90%(10)9 out of 10 people found this document helpful
This preview shows page 42 - 45 out of 92 pages.
Using the Green’s function of the diffusion equation,P(r,t|r0,0) =1(4πDt)3/2e-|r-r0|2/4Dtanalyze the time dependence of the electron density distribution.Is it correct to use this Green’s function for the case of the electron creation at the surface?
43Solution 4.3:For the density distribution at timetone has (by definition)n(r,t) =Zd3r1P(r,t|r1,0)n(r1,0).The solution is substantially simplified becauseP(r,t|r0,0)depends only on the difference(r-r1). Then, in general, one can use the Fourier method.However, this problem is even simpler since this function depends only one-|r-r1|/4Dt=e-(x-x1)2/4Dt·e-(y-y1)2/4Dt·e-(z-z1)2/4Dt,whilen(r0,0)∝e-x21/2σ·e-y21/2σ·e-z21/2σ.Thus we have to analyze only one integral,I(x)≡12π√2σDtZ∞-∞dx1e-(x-x1)2/4Dte-x21/2σ=1p2π(σ+2Dt)e-x2/2(σ+2Dt).As a result,n(r,t) =N[2π(σ+2Dt)]3/2exp•-|r|22(σ+2Dt)‚.This expression cannot be used for the case of surface excitation since the Green’s function forthat problem has different form.Problem 4.4:In the presence of electrical field, the electrons acquire a constant (drift) velocity,proportional to the applied electric field,vd=μE. The quantityμis called themobility.LetNelectrons be injected at the planex=0,t=0, and then diffuse in the presence of theelectrical field creating the drift velocityvdkx.Find the distribution of electrons at timet.Solution 4.4:The particle current in the presence of the drift isj=n(x)vd-Ddndx.Thus the diffusion equation acquires the form:dndt+d jdx=0→dndt+vddndx-Dd2ndx2=0.It is natural to introduce a new variable,ξ=x-vdtand search solution asn(ξ,t). Sincedn(ξ,t)dt=∂n(ξ,t)∂t-vd∂n(ξ,t)∂ξ.
44CHAPTER 4. DEFECTS AND DIFFUSIONOn the other hand,vddndx=vd∂n(ξ,t)∂ξ∂ξ∂x=vd∂n(ξ,t)∂ξ.In this way we return to the well known diffusion equation,∂n∂t=D∂2n∂ξ2having the solutionn(x,t) =n0√4πDtexp-(x-vdt)24Dt¶.Heren0≡N/Ais the initial surface density of injected particles.Problem 4.5:Frenkel defects:Show that the numbernof interstitial atoms in equilibriumwithnlattice vacancies in a crystal havingNlattice points andN0possible interstitial positionsis given by the equationEI=kBTln£(N-n)(N0-Nn)/n2/,whence forn¿N,N0, we haven’(NN0)1/2exp(-EI/2kBT). HereEIis the energy necessaryto remove an atom from a lattice site to an interstitial position.Solution 4.5:The internal energy of formation ofnFrenkel defects isU=nEI. The numberof ways to picknfromNisN!n!(N-n)!.The number of ways to putnatoms intoN0sites isN0!n!(N0-n)!.Hence, the entropy isS=kBlnN!n!(N-n)!+lnN0!n!(N0-n)!¶.Using the Stirling’s formula we getlnN!