Model Problems update Particle in box - all texts plus Tunneling, barriers, free particle - Tinoco (pp455-63), House Ch 3

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1 VI 15 Model Problems update Particle in box - all texts plus Tunneling, barriers, free particle - Tinoco (pp455-63), House Ch 3 Consider E-M wave 1st wave: E 0 e i(kx ωt) = E 0 [cos (kx - ωt) i sin (kx - ωt)] magnitude: k = π/λ ω = πc/λ =πν ν = c/λ moves in space and time traveling wave reflect at the node keeps the wave continuous (if not create an interference) is other cycle, at Δt=λ/c node if trap wave like violin string tied down at end standing wave (principle of laser light trap in cavity specific frequency / phase amplified) restriction - number wavelengths integral divisor of length integer representation of frequencies not continuous (analogy to Bohr orbit, wavelength must match path) Now think of traveling particle 1-D no forces V = 0 Hψ = Eψ = Tψ let V = 0, free moving particle Hψ= -h /m d /dx ψ(x) 15

2 VI 16 Solution: need some (wave)function that we can take derivatives twice and get function back choices: a) de ax /dx = ae ax derivative works d /dx e ax = a e ax (but energy, i.e. (ħ /m)d ψ/dx, must be positive, nd deriv. must be negative, so use: a = iα, e iαx wavefunction complex, sum: cos & sin) b) d /dx sin kx = -k sin kx (note: same for cos kx) (Note: e iαx = cos αx - i sin αx general form wave) No constraint traveling wave (but for particle) Solve Schroedinger Equation for free particle: -(h /m) d /dx ψ = Eψ if ψ = e iαx plug in -(h /m)(iα) e iαx = E e iαx from (b): α = k = (me) 1/ /h E = α h /m (all K.E. - positive, not quantized) no restrictions free particle, any energy or wavelength Boundary Conditions Restrictions on wave function to let fit postulates B.C. relate to continuous and finite properties wave/fct properties - both sides boundary--must match Note effect of momentum: (for ψ = e ikx ) pψ = -ih(ik) ψ Magnitude: p = hk signs direction p = hk (motion in +x) [opposite: ψ = e -ikx, (motion -x)] 16

3 VI 17 Particle in a box introduce potential energy in box V = 0 outside V = but not in SchEqn For finite E: particle must be in box, or definite E B.C. (also think of as F = -dv/dx, force at wall is ) -h /m d /dx ψ = Eψ try ψ = A sin αx + B cos βx B.C. ψ(0) = 0 restrict: B = 0 (since cos 0 = 1) ψ(l) = 0 restrict: α = nπ/l (since sin nπ = 0) idea continuous w/f, zero probability outside: ψ out = 0 for ψ(x) 0, must have: A 0, n 0 and n = 1,,3, (i.e. node both sides integral number internal lobes and be non-zero someplace- A 0 non-trivial solution) forms a standing wave -- quantized (recall - laser) -h /m d /dx (A sin nπx/l) = E (A sin nπx/l) (-h /m)(- n π /L ) = E n = n h /8mL recall h=h/π Expanding E-levels ~ n each increases number of nodes - curvature restricted energy levels lowest energy 0 (particle always moving) 17

4 VI 18 Probability distribution: ψ*ψ dx ψ*ψ dx = 1 (if normalize, multiply function by N) L 0 or 1/N = ψ*ψ dx = sin nπx/l, let y=nπx/l, dy=(nπ/l)dx L 0 L 0 = (L/nπ)[ ½ y+ ¼ sin y nπ 0 = L/ N=(/L) 1/ =A but plot ψ*ψ not uniform in x n = 1 more probable in middle n = zero probability at x = L/ 1 L/ as n inc. probability evens classical Orthogonal ψ m *ψ n dx = 0 if n m L/ sin(nπx/l) sin(mπx/l) dx=0 easiest seen graphically Amplitude? A = (/L) 1/ see normalization L A sin (nπx/l) dx = 1 A (L/) = 1 0 Probability b ψ*ψ dx probability between a + b a nπb/l From above: P ab = (L/nπ)[½ y + ¼ sin y nπa/l (y=nπx/l) Use for pib? Great model / see how B.C. quantization Application: polyenes π-system delocalize electrons move through π-bonds ( e - per level know) spectra e - could be in different levels ΔE = E n+1 E n = hν n n + 1 ΔE = (n+1) h /8mL n h /8mL = (n + 1) h /ml = hν 18

