CHM 671. Homework set # 4. 2) Do problems 2.3, 2.4, 2.8, 2.9, 2.10, 2.12, 2.15 and 2.19 in the book.

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1 CHM 67 Homework set # 4 Due: Thursday, September 28 th ) Read Chapter 2 in the 4 th edition Atkins & Friedman's Molecular Quantum Mechanics book. 2) Do problems 2.3, 2.4, 2.8, 2.9, 2.0, 2.2, 2.5 and 2.9 in the book.

2 2.3 See Fig. 2.. Wavefunction (a) Potential Wavefunction + reflections (b) Potential Wavefunction + reflections (c) Potential Figure : The wavefunction in the presence of various potentials. Exercise: Sketch the general form of the wavefunction for a potential with two parabolic wells separated and surrounded by regions of constant potential.

3 2.4 ψ = C cos kx + D sin kx ( e ikx +e ikx ) ( e ikx e ikx ) = C + D 2 2i = 2 (C id)eikx + 2 (C +id)e ikx = Ae ikx + Be ikx Therefore, A = 2 (C id), B= 2 (C +id).

4 2.8 Consider the zones set out in Fig. 2.5; impose the condition of continuity of ψ and ψ at each interface. I II III E V 0 L x Figure : The zones of potential energy used in Problem 2.8. ψ I = Ae ikx + Be ikx,k 2 =2mE/ h 2 ψ II = A e ik x + B e ik x,k 2 =2m(E V )/ h 2 ψ III = A e ikx [no particles incident from right] } γ = k/k () A + B = A + B, [fromψ I (0) = ψ II (0)] (2) A e ik L + B e ik L = A e ikl, [fromψ II (L)=ψ III (L)] (3) ka kb = k A k B, [fromψ I(0) = ψ II(0)] (4) k A e ik L k B e ik L = ka e ikl, [fromψ II(L)=ψ III(L)] From () and (3): A = 2 ( + γ)a + 2 ( γ)b; B = 2 ( γ)a + 2 ( + γ)b From (2) and (4) A = A e i(k k)l + B e i(k +k)l γa = A e i(k k)l B e i(k +k)l so Then 2 ( + γ)a = A e i(k k)l, 2 ( γ)a = B e i(k +k)l A e ikl { ( + γ) 2 e ik L ( γ) 2 e ik L } =4γA A /A =2γe ikl /{2γ cos k L i( + γ 2 ) sink L}

5 The transmission coefficient (or tunnelling probability) is P = A 2 / A 2 = A /A 2 =4γ 2 /{4γ 2 +( γ 2 ) 2 sin 2 k L}, γ 2 = E/(E V ) Exercise: Find the transmission coefficient for a particle incident on a rectangular dip in the potential energy. 2

6 2.9 Consider the zones set out in Fig. 2.. Impose the conditions of continuity of ψ and ψ at the single interface. ψ I = Ae ikx + Be ikx,k 2 =2mE/ h 2 ψ II = A e ik x, k 2 =2m(E V )/ h 2 Note that () A + B = A [from ψ I (0) = ψ II (0)] (2) ik(a B)=ik A [from ψ I (0) = ψ II (0)] (a) E<V,k = i(2m/ h 2 ) /2 (V E) /2 =iκ, κ real (b) E V,k =(2m/ h 2 ) /2 (E V ) /2,k real I II E V 0 x Figure : The zones of potential energy used in Problem 2.9. Therefore the second condition becomes: (2a) ik(a B)= κa,or iλ(a B)=A with λ = k/κ Hence B/A = (+iλ)/( iλ), so that R = B/A 2 =for E<V (2b) k(a B)=k A,orγ(A B)=A with γ = k/k. Hence B/A = ( γ)/( + γ), R =( γ) 2 /( + γ) 2 =(k k) 2 /(k + k) 2 { (E V ) /2 E /2 } 2 { } ( a) /2 2 = =, with a = V/E and E>V (E V ) /2 + E /2 +( a) /2

7 R V / E Figure 2: The reflection probability as a function of barrier height. Note that R is independent of m. IfE 0. evandv 0.0 ev, we have R The function R is plotted in Fig Exercise: Consider particles incident from the other side of the step with E>V. What is the reflection coefficient? 2

