ps1ans_1994

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 5.80 Small-Molecule Spectroscopy and Dynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS INSTITUTE OF TECHNOLOGY Chemistry 5.76 Spring 1994 Problem Set #1 ANSWERS 1. (a) Make the necessary conversions in order to fill in the table: Wavelength (Å) 420 Wavenumber (cm−1 ) 100 Energy (J) Energy (kJ/mole) 490 Frequency (Hz) 8.21 × 1013 Answer: 10−2 m 108 cm−1 1 1 · = = λ λ[Å] · 10−10 mÅ−1 1 cm λ[Å] hc 6.626 · 10−34 J s · 2.998 · 108 m s−1 = E= λ λ[Å] · 10−10 mÅ−1 1.988 · 10−15 J = λ[Å] 1.988 · 10−15 J/photon E= · 6.022 · 1023 photon/mole · 10−3 k J/ J λ[Å] 1 . 1 9 7 · 1 0 6 k J /m o l e = λ[Å] c ν is not angular frequency, but the reciprocal of the period of oscillation ν= ; λ 2.998 · 108 m s−1 2.998 · 1018 Hz = = λ[Å] · 10−10 mÅ−1 λ[Å] ν= ˜ (b) Name the spectral region associated with each of the last four columns of the table. Answer: XUV Far IR UV Mid IR Name for region of the spectrum 6 wavelength/Å 420 1 .0 0 ·1 0 2440 3 .6 5 · 1 0 4 −1 5 wavenumber/cm 2 .3 8 · 1 0 100 40900 2740 −18 −21 −19 energy/J 4 .7 3 · 1 0 1 .9 9 · 1 0 8 .1 4 · 1 0 5.44 · 10−20 energy/kJ mole−1 2850 1 .2 0 490 3 2 .8 frequency/Hg 7.14 · 10−15 3.00 · 1012 1.23 · 1015 8.21 · 1013 Problem Set #1 ANSWERS Spring, 1994 Page 2 2. A 100-W tungsten filament lamp operates at 2000 K. Assuming that the filament emits like a black­ body, what is the total power emitted between 6000 Å and 6001 Å? How many photons per second are emitted in this wavelength interval? Answer: This problem makes use of the Stefan-Boltzman Law, I = σT 4 . You can either derive it (problem 8a) or look it up. 100 W tungsten filament lamp operating at 2000 K. What is the total power emitted between 6000 Å and 6001 Å? We can calculate this quantity by integrating the Planck function from 6000 Å to 6001 Å, and comparing that the value of the Planck function integrated over all λ’s. v2 Power = v1 ∞ 0 ρ(v, T )d v ρ(v, T )d v × 100W ; v1 = c 6001Å , v2 = c 6000Å . ρ(v, T ) is approximately constant between 6000 Å and 6001 Å. d v = − λc2 d λ, d v ≃ λc2 |Δλ| v2 ρ(v, T )d v ≃ ρΔv v 1 hν 8πh 3 − kT ν e , hν ≫ k T c3 ∞ 3 ρ(v, T ) = 8cπ3h 0 hν d v ν 0 e kT −1 hν let x = kT , d ν = kh d ν T ρ≃ ∞ ∞ ρ(v, T )d ν = 0 8πh kT c3 h 4 = ρΔν ∞ 0 ρ(v)d v hν = ν3 e− kT Δν π4 15 kT 4 h = 8πh π c3 15 4 ∞ 0 kT h 4 x3 dx ex − 1 Stefan-Boltzman Law (4.996 × 1014 s−1 )3 (1.611 × 105 )−1 (8.326 × 10+10 s−1 ) (6.494)(4.168 × 1013 s−1 )4 = 3.29 × 10−6 3.29 × 10−6 × 100W = 329 µW 3.29 × 10−6 J s−1 3.313 × 10−19 J photon−1 = 