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作家:宋坤祐
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    晶體鍵結
    2017/12/13 11:43:31
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    Chap. 3 Crystal binding

    Principle types of crystalline binding

    1. crystal of inert gas(VIII A): van der Wals force; dipole-to-dipole interaction

    2. ionic crystal: NaCl electrostatic force between ions(IA & VIIA)

    3. covalent crystal: C, Si(IVA) quantum effect –spin repulsive or attractive by exchange force.

    4. metallic crystal: Na, K(IA, B) charge attraction between positive and negative charges.

    Crystal of inert gas

    不考慮thermal vibration if –q分布中心在中心,則場效應=0

    He

    Ne

    Ar

    Kr

    Xe

    hcp

    ( fcc )

    1s²

    1s²

    2s²

    2p

    (Ne)

    3s²

    3p

    (Ar)

    3d¹⁰

    4s²

    4p

    (Kr)

    4d¹⁰

    5s²

    5p

    if thermal vibration at 0K, zero-point energy = ½ℏω=½cu²=½mw²u²

    u²1/2平均距離, u:位移, c:彈性係數 k

    binding force in the inert gas crystal at instant t :

    the interaction energy between two diploes, p, p, separated by R is given by

    U(R)=(p₁∙p)/R³-3(p₁∙R)(p₂∙R)/R⁵ → U(R)=(p₁∙p₂)/R³-3(p₁∙p₂)/R³=-2(p₁∙p₂)/R³

    V=pR/R³, E=-V=-(pR/R³)=-p∙(R/R³)-(R/R³)p

    p=0, 與電荷密度有關∵r距離太遠

    i.e.

    (R/R³)=[R/R³+R(1/R³]=I/R³+R[-3R⁻⁴(R)]=I/R³+R[-3R⁻⁴(R/R)]

    where =(∂∕∂x)i+(∂∕∂y)j+(∂∕∂z)k= (∂∕∂x)e, r=xi+yj+zk=j xj ej ,

    r=[ (∂∕∂x)e]∙[∑j xj ej]=[∑j (xj ∕∂x)(eej )]=δj=

    =I, i.e. δj=1 if i=j or δj=0 if ij

    E=-V=-[I/R³-3R⁻⁴(RR/R)]p=-p/r³+3R(Rp/R)=-p/R³+3R²p/R⁵=2p/R³=|E| R// p

    the induced dipole, p: pEp₂=αE, : electronic polarizability電極化率

    p₂=α(2p/R³)=2αqr₀/R³ and U(R)=-2(p₁∙p₂)/R³=(-2/R³)rq∙(2αqr₀/R³)

    =-4α(rq)²/R⁶=-A/R⁶ → F=-U=-dU/dR  1/R

    Quantum derivation of van der Waals force

    total energy, H=(p₁²/2mcx₁²)+(p₂²/2mcx₂²)+H, H1: 靜電位能

    ex. H原子, E=(n+½)ℏω

    H₁=e²/R-e²/(r+x₁)-e²/(R-x)+e²/(R-x₂+x)

    x₁, x₂<<R → (1+R/r)⁻¹=1-x/R+(x/R)²-(x/R)³+... Taylor series

    H₁=e²/R-e²/R[1-x₁/R+(x₁/R)²-(x₁/R)³+...]-e²/R[1-x₂/R+(x₂/R)²-(x₂/R)³+...]+e²/r[1-(x₂-x₁)/R+(x₂-x₁)²/R²-(x₂-x₁)³/R³+...] → H₁≈-2e²xx₂/R³

    H=(p₁²/2mcx₁²)+(p₂²/2mcx₂²)-2e²xx/R³

    let x=1/√2(x₂+x), x=1/√2(x₁-x)x₁=1/√2(xₛ+xₐ), x₂=1/√2(xₛ-xₐ)