5 VI 19 Now see properties a) bigger n more separation higher frequency - hν b) bigger m less separation (but all same m e - electron) c) bigger L less separation (as square), experimental Sample dye problem: λ max (Å) Polyene N Obs Calc H H C C H H C C H C N C C H H ΔE = (N + 1) h /8mL L 0.81 N (in nm) λ = c/ν = hc/δe = (8m e c/h) (0.81x10-9 m) N /N + 1 units! m [to do: insert m e (for elect), h, c; when done convert to nm, 10-9 m] Note: trend is as expected N increase, λ increase (big boxes lower energy states) values off calc. change much faster than exper. -- box length approximate -- and evenness of V (real potential vary over bonds) (if use oscillating V(x) - potential in box, answer fits data) Ref: Jochen Autschbach, J.Chem. Ed. 84, 1840 (007) 19

VI 0 Dye S C N C H 5 C H H C C N H C C H 5 N + S N Obs Calc 450 380 3 5600 4540 4 6500 5800 5 7600 7060 6 8700 8330 7 9900 9600 Model does better (here use N+1 N+) and use different length, but still

6 VI 0 Dye S C N C H 5 C H H C C N H C C H 5 N + S N Obs Calc Model does better (here use N+1 N+) and use different length, but still λ ~ N /N type term (linear) Bio-connect -Vision: retinal undergoes cis-trans isomerization N (trans) HOMO LUMO difference Transition wavelength (λ=c/ν) Decrease with length increase with box length fit with oscillating V(x) potential 0

7 VI 1 Ionization potential measures energy of the orbital see dec. from ethylene butadiene (left peak lowest) Butadiene examples real spectra shift with length structure from vibrations, absorp. inc. ΔE, fluor. dec. ΔE 1

8 VI -D box example π-system expand energy, difference gets smaller big box, small energies Problems worked out in many books (Engel Ch 14.4, House Ch. 3) poly arene examples (in wavelength, so going to right, lower energy due to larger D boxes): 1 ring rings 3 rings 4 rings

9 VI 3 3-D Particle in box Separation of Variables (House Ch.3) Critical Method - need to solve atoms & molecules write: Hψ = h m x V = 0 0 < x <a 0 < y < b 0 < z < c V = outside the box + y + z ψ = Eψ ψ = -me/h ψ Note: a) / x only operate on x-dependent function b) H is a sum of terms each depend on 1 variable IN GENERAL can find solution -- product function form Ψ = X(x) Y(y) Z(z) where X(x) is only fct. of x, etc. AND energy also a sum: E = E 1 + E + E 3 Substitute: X ( x) Y ( y) Z( z) me XYZ = YZ + XZ + XY = x y z h XYZ divide by XYZ: me h = 1 X X x + 1 Y Y y + 1 Z Z z each term must be a constant since independent i.e. 1/X X/ x = α etc. α + β + γ = -me/h 3

10 VI 4 These individual 1-D equations are pib solutions again: n ψ (x,y,z) = sin x n π sin y n y sin z π z 8 abc π x a b c E = h n n x y n + 8m a b c z + = E 1 + E + E 3 Lowest state n x = n y = n z = 1 But 3 ways for next state n x =, n y = n z = 1, etc. Each of these could have different energies However, if a=b=c, then each has same energy degeneracy from symmetry Barriers (Engel Ch.14.9) Concept is goal, not derivation Now what if wall not so high or wide high wall wave must have zero amplitude ψ*ψ = 0 at wall reflect shorter wall wave can be penetrated also thin wall go through or tunneling (-h /m d /dx + V)ψ = Eψ if ψ = e iαx [h α /m + (V E)]ψ = 0 α = m (E V) h now x < x 0, V = 0 E V = (+) ψ = e iαx is complex exponential fct. wave 4

