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.
3.4 φ m lφ ml dφ =(/2π) e i(m l m l )φ dφ { } e 2i(m l m l )π =(/2π) (m l m l )i = ifm l m l [e 2inπ =,nan integer] (Note that when m l = m l the integral has the value 2π.) Exercise: Normalize the wavefunction e iφ cos β +e iφ sin β, and find an orthogonal linear combination of e +iφ and e iφ.
3.7 (a) Ψ=N (/m l!)e imlφ = N (/m l!)(e iφ ) m l m l = m l = = N exp(e iφ )=N exp(cos φ + i sinφ) For normalization, Ψ Ψ=N 2 e cos φ i sin φ e cos φ+i sin φ = N 2 e 2 cos φ ; Ψ 2 dφ = N 2 e 2 cos φ dφ =2πN 2 I (2) = [required] Therefore, N =/ {2πI (2)} =.2642 [I (2)=2.28, given]. The form of Ψ 2 is which is plotted in Fig. 3...6 Ψ 2 = N 2 e 2 cos φ =.698 e 2 cos φ.4 Ψ 2.2 π 2π φ Figure : The probability density for the wavepacket calculated in Problem 3.7. (b) φ = N 2 φe 2 cos φ dφ π = N 2 φe 2 cos φ dφ π [ same integration range but differently expressed = [φ odd under φ φ,cos φ even under φ φ] ]
sin φ = N 2 sin φe 2 cos φ dφ = [same argument as for φ] l z = N 2 e cos φ i sin φ l z e cos φ+i sin φ dφ =( h/i)n 2 e cos φ i sin φ (d/dφ)e cos φ+i sin φ dφ =( h/i)n 2 e cos φ i sin φ ( sin φ + i cos φ)e cos φ+i sin φ dφ = hn 2 cos φe 2 cos φ dφ =2π hn 2 I (2) = {I (2)/I (2)} h =.698 h l z < h because it is the weighted average value, and m l = contributes significantly to the sum. Exercise: Repeat the calculation for a wavepacket found in the same way but with the omission of m l =. 2
3. The expression for the Laplacian in cylindrical polar coordinates (Problem 3.2) is 2 = 2 r 2 + r r + 2 r 2 φ 2 + 2 z 2 Now substitute r = ρ sin θ,z = ρ cos θ,ρ =(r 2 + z 2 ) /2,θ= arctan(r/z) and later let ρ r for the conventional notation. As in Problem 3.2, whence 2 r 2 + 2 z 2 = 2 ρ 2 + 2 ρ 2 θ 2 + ρ 2 = 2 ρ 2 + 2 ρ 2 θ 2 + ρ However, from Problem 3.2, Therefore, 2 = 2 ρ 2 + ρ 2 2 = ρ 2 { ρ Now we let ρ r, and obtain ρ ρ + 2 r 2 φ 2 + r r = sinθ ρ + cos θ ρ θ 2 r θ 2 + 2 ρ ρ + ρ 2 sin 2 θ φ 2 + cot θ ρ 2 ( ρ 2 ) + ( sin θ ρ sin θ θ θ θ ) + sin 2 θ 2 φ 2 } with as required. Λ 2 = sin θ 2 = r 2 r r2 r + r 2 Λ2 ( sin θ ) + θ θ sin 2 θ 2 φ 2
3.4 From eqn 3.34, E = J(J +)( h 2 /2I)=(.3 22 J)J(J +) [Problem 3.]. Draw up the following Table, using degeneracy g J =2J +: J E/( 22 J) g J 2.6 3 2 7.8 5 E( ) = 2.6 22 J λ( ) = hc/ E( )=7.64 4 m=.764 mm (far infrared) Exercise: Calculate the same quantities for the deuterated species.
3.5 Refer to Fig. 3.. z { ll ( + )} m l θ /2 Figure : The vector diagram for a general angular momentum with quantum numbers l and m l. cos θ = m l /{l(l +)} /2, hence θ = arccos {l(l +)} /2 The minimum angle is obtained when m l takes its maximum value (m l = l). l θ min = arccos {l(l +)} /2 = arccos +/l lim l θ min = arccos = Draw up the following Table (angles in degrees): m l m l +3 +2 + 2 3 l =,θ= arccos(m l / 2) 45. 9 35. l =2,θ= arccos(m l / 6) 35.3 65.9 9 4. 44.7 l =3,θ= arccos(m l / 2) 3 54.7 73.2 9 6.8 25.3 5 Exercise: The chlorine nucleus has a spin I = 3 2. What values of θ are allowed?
3.6 See Fig. 3.. From Problem 3.5, when l = 3 and m l =,θ =9, and the angular momentum vector lies in the equatorial plane; therefore Y 2 will have maxima on the z-axis, as seen in Fig. 3.. We also see from Problem 3.5 that as m l increases, the deviation of θ from 9 increases; as the projection of the angular momentum vector on the z-axis increases, Y 2 becomes larger in the equatorial plane. m l = m l = ± m l = ±2 Figure : A representation of the wavefunctions and the location of the angular nodes for a particle on a sphere with l =2. Exercise: Draw the corresponding diagrams for l =4.