    H=(p₁²/2mcx₁²)+(p₂²/2mcx₂²)-2exx/R³

    ={[1/√2(p+p)²/2m]+½c[1/√2(xₛ+xₐ)]²}+{[1/√2(p-p)²/2m]+½c[1/√2(xₛ-xₐ)]²}-2e²[1/√2(xₛ+xₐ)][1/√2(xₛ-xₐ)]/R³

    =[½(p²+2pp+p²+p²-2pp+p²)]/2m+c/2[½(xₛ²+2xₛxₐ+xₐ²)+½(xₛ²-2xₛxₐ+xₐ²)]-e²(xₛ²-xₐ²)/R³=p²/2m+p²/2mcxₛ²+½cxₐ²-e²(xₛ²-xₐ²)/R³

    =[p²/2m+½(c-2e²/R³)xₛ²]+[p²/2m+½(c+2e²/R³)xₐ²]

    ∵為Simple Harmonic motion E=(n+½)ℏω₀

    if H=H0+H1 H=E (H0+H1)=E H0+H1=E

    *The zero point energy(n=0) of the system

    ½ ħ(ωs+ωa)—with dipole-to-dipole interaction, H0+H1(Ep.p.+ E0-0)

    ½ ħ(ω0+ω0)—without D-D interaction(E0-0)

    U=½ ħ(ωs+ωa)-½ ħ(ω0+ω0)

    ω0=(c/m)1/2, ωs=[(c-2e²/R³)/m]1/2=(c/m)1/2[1-½(2e²/cR³)-(2e²/cR³)²+...]

    ωa=[(c+2e²/R³)/m]1/2=(c/m)1/2[1+½(2e²/cR³)-(2e²/cR³)²+...]

    U=½ ħ(ωs+ωa)-½ ħ(ω0+ω0) ½ ħ[ω0+ω0-¼ω0 (2e²/cR³)²]-½ ħ(ω0+ω0)

    =-ℏω₀/8(2e²/cR³)²

    Repulsive interaction—due to the overlapping of electron orbit[Pauli exclusion principle]

    Empirical form B/R¹²

    the total potential energy of two inert gas atoms at separation R is

    U(R)=B/R¹²-A/R⁶=4ε[(σ/R)¹²-(σ/R)⁶]

    Equilibrium lattice constant

    Uₜₒₜₐₗ=½ j Uijj (Uij)=(N/2)jUij

    now Uij=4ε[(σ/Rij)¹²-(σ/Rij)⁶ ] let Rij=Pij R → Uij=4ε[(1/Pij)¹²(σ/R)¹²-(1/Pij)⁶(σ/R)⁶]

    Uₜₒₜₐₗ=½N4ε[∑j(1/Pij)¹²(σ/Rij)¹²-∑j(1/Pij)⁶(σ/Rij)⁶]

    For inert gas crystal, in addition to He.

    1. nearest neighbors: 12, Pij=1 and R=a/2

    2. second nearest neighbors: 6, Pij=2 and R’=2R=a

    3. third nearest neighbors: 12, Pij=2 and R”=2R=2a….

    j(1/Pij)¹²=121⁻¹²+6∙√2⁻¹²+122⁻¹²+...=12.1388

    j(1/Pij)⁶=121⁻⁶+6∙√2⁻⁶+122⁻⁶+...=14.45392

    Uₜₒₜₐₗ=2Nε[12.1388(σ/R)¹²-14.45392(σ/R)⁶]

    R0=? at equilibrium dU/dR|R=R=0=-2Nε[12.13(12σ¹² /R¹³)-14.45(6σ⁶/R⁷)]

    R0/σ=1.09

    Cohesive energy游離能

    Utotal=-8.60N理論值, at equilibrium and 0 K, 只考慮位能

    K.E. into account>i.e. atom are at rest.

    Inert gas

    Ne

    Ar

    Kr

    Xe

    R0/

    1.14

    1.11

    1.10

    1.09

    ε(erg)1016

    50

    167

    225

    320

    Cohesive energy(KJ/mole)

    -2.589

    -8.646

    -11.649

    -16.567

    Experimental cohesive energy

    -1.88

    -7.74

    -11.20

    -16.0

    Cohesive energy reduced by quantum effect

    28%

    10%

    6%

    4%

    modification

    -1.864

    -7.781

    -10.950

    -15.904

     

    e.g. UNe=-(8.606.021023510-16)/(107103)

     

    Ionic crystal

    Two common crystal structure: 1. NaCl(f.c.c.), 2. CsCl(b.c.c.)

