Classifica(on  of  Solids  Many possible ways to classify solids. We’ve discussed BLs and symmetry. Could consider e.g., metallic, insulating, semiconducting, etc. Here we choose to classify according to bonding forces (related to configuration of valence electrons). So, first review a little atomic physics

Spectroscopic notation n = Princ. Q# : 1, 2, 3, …. l = Orbital Q# : 0, 1, 2, ….. l < n s, p, d, f, g, … m = Azimuthal Q# : -l ≤ m ≤ +l (2l + 1 values) σ = spin Q# : ± 1 (2 values)

Periodic table Generally quantum states of electrons in atom specified by four quantum numbers (Q#s) (hydrogen atom). We begin with independent electron picture.

A given energy level Enlmσ depends rather strongly on n and l, but much more weakly on m and σ. For a given l,m,σ the most tightly bound electron is the one with the lowest possible value of n. Similarly, for given n,m,σ electron with the lowest possible l is bound the most “tightly”. This behavior together with Pauli exclusion principle, permits one to establish Rules for constructing atomic states. These rules (with some exceptions) allow one to determine atomic ground states of the elements. (S-O interaction, exchange)

The Hydrogen Atom  

urVEmur

ur

rurr

)]([2sin1)(sin

sin1)(1

22

2

2222

2

−=∂∂

∂∂

∂∂

−∂∂

−φθθ

θθθ

Schroedinger Eq (one elec. In Coulomb Pot.).

)()(),(),,()(),,( φθφθφθφθ ΦΘ== YandYrRruwhere

Eigenfunct. of L2  

)(,)(cos),(),,()1(),(ˆ 22 mlePYandYllYL imml ≥=+= φθφθφθφθ

The Radial Eq.  rR

rmllrVEmrR

drd

r

]2

)1()([2)( 2

2

22

2

+−−−=

Centrifugal Pot.

E  

r  

Centrifugal  Pot.  

Coulomb  Pot.=  -­‐  e2/r.  

Effec9ve  Pot.  

eVnZEn 2

2

6.13−=

Q#’s  -­‐-­‐  n,l,m  

Hydrogen Atom Degeneracies  

NumberQuantumprincipalorradialnwherenemE r

n ≡−= ,2 22

4

Energy Eigenvalues

From the Series Solution .expint,1 ansionseriestheinegeranisqwhereqln ++=

In General, ).(#, rRintermsoftheistwheretln +=

Clear that there are many possible combinations of l and t yielding a given value of n -- degenerate states.

•  Example -- l = 1 with two term radial solution (n = 3) has same energy as l = 2 with a 1 term radial solution (n = 3)

(u310 or u31±1 compared with u320, u32 ±1 or u32 ±2 in the previous slide) The “extra” energy in the former case comes from k.e. (2 terms in radial w-f -- The “Wiggle” effect), while in the latter case it comes from the larger centrifugal potential (l = 2).

Atomic Physics Identical Particles - How do we deal with many electrons in atoms?

Quantum Particles are “Indistinguishable Example -- Collision of two Electrons

A 1

2 A) Initially 1 and 2 far apart

can distinguish B) During collision (A)

Indistinguishable C) After Collision

Indistinguishable

Interaction

1 or 2 ?

2 or 1 ?

“Two States differing only in the interchange of Identical Particles are one and the same state Quantum Mechanically”

Mathematically -- N particles -- symmetry of wavefunction under exchange utot (!r1,!r2,!r3 ,.., !rN ) = uT (1, 2, 3 ,..., N ) (Shorthand notation)

Interchange particles 1 and 2 ! uT (2, 1, 3 ,..., N )

uT (2,1,3,.., N ) = AuT (1, 2, 3 ,..., N ): exchange again, uT (1, 2, 3 ,..., N ) = A2uT (1, 2, 3 ,..., N )For indistinguish. Particles – must be same to within a phase factor

A = +1 symmetric; A = -1 Antisymm. Same state, so A2 = 1

Identical Particles (cont.) Schroed. Eq. For several particles (1st approx: no interaction among particles)

!!2

2m"1

2 +"22( )+V (1)+V (2)

#

$%

&

'(u(1, 2) = ETu(1, 2)

separate variables (no coupling), uT (1, 2) = ua (1)ub(2)

(Initially label particles and introduce indisting. later)

Subst. in, divide by ua(1)ub(2)

baTbbb

aaa

EEEwhereuEuVm

uEuVm

+==⎥⎦

⎤⎢⎣

⎡+∇−

=⎥⎦

⎤⎢⎣

⎡+∇−

),2()2()2(2

)1()1()1(2

22

2

21

2

(Except for labels these are same eq. ; e- f belong to same set for each case.)