9.93 × 1012 photons s−1 = Problem Set #1 ANSWERS Spring, 1994 Page 3 3. (a) What is the magnitude of the electric field for the beam of a 1−mW helium-neon laser with a diameter of 1 mm? Answer: Use Bernath (1.4b) E = 27.4I 1/2 = 988 V/m (b) How many photons per second are emitted at 6328 Å? Answer: 1 mW = 1 × 10−3 J s−1 @ 6328 Å 1.988 × 10−15 J photon−1 −3 −1 = 1 × 10 J s 6328 = 3.18 × 1015 photons s−1 (c) If the laser linewidth is 1 kHz, what temperature would a blackbody have to be to emit the same number of photons from a equal area over the same frequency interval as the laser? Answer: The HeNe beam has d = 1mm, A = A = AHeNe = 0.785mm2. c A I (v, T )d v ≃ Aρ(6328Å, T ) 4 Δv = 1mW hv 1 π ρ(6328Å, T ) = 8λ3h e kT λ − 1 c Aρ(λ, T ) 4 Δv = P hv 1 A 8πh c Δv = e kλ T − 1 P λ3 4 π 4 mm2 = 0.785mm2. For calculation of T bb , −1 A 8π hc 0.785 × 10−6 m2 (8π)6.626 × 10−34 J s 3.000 × 108 m s−1 × 103 s−1 = × × −3 J s−1 −7 m)3 1 × 10 (6.328 × 10 4 P λ3 4 = 396 ex ≃ 1 + x, x ≪ 1 hc 1 ≃ 3.86 × 10−6 hλ T 6.626 × 10−34 J s × 3.00 × 108 m s−1 1 × 1.381 × 10−23 J K −1 × 6.328 × 10−7 m 3.86 × 106 = 5 .8 9 × 1 0 9 K T= Problem Set #1 ANSWERS Spring, 1994 Page 4 4. The lifetime of the 32 P3/2 → 32 S1/2 transition of the Na atom at 5890 Å is measured to be 16 ns. Na 32 P3/2 ← 32 S1/2 λ = 5 8 9 0 Å, τ = 1 6 n s (a) What are the Einstein A and B coefficients for the transition? Answer: The radiative lifetime of the state is related to the Einstein spontaneous emission coefficients τ−1 = Ai j i j The 32 P3/2 state can radiate only to 32 S1/2 (the ground state), so A= 1 1 = = 6.3 × 107 s−1 τ 16n s From Eq. (1.22) λ3 A (5.890 × 10−7 m)3 = 8πh 8π6.626 × 10−34 J s(16 × 10−9 s) = 7.67 × 1020 J −1 m3 s−2 B= (b) What is the transition dipole moment in debye? Answer: from Eq. (1.52) 16π3 ν3 2 A10 = µ = 3.136 × 10−7 ν3 µ20 ˜1 3 10 3ǫ0 hc 1 5890 Å 1 = −4 3.136 × 10−7 16 × 10 108 Å cm−1 |µ10 | = 6.38 D µ20 1 ν = [cm−1 ] ˜ −3 = 4 0 . 7 D2 (c) What is the peak absorption cross section for the transition in Å2 , assuming that the linewidth is determined by lifetime broadening? Answer: Lorentzian line profile is due to only lifetime broadening g(ν − ν0 ) 4 1 ;γ = γ τs p g(0) = 4τs p g(0) = Problem Set #1 ANSWERS Spring, 1994 Page 5 Answer: 4c, continued From Eq. (1.57) Aλ2 g(ν − ν10 ) λ2 g(ν − ν10 ) 1 σ = = ;A = 8π 8πτs p τs p 2 2 λ 4τ s p λ σmax = = ; ν = ν0 8πτs p 2π (5.890 × 10−5 cm)2 = 5.52 × 10−10 cm2 = 2π 5. (a) For Na atoms in a flame at 2000 K and 760 Torr pressure calculate the peak absorption cross section (at line center) for