    What forces contribute to the crystal binding?

    1. electrostatic force: coulomb’s force, attractive force.

    2. exchange force: repulsive force due to the overlapping of electron orbits.

    3. van der Waal’s force: dipole-to-dipole interaction效果約1%忽略之

    Uₜₒₜₐₗ=½ j Uij, energy includes(i)coulomb: ±q²/rij, (ii)exchange: λe-rij/ρ

    Uₜₒₜₐₗ=½2NjUij=NjUij=Nj(λe-rij/ρ±q²/rij)

    rij=PijR, Uₜₒₜₐₗ=Njλe-rij/ρ-Nj(±1/Pij)(q²/R), ∑j(±1/Pij)=Modelung constant

    Uₜₒₜₐₗ=Nzλe-R/ρ-Nj(±1/Pij)(q²/R) , define αj(±1/Pij)=modelung constant

    NaCl crystal(f.c.c.): pick Cl- as a reference ion

    Nearest ion: 6Na+, R0

    2nd nearest ions: 12 Cl-, 2R0

    3rd nearest ion: 8 Na+, 3R0

    α=[61⁻¹+12∙√2⁻¹²+8∙√3⁻¹+...]=1.747565

    CsCl crystal(b.c.c.) : pick Cs+- as a reference ion

    Nearest ion: 8Cl-, R0=3a/2

    2nd nearest ions: 6Cs+, R=a=2R0/3

    3rd nearest ion: … α=1.762675

    Uₜₒₜₐₗ=Nzλe-R/ρ-Nα(q²/R)

    determination of , : characteristic force range

    bulk modulus, B-VdP/dV, definition for a crystal of f.c.c. structure

    V=2NR³ dV=6NR²dR B=-2NR³dP/(6NR²dR)=-(R/3)(dP/dR)

    dU=TdS-PdV=-PdV=-P6NR²dR at 0 K and R=R0

    P=-(1/6NR²)dU/dR, dP/dR=-(1/6NR²)d²U/dR²+(1/3NR³)dU/dR

    B=-(R/3)(dP/dR)=-(R/3)[-(1/6NR²)d²U/dR²+(1/3NR³)dU/dR]

    =(1/18NR)d²U/dR²+(1/9NR²)dU/dR]

    at equilibrium state, R=R0 dU/dR=0, B=(1/18NR)d²U/dR²|R=R

    Uₜₒₜₐₗ=Nzλe-R/ρ-Nα(q²/R) → dU/dR=N[-zλe-R/ρ/ρ+αq²/R²]=0, R0²=ραq²/(zλe-R/ρ)

    d²U/dR²=N[zλe-R/ρ/ρ²-2αq²/R³]|R=R=N[αq²/ρR²-2αq²/R³]

    B=(1/18NR)N[αq²/ρR²-2αq²/R³]=αq²/18R₀⁴(R/ρ-2)

    R/ρ=18R₀⁴B/αq²+2, ρ=R(18R₀⁴B/αq²+2)⁻¹

    ex. NaCl B=2.401011, R0=2.820Å, =1.747565, q=1.610-19coul=4.810-10 statcoulombs(1coul=3109 statcoulombs)

    =0.321Å

     

    U(R)=Nzλe-R/ρ-Nα(q²/R)=N[zλe-R/ρ-α(q²/R)]|R=R,

    dU/dR=N[-zλe-R/ρ/ρ+αq²/R²]=0, R0²e-R/ρ=ραq²/(zλ) when R=R0

    U(R0)=N[αq²ρ/R²-αq²/R]=Nαq²/R₀[ρ/R₀-1]

     

     

     

     

     

     

     

     

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