Therefore, total w-f for system (2 particles) can be constructed from two indep w-f for a single particle (product) -- energy is sum of 2 single particle energies

Impose indistinguishability (w-f must be either symm. or anti-symm.

[ ]

[ ])1()2()2()1(21)2,1(

)1()2()2()1(21)2,1(

Symmetric

babaA

babaS

uuuuu

symmetricAnti

uuuuu

−=

−

+=Note: in each w-f, one particle is in state a (at x1 or x2) and one is in state b (at x1 or x2), but no way of telling which is which!

Identical Particles (cont.)

Spin and the Pauli Exclusion Principle (all particles with half-integral spin must have antisymmetric total w-f: elec., protons, neutrons,...)

Note: uA vanishes when both particles in same state, e.g.

[ ] !!!0)1()2()2()1(21)2,1( ≡−= aaaaA uuuuu

No two spin-1/2 particles can be in the same QM state (identical Quantum numbers)

Pauli Exclusion Principle Anti-symmetry requirement on total w-f means

•  Spatial part symmetric when spin part antisymmetric •  Spin part symmetric when spatial part antisymmetric

Example (2-electrons): [ ] [ ]

[ ] [ ]

[ ] [ ]

[ ] [ ]21

2121

21

2121

)()()1()2()2()1(21

)()()()(21)1()2()2()1(

21

)()()1()2()2()1(21

)()()()(21)1()2()2()1(

21

ssbaba

ssssbaba

ssbaba

ssssbaba

uuuu

uuuu

uuuu

uuuu

−−•−

+−+−+•−

++•−

+−−−+•+Singlet (Stot = 0) Deg. = 1

Triplet (Stot = 1) Deg = 3

AntiSymm Symm Addition of two spin angular momenta Stot =0, Sztot=0 Stot = 1, Sztot=1, 0, -1

Identical Particles (cont.) Exchange AS requirement new kind of “force”; electrons must move to keep total w-f AS different energy for singlet and triplet states How ? Symm. spatial w-f -- particles closer together on average than for Antisymm. Spatial w-f larger Coulomb repulsion energy. Can show that Called Exchange because terms that contribute to difference in energy between Symm and Antisymm states involve exchange of particles in the integral, e.g. terms like

uAS (r2 ! r1)2uAS !"!! uS (r2 ! r1)

2uS

)1()2()2()1(12

2

baba uurr

euu ⎟⎟⎠

⎞⎜⎜⎝

⎛

−

Spin Dependence comes from total antisymmetry requirement -- symmetric spatial part w-f can only occur with antisymmetric spin part and vice versa.

For a given pair of single particle w- f’s (ua and ub) symmetric spatial state (Singlet) has higher (Coulomb) energy than antisymmetric spatial state (Triplet). What about states involving same single particle w-f’s?

Periodic Table Ionization Potentials of the Elements

Second Ionization Potential

First Ionization Potential

Large Dips in first ionization potential at Z = 3, 11, 19, 37 and 55

Li Na Cs K Rb

Each time we have to put an electron in the next n state, the energy increases substantially, and the first ionization potential decreases.

n=2

n=1

n=3

1s (2)

3s,3p,3d (18)

2s, 2p (8)

Large Maxima in second ionization potential at Z = 3, 11, 19, 37 and 55

and Dips at Z = 4, 12, 20, 38 and 56

Ba Sr Ca Mg Be

Label electrons by n,l,m,s (indep. elect.)

Remember -- changing # of pos. charges AND # of electrons! •  Main effect (large dips) is related to n (principle quantum #) Shells •  Lowest order effect within a shell is additional Coulomb attraction. = 2 +6 + 10

Periodic Table (cont.) Hund’s Rules (What happens within a shell? -- ground state configuration)

Given energy level, Enlms , depends strongly on n and l, but much less strongly on m and s. l-dependence comes from “screening’ of the attractive Coulomb potential by other electrons. (doesn’t exist in hydrogen: e.g. 3s, 3p, 3d states are degenerate except for fine structure.)

•  For a given l, m, s the most tightly bound electron has the lowest n.

