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Infinite Potential Well … bottom line d ψ 2m 2 (E V)ψ 0 2 dx Schrodinger Equation 2 V(x) V= 0 Energies are quantized, defined by one single quantum number, n = 1, 2, 3, 4 … n 2h 2 En 8ma Electron V= V=0 0 a x Tunneling “ … If an electron comes up a potential barrier greater than its energy … there is a finite probability that it will “pass” through the barrier…” D C A B E … for an electron ! We place an electron in region I… with energy E less than VO (E<VO) … what is the probability the electron will be in I … II … III ?? V(x) Vo I III II x=0 x=a x How do we calculate the probability ?? … we need to solve Schrodinger’s equation … apply boundary conditions etc. Tunneling d 2 ψ 2m 2 (E V)ψ 0 2 dx V(x) Vo ψ I (x) A1e jkx A 2 e -jkx ψ II (x) B1ex B2 e -x ψ III (x) C1e jkx C2e -jkx I III II x=0 x=a 2m(VO - E) 2mE 2 with k 2 and α 2 2 We now need to apply BC’s at x=0 and x=a … The properties of ψ require that it be continuous and single valued x Tunneling ψI (x) ... ψII (x) ... ψIII (x) V(x) Vo I A1 A2 yII Incident yIII Reflected I III II x=0 x=a x Therefore … the solution suggests that the electron can be found beyond the barrier VO … EVEN THOUGH its energy E is less than VO! Tunneling T ψIII (x) ψI (x) 2 2 2 VO 1 T where D 2 1 Dsinh (α a) 4E(VO - E) 2m(VO - E) α 2 2 What are the important factors that influence the tunneling probability ?? … the energy of the electron… the width and height of the barrier For a wide or high barrier … α a 1 and T TO e V(x) Vo I ( 2 α a) 16E(VO E) with TO 2 VO A1 A2 yII Incident yIII Reflected I III II x =0 x=a x Application of Tunneling Metal (a) y(x) Vacuum Second Metal Itunne Probe Vo V(x) Scan x Itunnel Material surface l y (Å) x (Å) (b) Tunneling current gray scale value (nA) x The Potential “Box” If you confine an electron in a box … what would you expect the wave-function to be? Think of it as a combination of 3 one-dimensional infinite potential wells… and therefore the general solution will have the form of: ψ(x, y,z) Asin(kx x)sin(k y y)sin(kz z) n3 π n1π n2 π where kx , ky , kz a b c V= c V =0 V= x z 0 a V= V= y b The Potential “Box” 2 E n1 , n 2 , n 3 h 2 2 2 (n n n 1 2 3) 2 8ma The solution to the electron in a “box” problem results in 3 quantum numbers A specific solution or eigenfunction i.e. ψ1,1,2 or ψ2,1,2 or ψ1,4,2 or ψ2,1,3 etc. is called a state … Note that the electron energy is quantized and depends on 3 quantum numbers The H-Atom … An Overview Describe the H-atom … i.e. what does the nucleus look like? how much charge is there at the nucleus? i.e. Z=1 how many electrons? The H-atom represents the simplest system we can use to have a look at a real quantum physics example The H-Atom … Force & PE Obviously the electron is being attracted to the nucleus because of the … … Coulombic attraction between two opposite charges! The force between two charges is: Q1 Q2 F 4π εo r 2 and the potential energy is given by: - e2 V(r) 4π εor The H-Atom … Spherical Coordinates Due to the spherical symmetry of the H-atom … it makes sense to work in the spherical coordinate system instead of the cartesian one … i.e. x, y, z r,θ,φ z P(r,q,f) e q r Nucleus +Ze x y f V(r) r - e2 V(r) 4π εor Ze2 V(r) = 4peor +Ze The H-Atom … Wavefunction No need to go through the solution in detail … … we do however need to understand the origin of certain parameters and functions! Obtaining the wavefunction for the H-atom electron can be done by solving … … in 3-dimensions i.e. one would expect to get … 3 quantum numbers! And the general wave function looks like: ψ n,l,m l (r, θ, φ) R n,l (r) Yl,m l (θ , φ) The H-Atom … Quantum Numbers ψ n,l,m l (r, θ, φ) R n,l (r) Yl,m l (θ , φ) Two functions … R a function of r and Y a function of θ and φ Three quantum numbers! … n, l, ml The spherical part i.e. R depends on n and l … while the angular or spherical one, Y depends on l and ml n=1,2,3,4,…… is the Principal Quantum Number l=0,1,2,3 ……(n-1) is the Orbital Angular