[閒聊] Debye Huckel
Poisson Eq. ▽^2Φ = -ρ/ε0εr , when
▽^2 = δ^2/δx^2 + δ^2/δy^2 + δ^2/δz^2
= (1/r^2)(δ/δr)(r^2)(δ/δr) + (cosθ/(r^2)sinθ)(δ^2/δ^2θ)
+ (1/(r^2)sinθ)(δ^2/δψ^2)
(後兩項不確定,反正不重要,Φ只和r有關)
ρ = Σci'Qi = ΣciQi*exp(-QiΦ/kT)
(ci'=ci*exp(-QiΦ/kT by Boltzmann distribution)
= ΣciQi(1 - QiΦ/kT) (when kT>>QiΦ)
= ΣciQi - (Φ/kT)Σci(Qi)^2
(電中性ΣciQi=0 when cations and anions are equilibrium)
= (-e^2Φ/kT)Σci(Zi)^2 (when Qi=Zi*e)
∴ ▽^2Φ = (e^2Φ/ε0εrkT)Σci(Zi)^2
= ((e^2)I/ε0εrkT)Φ (when I=Σci(Zi)^2 ionic strength)
= b^2Φ (b:1/debye length)
so that (1/r^2)(δ/δr)(r^2)(δ/δr) = b^2Φ
(δ^2/δ^2r)(r^2Φ) = b^2Φ
set y=rΦ substitude , that (δ^2/δ^2r)y = (b^2)y
so we get y = A*exp(-br) + B*exp(br)
but r→∞ but we want y not →∞ , that B=0
∴ y = rΦ = A*exp(-br) , Φ = (A/r)exp(-br)
∴ -ρ/ε0εr = b^2Φ = (Ab^2/r)exp(-br)
ρ = -(A(b^2)ε0εr/r)exp(-br)
又平衡時淨電荷為0 , so ∫ρdV + Zie = 0 (dV=(r^2)sinθdθdψ)(V:volume)
∫ρdV = ∫(a→∞) (r^2)dr*∫(0→π) sinθdθ*∫(0→2π) dψ ρ(r)
= ∫(a→∞) 4π(r^2)ρ(r)dr
= -4πA(b^2)ε0εr*∫(a→∞) r*exp(-br)dr
= -4πA(b^2)ε0εr)*[(ab+1)/(b^2)]exp(-ba)
= -Zi*e
∴ A = (Zi*e/4πε0εr)(exp(ba)/(ab+1))
∴ Φ = (A/r)exp(-br)
= (Zi*e/4πrε0εr)(exp(-b(r-a))/(ab+1)
= Φ' + +Zi*e/4πrε0εr
∴ Φ' = (Zi*e/4πrε0εr)[exp(-b(r-a))/(ab+1) - 1]
= (Zi*e/4πaε0εr)(-ab/(ab+1)) (when r=a)
= -Zi*eb/(4πε0εr(1+ba))
= -Zi*eb/4πε0εr (when ab<<1)
= -Qi*b/4πε0εr
--
※ 發信站: 批踢踢實業坊(ptt.csie.ntu.edu.tw)
◆ From: 61.223.19.117
推
推140.112.244.136 06/21, , 1F
推140.112.244.136 06/21, 1F
推
推 218.166.72.108 06/21, , 2F
推 218.166.72.108 06/21, 2F
推
推 61.223.19.117 06/21, , 3F
推 61.223.19.117 06/21, 3F