•  For a given n, m, s the most tightly bound electron has the lowest l.

•  Combine with Pauli principle (for a given n and l, the “triplet” (symmetric spin) states have the lowest energy. Therefore, fill l-shells initially with all “spin-up” and then with “spin-down”. Maximize M = Σmi . No two states can have the same set of quantum #’s, nlms) Example: Consider a particular (l = 2) subshell (2l + 1)x2 = 10 states (holds 10 electrons) (l = 0, 1 subshells are at lower energy)

2 electrons L = 3, S = 1 (M = m1 + m2 = 3; Ms = 1/2 + 1/2 = 1)

5 electrons L = 0, S = 5/2 (M = 2 + 1+ 0 -1 -2 = 0; Ms = 5x(1/2) = 5/2)

10 electrons L = 0, S = 0 (M = 0 + 0 = 0; Ms = 5/2 - 5/2 = 0)

Atomic and Molecular Spectroscopy States of Two-electron Atoms (Two electrons outside closed shell)

Ideal LS coupling Single particle labeling (e.g., 4s5p) inadequate: S-O interaction, electrostatic/Pauli, “orbit-orbit” interactions not accounted for.

Need to combine ang. mom. of 2 electrons to form resultant 2211

ˆˆˆˆˆ SLSLJ

+++=

Two logical ways: 1) LS (Russel-Saunders) Coupling (lighter elements, small S-O int.)

2) j- j coupling (heavier elements)

SLJSSSLLL

ˆˆˆˆˆˆ;ˆˆˆ2121

+=

+=+=

21

222111ˆˆˆ

ˆˆˆ;ˆˆˆ

JJJSLJSLJ

+=

+=+=

“ ”

“Straight” coul.repulsion

IS included

Exchange - symm, AS

Largest L1•L2 ∝ L Lowest energy

Spin-Orbit -- J; smallest j – lowest energy

Atomic Physics (2-Elec. Atoms Cont.) Various Terms

u  Largest -- Exchange (S =0, S = 1) - Singlet -- Symm spatial w- f. On average electrons closer together -- larger Coulomb repulsion -- larger energy (relative to direct coulomb repulsion) - Triplet -- AS spatial w-f. On average electrons farther apart -- smaller coulomb repulsion -- lower energy (relative to direct Coulomb repulsion)

u  Next Largest -- Orbit- Orbit Coulomb effect also -- ave. distance between electrons depends on relative orientation of orbits (L1•L2 = (1/2)[L2 - L1

2 - L22]) -- therefore depends on L. Largest

L has smallest energy -- keeps electrons farthest apart (not obvious).

u  Smallest -- Spin-Orbit (E = λL •S) (L•S = (1/2)[J2 - L2 - S2]) so splitting depends on J (J = L+ S, L+ S - 1, ….|L - S|). Sign of λ -- turns out to be positive -- largest J has largest energy.

Notation (usually specify by 7 Q #’s -- n1 , n2 , l1 , l2 , S, L and J)

In spectroscopic notation

2S+1LJ Multiplicity

Orbital Ang. Mom. in spectroscopic

notation

Total angular momentum e.g.,4p4d 3F2

Classifica(on  of  Solids  Pauli Principle: Remember it has to do with anti-symmetry requirement of total wavefunction for Fermions, but for present purposes, just use it in the usual elementary form: no more than one electron can occupy the same state as specified by the Q#s (n, l, m, σ).

Rules for constructing atomic configuration (shells) : Verified  experimentally  

and  theore9cally  

I.  Make table: n + l as column increasing downward, and l as row increasing to right.

II.  Allow correct number of blocks for each value of n = l(# of ways can get n + l = constant) e,g,, n = l = 5

l = 3, n = 3 l = 1, n = 4; l = 0, n = 5

III.  Fill in the blocks starting always at the top (lowest n + l value and proceeding right to left in filling blocks as we go down (highest l for a given n+l is filled first – lowest energy config.)

#  of  elec.  That  can  be  accommodated;    for  each  value  of  l  is  2(2l+1).    m  can  take  on  values    -­‐l,  -­‐l  +  1,  -­‐l  +  2,…0,  1,  2,  …+l,  

and  for  each  l  and  m,  σ  has  two  possible  values.    