Momentum Quantum Number ml=-l, -(l-1), -(l-2), ……-2, -1, 0, 1, 2, …+l is the Magnetic Quantum Number … for now MEMORIZE these! The H-Atom … Quantum Numbers l n 0-> s 1-> p 2-> d 3-> f 4-> g 1 1s 2 2s 2p 3 3s 3p 3d 4 4s 4p 4d 4f 5 5s 5p 5d 5f 5g Let’s check the validity of these Energy States 3,2,2 2,3,1 2,3,0 1,2,4 2,3,-1 2,0,1 1,1,0 1,2,3 4,1,2 1,0,0 Energy ! Electron energies depend on n only … given by: 4 2 me Z (13.6eV)Z En 2 2 2 2 8ε o h n n 2 What does this energy represent ? … the energy required to remove the electron from the n=1 state (i.e. to free the electron) also known as the ionization energy … E1 13.6 eV Electrons prefer to minimize their energy … therefore most likely to be found in n=1 state known as the ground state! An electron with velocity 2.1E6 m/s strikes a H atom. Find the n th energy level the electron will excite to. Calculate the wavelength of the light as the electron returns to ground state. K.E. = 12.5eV; n = 3.51 -> 3; ΔE = 12.09 eV; λ = 102.6 nm Energy ! Electron energy, En. 0 0.54 0.85 1.51 5 4 3 3.40 2 n= Excited states E = KE Continuum of energy. Electron is free 5 Ionization energy, EI 10 13.6 eV 15 1 Ground state n n=1 Orbital Angular Momentum Just like energy (En) … angular momentum L is also quantized… by ‘l’ Ll [l(l 1)] 1/2 … what happens when l=0 ? Bexternal z Lz q L y x Orbiting electron Orbital Angular Momentum For l=2 … ml would be … …-2, -1, 0, 1, 2 LZ ml z Bexternal Bexternal z ml 2 1 l =2 Lz L= 2(2+1) q L 0 y 1 2 x Orbiting electron Selection Rules … Electron has momentum … also photons have intrinsic momentum When photons are absorbed … in addition to the energy conservation momentum must also be conserved … Selection rules: Δl=±1 … Δml=0, ±1 i.e. if electron is in ground state 1,0,0 … (n,l,ml) If enough energy is gained to move up to n=2 then what are l and ml? l …0, 1 … and ml … -1, 0, 1 Therefore … n=2, l=1, and ml=-1,0,1 Selection Rules … Energy 0 l l =1 l =2 l =3 n 5 5s 5p 5d 5f 4 4s 4p 4d 4f 3s 3p 3d 2 2s 2p 3 13.6eV =0 1 1s Photon l Selection Rule Example An electron in State (3,2,-2). What are the energy states in Shell N, this electron can jump to? Spin … (intrinsic angular momentum) Spin: last quantum number required to fully describe an electron! S [s(s 1)] 1/2 with s 1/2 SZ m s with m s 1/2 The component of the spin along a magnetic field is also quantized (i.e. if B-field is in Z-direction) The Quantum Numbers Radial Probability ψ n,l,m l (r, θ, φ) R n,l (r) Yl,m l (θ , φ) n=1 n=2 R 1,0 n=1 R 20 n=2 r 2 |R 2,0 |2 r 2 |R 1,0 |2 2s 2s 0 1s 0 R 21 1s 0 .2 r (nm) 0 .4 0 0 2p 2p 0 0 r 2 |R 2,1 |2 0 .2 0 .4 0 .6 0 .8 0 0 .2 r (nm) 0 0 .4 0 0 .2 0 .4 0 .6 r (nm) r (nm) Bohr Radius … the radial distance where the radial probability is maximum 0 .8 “Angular” Probability z z y y x x 2 |Y|2 for a 1 s orbital |Y| for a 2 px orbital z z y y x |Y|2 for a 2 py orbital s-states symmetrical p-states directional x |Y|2 for a 2 pz orbital (m l = 0) Multi electron atom : He -e Electron 1 r1 r12 r2 Nucleus +Ze -e Electron 2 A helium-like atom. The nucleus has a charge of +Ze, where for He Z = 2. If one electron is removed, we have the He+ ion which is equivalent to the hydrogenic atom with Z = 2. Energy states for multi electron atom O Energy 5g 5f N 6p 4f 4d M 3d 5d 6s 5p 5s 4p 4s 3p 3s L Energy depends on both n and l 2p 2s K 1s 1 2 3 4 5 6 n Pauli Exclusion Principle & Hund’s Rule NO two electrons can have the same set of quantum numbers … i.e. if one electron in ψ1,0,0,1/2 then a second electron in the same system will have … ψ1,0,0,-1/2 Electrons in the same n, l orbital “like” to have “parallel" spins … -1 L (n=2) K (n=1) 1 = ml p s K (n=1) L (n=2) 0 H He Be B Li s p Electronic configurations for the first five elements. Each box represents an orbital y (n, l, ml ). C O N p L s K s F Ne p L s K s Electronic configurations for C, N, O, F and Ne atoms. Notice that Hund's rule forces electrons to align their spins in C, N and O. The Ne atom has all the K and L orbitals full.