the 32 P3/2 − 32 S1/2 transition at 5890 Å. Use 30 MHz/Torr as the pressure-broadening coefficient and the data in Problem 4. Answer: Na atoms, T = 2000K, p = 760 torr. Calculate peak absorption cross-section for 32 P3/2 − 32 S1/2 5890Å. Δν1/2 = (30 MHz/torr) p 3 0 M Hz = (760 torr)(29979 MHz/cm−1 )−1 torr = 0.76 cm−1 From Eq. (1.75), for a Lorentzian lineshape Δν1/2 = γ 2π ; γ = 2πΔν1/2 From Eq. (1.57), 4 Aλ2 g(0) Aλ2 = , g(0) = 8π 2πγ γ 2 Aλ 1 (5.890 × 10−5 cm)2 = = (2π)2 Δν1/2 16 × 10−9 s (2π)2 (22900 × 106 s−1 ) = 2.41 × 10−13 cm2 σmax = Problem Set #1 ANSWERS Spring, 1994 Page 6 (b) If the path length in the flame is 10 cm, what concentration of Na atoms will produce an absorption (I /I0 ) of 1/e at line center? Answer: N1 ≈ 0, N0 ≈ N ln N N0 169.78 = 4 e− 1340 = 1.0 × 10−5 2 I = −σN ℓ = −1 I0 N = [2.41 × 10−13 cm2 × 10cm]−1 = 4.15 × 1011 cm−3 (c) Is the transition primarily Doppler or pressure broadened? Answer: −7 Δν0 = 7.1 × 10 ν0 ˜ ˜ T M 1/2 ν0 = 16978 cm−1 ˜ T = 2000 K M = 23 amu Δν0 = 0.11 cm−1 ˜ This compares to the pressure broadened linewidth of 0.76 cm−1 , as determined in part (a). The transition is primarily pressure broadened. NOTE: lifetime broadening contributes < 0.001 cm−1 to homogeneous broadening. The fact Δν L ˜ that ΔνD ≃ 7 means that the cross-section calculated in part (a) is slightly overestimated. ˜ (d) Convert the peak absorption cross section in cm2 to the peak molar absorption coefficient ǫ . No Answer given Problem Set #1 ANSWERS Spring, 1994 Page 7 6. For Ar atoms at room temperature (20◦ C) and 1–Torr pressure, estimate a collision frequency for an atom from the van der Waals radius of 1.5 Å. What is the corresponding pressure-broadening coefficient in MHz/Torr? Answer: Collision rate was determined using equations from P. W. Atkins, Physical Chemistry, 2nd edition. From Bernath, Eq. (1.79) 1 ; T 2 is the average time between collisions πT 2 T 2 = z−1 ; z = collision rate √ N z = 2σ c V σ = πr2 = π[1.5 × 10−8 cm]2 = 7.07 × 10−16 cm2 Δν1/2 = T 1/2 cm s−1 = 3.80 × 104 cm s−1 M 1torr × 710 atm torr−1 × 10−3 ℓcm−3 p p/torr N 6 = = × 6.022 × 1023 molecules · mole−1 V RT T /K 1 K × 0.08206 ℓ atm mole−1 K −1 p molecules cm−3 = 9.656 × 1018 T c = 14551 Δν1/2 z 1 √ 1 T 1/2 9.656 × 1018 2 = = p = 2(πr )14551 π T2 π π M T 1/2 √ 9.656 × 1018 −8 2 293 14551 p = 2(1.5 × 10 cm) 40 293 = 0.41 MHz/torr × p/torr 7. Solve the following set of linear equations using matrix methods 4 x − 3y + z = 11 2 x + y − 4z = −1 x + 2y − 2z = 1. Answer: 11 4 −3 1 2 1 −4 x = −1 1 1 2 −2 Problem