See  next  page  

Classifica(on  of  Solids  and  Periodic  Table  Energy Ordering :

n.l   n  +  l   Designa(on   Capacity  -­‐2(2l+1)  

6,2   8   6d   10  (trans.  Elements)  

5,3   8   5f   14  (ac9nides)  

7,0   7   7s   2  

6,1   7   6p   6  

5,2   7   5d   10  (trans.  Elements)  

4,3   7   4f   14  (rare  earths)  

6,0   6   6s   2  

5,1   6   5p   6  

4,2   6   4d   10  (trans.  elements)  

5,0   5   5s   2  

4,1   5   4p   6  

3,2   5   3d   10  (trans.  elements)  

4,0   4   4s   2  

3,1   4   3p   6  

3,0   3   3s   2  

2,1   3   2p   6  

2,0   2   2s   2  

1,0   1   1s   2                                Shell  

E  

Shell  

Shell  

Lowest  n  +  l  first,  Then  highest  l  for  

a  given  n  +  l  

Classifica(on  of  Solids  and  Periodic  Table  

n  +  l  \l   0  (2)  

1  (6)  

2  (10)  

3  (14)  

4  (18)  

1   [1s2]  He  [1s1]  H  

2   [2s2]Be  [2s1]Li  

3   [3s2]Mg  [3s1]Na  

 [2s2,2p1]B  

4    [4s2]Ca  [4s1]K  

 [3s2,2p2]Si  [3s2,3p1]Al  

5    [5s1]Rb  

 [3d10,4s2,  4p1]Ga  

   [4s2,3d1]Sc  

6    [6s1]Cs  

 [4d10,5s2,  5p1]In  

   [4d1,5s1]Sc  

7    [7s1]Fr  

 [5f14,5d10,  6s2,6p1]Tl  

   [5d1,6s2]La  

Ordering of Configurations - start at upper left (lowest energy) proceeding right to left as go down These are lowest energy configurations

2s  

1s  

3s  

4s  

6s  

5s  

7s  

One  block:  n=1,  l  =  0  

One  block:  n=2,  l=0   Two  blocks:  

n=3,  l=0  n=2,  l=1  

Two  blocks:  n=4,  l=0  n=3,  l=1  

2p  

3p  

[nlN]  

Nota9on  (refers  to  outermost  

occupied  shell)  #  of    Electrons  in  lth  state  

Spectroscopic  nota9on  

4p  

 Three  blocks:  

n=6,  l=0  n=5,  l=1  n=4,  l=2  

 

 Three  blocks:  

n=5,  l=0  n=4,  l=1  n=3,  l=2  

 

3d  

Now  make  up  usual  periodic  table  by  “plohng”  

n  down,  and  #  of  val.  electrons  more  than  a  filled  

shell  to  the  right  

n  #   …  8  columns  S  (2)  p  (6)  

5p   4d  

6p   5d   4f  

 Four  blocks:  n=7,  l=0  n=6,  l=1  n=5,  l=2  n=4,  l=3  

 

Classifica(on  of  Solids  and  Periodic  Table  Contracted  Periodic  Table  

Lej  out  transi9on  elements,  rare  earths  and  ac9nides  for  simplicity    

                     #        n  

I   II   III   IV   V   VI   VII   VIII  

1   H1s   He1s2  

2   Li2s   Be2s2   B2p1   C2p2   N2p

3   O2p4   F2p5   Ne2p6  

3   Na3s   Mg3s2   Al3p1   Si33p2   P3p3   S3p4   Cl3p5   Ar3p6  

4   K4s   Ca4s2   Ga4p1   Ge4p2   As4p3   Se4p4   Br4p5   Kr4p6  

5   Rb5s   Sr5s2   In5p1   Sn25p2   Sb5p3   Te5p4   I5p5   Xe5p6  

6   Cs6s   Ba6s2   Tl6p1   Pb6p2   Bi6p3   Po6p4   At6p5   Rn6p6  

7   Fr7s   Ra  

He  +  

Ne  +  

Ar  +  

Kr  +  

Xe  +  

Rn  +  

#  of  elec.  Beyond  

filled  shell  

Discussion:    He,  Ne,  Ar,  Kr,  Xe  and  Rn  are  closed  shell  atoms  (inert  gases);  outer  elec.  are  9ghtly  bound  –    large  ioniz.  energies.  Adding  one  elec.  (and  a  proton)  to  these  atoms  leads  to  weak  binding  (Li,  Na,  K,  Rb,    Cs,  Fr  have  one  elec.  outside  closed  shell  –  easy  to  detach    -­‐-­‐-­‐  very  reac(ve  materials.      Conversely,  atoms      with  one  elec.  less  than  closed  shell    (Column  VII  –  F,  Cl,  Br,  I,  At)  will    bind  an  addi(onal  electron  very    (ghtly.      (also  very  reac9ve)  