8 2.0 Use the normalized wavefunctions in eqn 2.3: ψ n =(2/L) /2 sin(nπx/l); also use sin 2 axdx = 2x (/4a) sin 2ax (a) P n = 2 L 0 ψndx 2 =(2/L) 2 L 0 sin 2 (nπx/l)dx = 2 for all n (b) P n = 4 L 0 ψndx 2 =(2/L) 4 L 0 sin 2 { (nπx/l)dx = 4 (2/πn) sin( 2 nπ)} P = 4 { (2/π)} = (c) 2 L+δx 2 L+δx P n = ψ ndx 2 =(2/L) sin 2 (nπx/l)dx 2 L δx 2 L δx =(2/L){δx (L/2πn) cos(nπ) sin(2nπδx/l)} =(2/L){δx ( ) n (L/2πn) sin(2nπδx/l)} P =(2/L){δx +(L/2π) sin(2πδx/l)} 4δx/L when δx/l Note that lim P n = (a) n 2, (b) 4, (c) 2δx/L the last corresponding to a uniform distribution (the classical limit). Exercise: Find P n (and P ) for the particle being in a short region of length δx centred on the general point x.

9 2.2 Since E n = n 2 h 2 /8mL 2 [eqn 2.3], F = (de n /dl)=n 2 h 2 /4mL 3 For an electron (m = m e ) with n=, F = h 2 /4m e L 3, hence L =(h 2 /4m e F) /3 = pm/(f/n) /3 hence, when F =.0 N,L =0.49 pm. Exercise: Consider the case of N particles in a cubic box. Find an expression for the product pv (p: pressure, V : volume). Discuss the relation of this result with the perfect gas equation of state.

10 2.5 Intuitive solution: p n = 0 because the wavefunction is a standing wave. Elegant solution: p = n p n = n p n [hermiticity] = n p n = n p n [p = p]. Therefore since p = p, p = 0. Straightforward solution: p n =( h/i)(2/l) L 0 = (2 h/il)(nπ/l) sin(nπx/l)(d/dx) sin(nπx/l)dx L 0 sin(nπx/l) cos(nπx/l)dx =0 p 2 n =2mE n = n 2 h 2 /4L 2 p n = { p 2 n p 2 } /2 n = p 2 /2 = nh/2l Using the value of x n from Problem 2.4: x n p n =(L/2 3){ (6/n 2 π 2 )} /2 (nh/2l) =(n/4 3){ (6/n 2 π 2 )} /2 h =(nπ/ 3){ (6/n 2 π 2 )} /2 ( h/2) x p =(π/ 3){ (6/π 2 )} /2 ( h/2)=.357( h/2) > h/2 as required. Exercise: Repeat the calculation for the mixed state ψ cos β +ψ 2 sin β. What value of β minimizes the uncertainty product? n

11 2.9 E n = n 2 [h 2 /8m e L 2 ) [eqn 2.3, m = m e ] E n+ E n =(2n + )(h 2 /8m e L 2 ); hc/λ = E n+ E n Hence, λ =8m e cl 2 /(2n +)h Because n = 2 N and L =(N )R CC [N is even] λ =(8m e cr 2 CC/h)(N ) 2 /(N +) λ/nm= (R CC /pm) 2 (N ) 2 /(N +) Taking R CC = 40 pm and N = 22 gives λ 240 nm. Exercise: The colour of tomatoes is due to a molecule resembling β- carotene, but in which the rings have undergone scission. This enables the two terminal double bands to conjugate. What colour are tomatoes if carrots are orange?

CHM 671. Homework set # 6. 2) Do problems 3.4, 3.7, 3.10, 3.14, 3.15 and 3.16 in the book.

CHM 671. Homework set # 6. 2) Do problems 3.4, 3.7, 3.10, 3.14, 3.15 and 3.16 in the book. CHM 67 Homework set # 6 Due: Thursday, October 9 th ) Read Chapter 3 in the 4 th edition Atkins & Friedman's Molecular Quantum Mechanics book. 2) Do problems 3.4, 3.7, 3., 3.4, 3.5 and 3.6 in the book.