 rr4ee

−−−

+oπε

Helium atom

Each electron sees not only the attractive Helium nucleus but also the other electron.

Repulsive interaction weakens the attractive potential.

Is the charge of nucleus +2e from electron’s point of view ?

Inner electrons screen the positive charge of nucleus and hence, reduce the effective Z and the attractive potential.

smaller negative potential energyEspecially, for multi-electron atoms.

Energy is lower for lower l/ more-elliptical orbits.Fill electrons to states : according to the lowest value of n+l

with “preference” given to n

Build a multiple electron atomChemical properties of an atom are

determined by the least tightly bound electrons.

Occupancy of subshellEnergy separation between the subshell and the next higher subshell

s shell l=0

p shell l=1

n

1

2

3

Helium and Neon and Argon are inert…

their outer subshellis closed.

Hydrogen: (n, l, mℓ, ms) = (1, 0, 0, ±½) in ground state.In the absence of a magnetic field, the state ms = ½ is degenerate

with the ms = −½ state.

Helium: (1, 0, 0, ½) for the first electron.(1, 0, 0, −½) for the second electron.

Electrons have anti-aligned spins (ms = +½ and ms = −½) .

The principle quantum number has letter codes.n = 1 2 3 4...Letter = K L M N… n = shells (eg: K shell, L shell, etc.)

nl = subshells (e.g.: 1s, 2p, 3d)

How many electrons may be in each subshell?

l = 0 1 2 3 4 5 …Letter = s p d f g h …

2(2l + 1)For each l : (2 l + 1) values of mℓ

2For each mℓ: two values of ms

Total

1s

1s2

1s22s1s22s2

1s22s22p1

1s22s22p2

1s22s22p3

1s22s22p4

1s22s22p5

1s22s22p6

1s22s22p63s

1s22s22p63s2

Energies of orbital subshells increase with increasing l.

The lower l values have more elliptical orbitsthan the higher l values.

Electrons with higher l values are more shielded from the nuclear charge.

Electrons lie higher in energy than those withlower l values.

4s fills before 3d.

1

23

34

8

455

56667

78

n+l

1s2s 2p

7s 7p 7d 7f 7g 7h 7i 6s 6p 6d 6f 6g 6h

3s 3p 3d4s 4p 4d 4f5s 5p 5d 5f 5g

Groups:Vertical columns.Same number of

electrons in an l orbit.Can form similar

chemical bonds.

Periods:Horizontal rows.Correspond to filling of the subshells.

Inert Gases:Last group of the periodic tableClosed p subshell except heliumZero net spin and large ionization energyTheir atoms interact weakly with each other

Alkalis:Single s electron outside an inner coreEasily form positive ions with a charge +1eLowest ionization energiesElectrical conductivity is relatively good

Alkaline Earths:Two s electrons in outer subshellLargest atomic radiiHigh electrical conductivity

Halogens:Need one more electron to fill outermost Need one more electron to fill outermost subshellsubshellForm strong ionic bonds with the alkalisForm strong ionic bonds with the alkalisMore stable configurations occur as the More stable configurations occur as the pp subshellsubshell is filledis filled

Transition Metals:Three rows of elements in which the 3Three rows of elements in which the 3d, 4, 4d, and 5, and 5d are being filledare being filledProperties primarily determined by the Properties primarily determined by the ss electrons, rather than by the electrons, rather than by the d subshellsubshell being filledbeing filledHave Have d--shell electrons with unpaired spinsshell electrons with unpaired spinsAs the As the d subshellsubshell is filled, the magnetic moments, and the tendency is filled, the magnetic moments, and the tendency for neighboring atoms to align spins are reducedfor neighboring atoms to align spins are reduced

Lanthanides (rare earths): Z=57~Z=70Have the outside 6s2 subshell completedAs occurs in the 3d subshell, the electrons in the 4f subshell have unpaired electrons that align themselvesThe large orbital angular momentum contributes to the large ferromagnetic effects

Actinides: Z=89~Z=102Inner subshells are being filled while the 7s2 subshell is completeDifficult to obtain chemical data because they are all radioactiveHave longer half-lives

X-ray spectra and Moseley’s lawMoseleyTarget :W

1887~1915

series of sharp lines: characteristic spectrumCharacteristic of the target material

1913 Moseley measured characteristic X-ray spectra of 40 elements.Observed “series” of X-ray energy levels

called K, L, M, … etc.Analogous to optical series for Hydrogen (e.g. Lyman, Balmer, Paschen…)

Rigaku Miniflex X-ray tube Target : Cu

Scattered incident electronIncident

electron

Ejected K-shell electron

Hole in K-shellThe characteristic spectrumdiscovered by Bragg systematized by Moseley

The characteristic peak is created when a hole in the inner shellcreated by a collision event is filled by an electron from higher energy shell.

Hole in L-shell

X-ray

When a K-shell electron be knocked out, the vacancy can be filled by an electron from L-shell (Kα radiation) or

the M-shell (Kβ radiation).

γ Lδ

Nα Nβn=5

K

L

MMα Mβ Mγ

On=4N

n=3

Lα Lβ Ln=2

Kα K

n=1

β Kγ Kδ Kε

Moseley plot

K series : n=2,3,… to n=1

( )22K 1

n11V6.31E −

−== zehc

λ

L series : n=3,4,… to n=2

( )222L 4.7

n1

21V6.31E −

−== zehc

λ

L series

K series

Kα

Kβ

Lα

Tc

Pm

Re

α：n+1 nβ：n+2 nγ：n+3 nδ : n+4 nε : n+5 n

Mo, z=42( ) keV15.17142

211eV6.31E 2

2K =−

−=

α

m1023.7 11−×=λPhil. Mag. (1913), p. 1024