Classifica(on  of  Solids  -­‐  bonding  Forma(on  of  Crystals  in  terms  of  decreasing  bonding  energy  Ionic  Crystals:    e.g,,  LiF,  NaCl,  etc.  In  such  crystals  Alkali  atom  (Li,  Na,  K,  …  )  from  Column  I  gives  up  an  elec.  to  the  halogen  atom  (Column  VII),  e.g.,  F,  Cl,  Br,  …).  

 Leaves  (Alkali)+  and  (Halogen)-­‐  ions:    direct  Coulomb  forces  hold  crystal  together.  

V(r)  

r  

Core    repulsion  

Coulomb  anrac9on    

Due  to  Pauli  princ.  and  stable  closed  shell  config.  of  ions.  requires  excess  charge  distribu(on  introduced  in  

neighborhood  of  each  ion  by  neighbor  be  accommodated  in  unoccupied  level.  Large  gap  between  occ.  and  unocc.  Levels  –  costs  a  lot  of  energy  to  force  ions  together  

Covalent  crystals:    e.g.,    Ge,  Si,  etc.  Neighboring  atoms  tend  to  share  electrons.  Substan9al  electron  density  between  atoms.  Shared    electrons  must  sa9sfy  the  Pauli  principle.    H2  molecule  simplest  example  of  covalent  bond.    Covalent  bonds  very  direc9onal  and  show  strong  angular  proper9es;  e.g.,  diamond  structure  (Si  and  Ge)  -­‐  nearest  neighbors  in  tetrahedral  coordina9on  (corner  atom  and  4  nn  along  body  diagonals.  (look    at  diamond  structure.    Bonds  account  for  large  resistance  ot  shear  forces  in  these  materials.    Also  usually  Poor  conductors  (electrons  are  localized  and  paired  off).  Each  atom  shares  one  of  its  outermost  elec.    with  each  of  its  4  nns.  Therefore  surrounding  any  atom,  electrons  form  a  closed  shell  (s2,  p2  config.  –  sp    Hybrid  wavefunc9ons)  .  

Classifica(on  of  Solids  -­‐  bonding  Metallic  Crystals:  e.g.,  Na,  Al,  Au,  Fe…)  Conduc9on  electrons  outside  closed  shell  config.  (Na  has  one  electron  outside)behave  as  nearly  free  elec.  Become  detached  from  atom,  and  “wander  throughout  the  crystal.    These  electron  act  as  nega9vely    charged  “glue”  that  holds  posi9ve  ions  (on  lahce  sites)  together.  (A  VERY  SIMPLE  MINDED  AND    QUALITATIVE  PICTURE.    Usual  bonding  (ionic,  covalent)  is  of  no  importance  in  such  cases).      

More  or  less  uniformly    distributed  electron  density  -­‐  The  nega9vely  charged  “glue”  

+  +   +  +   +  +  

+   +   +   +  +   +  

+   +  +   +  +   +  +  

+  +   +  +   +   +  

+   +  +  +  +   +  

Simple  picture  gives  an  inkling  of  why  metals  (mostly)(  are  duc9le  and  have  high    electrical  conduc9vi9es  (more  about  this  later).  

Classifica(on  of  Solids  -­‐  bonding  

Molecular  crystals:  e.g.,    A  C  H4  (bonding  energies  about  0.1  eV/Molecule)  Bonding  forces  are  Van  der  Waals  forces  –  fluctua9ons  in  9me  of  electric  dipole  moment  of  one  molecule    induces  dipole  m0oment  in  neighboring  molecule,  and  these  anract  one  another;  ave.  dip.  Mom.  is  zero,  But  at  any  instant  there  may  be  a  net  dipole  moment.  Interac9on  energy  of  tow  neighboring  dipoles  is  propor9onal  to  the  product  of  the  moments  divided  by    cube  of  distance  separa9ng  them.    Very  weak  anrac9ve  force.    Simple  Van  der  Waals  crystals  are  rare  gases    (inert  gases  –  Ar,  Ne,…)  at  very  low  temperatures.    

Students  read  this  sec9on  in  A&M