Magnetic Field Lines for a Current Loopocw.nctu.edu.tw/upload/classbfs12090325531508.pdf ·...
Transcript of Magnetic Field Lines for a Current Loopocw.nctu.edu.tw/upload/classbfs12090325531508.pdf ·...
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