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