Set #1 ANSWERS Spring, 1994 Page 8 Answer: #7, continued A x = b x = A−1 b The difficult part of this problem is calculating A−1 . Bernath, Eq. (3.28) (A−1 )i j = | | = 4 A 2 −4 2 1 1 −4 − (−3) + (1) 2 −2 1 −2 1 2 = 4[−2 − (−8)] + 3[−4 − (−4)] + [4 − 1] = 27 M11 = (−1)2 M12 = (−1)3 1 −4 = 6 2 −2 2 −4 = 0 1 −2 M13 = (−1)4 2 1 = 3 1 2 M21 = (−1)3 −3 1 = −4 2 −2 M22 = (−1)4 4 1 = −9 1 −2 M23 = (−1)5 4 −3 = −11 1 2 M31 = (−1)4 −3 1 = 11 1 −4 M32 = (−1)5 4 1 = 18 2 −4 4 −3 M33 = (−1)6 = 10 1 1 6 −4 11 1 −1 0 −9 18 A b= 27 3 −11 10 6 −4 11 11 3 1 −1 0 −9 18 −1 = 1 A b= 27 2 3 −11 10 1 3 x = 1 2 M ji |A| Problem Set #1 ANSWERS 8. Spring, 1994 Page 9 (a) Find the eigenvalues and normalized eigenvectors of the matrix A= 2 4−i . 4 + i −14 Answer: It will be much easier to find the eigenvalues and eigenvectors if we rewrite A as follows: A= ǫ◦ V ⋆ V −ǫ ◦ + + B 1 ǫ = (2 − 14) 2 ◦ ǫ =8 ǫ0 0ǫ ǫI = −6 V ⋆ V = |V |2 = (4 + i)(4 − i) = 17 Eigenvalues of A are found by solving d et[B − E I] = 0 and adding ǫ d et[B − E I] = (ǫ ◦ − E )(−ǫ ◦ − E ) − |V |2 = 0 E 2 = (ǫ ◦ )2 + |V |2 E = ±9 , ǫ = −6 The eigenvalues of A are +3 and −15. Determine the eigenvectors of A by substituting the appropriate values of ǫ into A − ǫ I ǫ1 = +3 A − ǫ1 I = −1 4 − i 4 + i −17 Normalize it to unit length 1 4−i ǫ1 = +3, u1 = √ 18 1 ǫ2 = −15 A − ǫ2 I = 17 4 − i 4+i 1 Same as for ǫ1 , but 1 −1 ǫ2 = −15, u2 = √ 4+i 18 Problem Set #1 ANSWERS Spring, 1994 Page 10 (b) Construct the matrix X that diagonalizes A and verify that it works. Answer: X, which diagonalizes A, consists of the column vectors u1 and u2 . X−1 AX = Λ 1 −1 4 − i X= √ 1 18 4 + i −1 ⋆T X = X = (X ) Let a = 4 + i, a⋆ = 4 − i, a⋆ a = 17 1 18 1 = 18 1 = 18 X−1 AX = −1 a⋆ 2 a⋆ −1 a⋆ a 1 a −14 a 1 −1 a⋆ 15 3a⋆ a 1 −15a 3 −15(1 + 17) 0 −15 0 = 0 3(17 + 1) 0 3 X diagonalizes A. 9. Given the matrices A and B as −1 3 A= 2 3 √ 2 3 2 3 0 1 √ 3 √ 2 3 1 √ 3 2 −3 5 3 √ B = 16 1 − √ 32 1 √ 6 3 2 1 √ 23 1 − 3 √2 1 √ . 2 3 11 6 Show that A and B commute. Find their eigenvalues and eigenvectors, and obtain a unitary trans­ formation matrix U that diagonalizes both A and B. Answer: A= −.33333 .81650 .4 7 1 4 0 .8 1 6 5 0 .0 0 0 0 0 .5 7 7 3 5 .47140 .57735 −.66667 B= 1.66667 .40825 −.23570 .4 0 8 2 5 1 .5 0 0 0 0 .2 8 8 6 8 −.23570 .28868 1.83333 Problem Set #1 ANSWERS Spring, 1994 Answer: #9, continued COMMUTATOR (AB − BA) : .0 0 0 0 0 .0 0 0 0 0 .0 0 0 0 0 .0 0 0 0 0 .0 0 0 0 0 .0 0 0 0 0 .0 0 0 0 0 .0 0 0 0 0 .0 0 0 0 0 MATRIX BEFORE DIAGONALIZATION: |1 |2 |3 1| = 1.66667 .40825 −.23570 