11 but for : x > x 0, V > E E V = ( ) so α = i m (V E) h VI 5 = iκ ψ'(x) = e -Κx real, decaying fct. At wall, x-x 0 ψ(x 0 ) = ψ'(x 0 ) i.e. must be continuous If non-zero in wall, then ψ must decay as move +x On other side: ψ'(x 1 ) = ψ''(x 1 ) (contin. go out: ψ'' < ψ) equation 9.10 Atkins: Tunneling probability, T T 16ε (1 - ε) e -ΚL where : ε = E/V E or V T Κ=[m(V-E)] 1/ /h and L=x 1 x 0 inc. V or L, T Solution (extra-repeat): Look at just the barrier: H A = -h /m d /dx = H C H B = -h /m d /dx + V solve each region separately: ψ A = Ae ikx + Be -ikx k = (me/h) 1/ ψ B = A'e ik'x + Be -ik'x k' = [m (E - V)/h] (in the barrier) ψ C = A''e ik x + B''e -ik x k = (me/h) 1/ = k Note: if E < V, then k' = imaginary let k' = iκ, Κ = [m (E - V)/h] 1/ ( = real) ψ B = A'e -Κx + B'e +Κx exponentially decreasing or increasing function 5

12 VI 6 no oscillation in barrier amplitude: ψ*ψ 0 in barrier, thus can tunnel probability non-zero of in and other side barrier damping ~ mass heavy don t penetrate classic low energy don t penetrate tunnelling --skip, read i.e. w/f okay if bound in area of wall must be thin to solve for A, B s must set up simultaneous equation based on: boundary constraints ψ A (0) = ψ B (0) A + B = A' + B' ψ B (l) = ψ C (l) A'e -Κl + B'e +Κl = A''e ik l + B''e -ikl and continuous slopes ψ A / x 0 = ψ B / x 0 ika ikb = -ΚA' + ΚB' ψ B / x l = ψ C / x l -ΚA'e Κl +ΚB'e Κl = ika''e ik l +ikb''e -ikl Then consider structure as: B = 0, A 0 (come from left) then B'' = 0 and A'' ~ transmission B ~ reflection Probability of tunneling: A'' / A P = 1/(1 + G) G = (e Κl e -Κl ) 4 (E / V) (1 E / V) Note: P non zero, K > 0 E increased, G decreased, P increased 6

13 Particle on a ring: Circumference = πr (fixed distance) B.C. ψ(φ) = ψ(φ + π) (vary angle) continuous but not zero (no wall) h Hψ = ψ = Eψ mr φ r dr 7 unit length ~ r dφ VI 7 ψ = Ae iαφ + Be -iβφ r=xi+yj, x +y =1 x= r cosφ y= r sinφ or φ = cos -1 x/ r B.C. e iαφ = e iα(φ + π) e iα(π) = 1 α = n = 0, ±1, ±, nd term (B-dependent) redundant E n = h n /mr = h n / I I = mr moment of inertia Note: levels degenerate for ±n no zero-point E E 0 = 0, φ unknown on ring spacing ~n same pattern (OK uncert.) bigger ring lower E n Angular Momentum J = r x p in general J z = r p (1-D z out of plane) I = mr moment of inertia E = p /m = J z /mr = J z /I from de Broglie p = h/λ λ = πr/n (int. # waves ring) E n = p n /m = (h/π) n /mr = E n from above E n = n h /I E = J z /I J z = nh get quantized solution for Energy and angular momentum why? [H,L] = 0, commuting operators, simul. eigenfct. This form works for molecular rotation / atom, add dimen.

14 VI 8 Consider particle in box - short side (finite well) (Engel 14.5): V = 0 0 < x < L; V = V 0 0 < x < L E n : Energy no longer ~n (spacing will get closer with n) ψ: Solution to this more complex but have new property ψ(0) & ψ(l) 0 -- since V hence w/f non zero inside wall -- from B.C. (turns out to be exponential e -βy, i.e. decay function where y = x L, x > L ;y = -x, x<0) Imagine boxes side by side: as (L M) 0 wave functions will overlap, then ψ*ψ will be non zero in other box and particle will tunnel Additional property as E V 0, levels must close in together E > V 0 levels continuous 8

Model Problems 09 - Ch.14 - Engel/ Particle in box - all texts. Consider E-M wave 1st wave: E 0 e i(kx ωt) = E 0 [cos (kx - ωt) i sin (kx - ωt)]

VI 15 Model Problems 09 - Ch.14 - Engel/ Particle in box - all texts Consider E-M wave 1st wave: E 0 e i(kx ωt) = E 0 [cos (kx - ωt) i sin (kx - ωt)] magnitude: k = π/λ ω = πc/λ =πν ν = c/λ moves in space