2| = .40825 1 .5 0 0 0 0 .2 8 8 6 8 3| = −.23570 .2 8 8 6 8 1 .8 3 3 3 3 EIGENVALUES AND EIGENVECTORS: #1 #2 #3 Value: 1.0000 2 .0 0 0 0 2 .0 0 0 0 Vector: 1| = .57735 −.70711 .4 0 8 2 5 2| = .22116 −.34588 −.91184 3| = .78597 .61674 −.04331 MATRIX BEFORE DIAGONALIZATION: |1 |2 |3 1| = −.33333 .8 1 6 5 0 .4 7 1 4 0 2| = .81650 .0 0 0 0 0 .5 7 7 3 5 3| = .47140 .57735 −.66667 EIGENVALUES AND EIGENVECTORS: #1 #2 #3 Value: −1.0000 −1.0000 −1.0000 Vector: 1| = −.22217 −.34509 .9 1 1 9 0 2| = .78569 −.61718 −.04214 3| = .57735 .70711 −.40825 Page 11 Problem Set #1 ANSWERS Spring, 1994 Page 12 10. Obtain eigenvalues and eigenvectors of the matrix 1 2α 0 2 3α H = α 2 + α 0 3α 3 + 2α to second order in the small parameter α. Answer: 1 2α 0 3α H = 2α 2 + α 0 3α 3 + 2α H = H◦ + H′ 1 0 0 H◦ = 0 2 0 0 0 3 Energy corrected to second order: 0 2 0 , H′ = α 2 1 3 032 E i = E i◦ + i|H′ |i + 1 |H ′ |1 1 |H ′ |2 1 |H ′ |3 2 |H ′ |2 2 |H ′ |3 3 |H ′ |3 | i |H ′ | j |2 E i◦ − E ◦ j i� j =0 = 2α =0 =α = 3α = 2α (2α)2 = 1 − 4α2 1−2 (2α)2 (3α)2 E 2 = 2 + α + + = 2 + α − 5α2 2 − 1 2 − 3 (3α)2 E 3 = 3 + 2α + = 3 + 2α + 9α2 3−2 Hi j |j ◦ | = |i ◦ + i E i◦ − E ◦ j i� j E1 = 1 + 0 + |1 = |1 ◦ − 2α |2 ◦ |2 = |2 ◦ + 2α |1 ◦ − 3α |3 |3 = |3 ◦ + 3α |2 ◦ ◦ Problem Set #1 ANSWERS Spring, 1994 Page 13 Answer: # 10, continued Checking orthogonality: 1 |2 = 0 1|3 = −6α2 2 |3 = 0 ; α ≪ 1, so approaches zero 11. A particle of mass m is confined to an infinite potential box with potential ∞, x < 0, x > L, x k 1 − L , 0 ≤ x ≤ L. V ( x) = Calculate the ground and fourth excited-state energies of the particle in this box using first-order perturbation theory. Obtain the ground and fourth excited-state wavefunctions to first order, and sketch their appearance. How do they differ from the corresponding unperturbed wavefunctions? Answer: Particle of mass m confined a perturbed square well potential. ψ◦ ( x) n 2 = L 1/2 sin ∞, x k 1− L x H′ = k 1 − L V ( x) = nπ x L x < 0, x > L , 0≤x≤L k 2 L 2 nπ x sin2 xd x L L nπ 0 2k (nπ)2 1 =k =k− (nπ)2 4 2 L 2k nπ mπ = k ψ◦ |ψ◦ − 2 x sin x sin x dx nm L0 L L ψ◦ |H′ |ψ◦ = k − n n ψ◦ |H′|ψ◦ n m sin A sin B = 1 [cos(A − B) − cos(A + B)] 2 L L −k (n − m)π cos(n + m) x cos x xd x − xd x 2 L L L 0 0 (n−m)π (n+m)π L2 L2 −k x cos xd x − x cos xd x =2 L (n − m)2 π2 0 (n + m)2 π2 0 ψ◦ |H′ |ψ◦ = n m n ± m = even, ψ◦ |H′|ψ◦ = 0 n m Problem Set #1 ANSWERS Spring, 1994 Page 14 Answer: #11, continued n ± m = odd, ◦ ′ ◦ +2k 1 1 2k 4nm − = 2 ψn |H |ψm = 2 π (n − m)2 (n + m)2 π (n + m)2 (n − m)2 2k 8 ψ◦ |H′ |ψ◦ = 2 1 2 π9 2k 16 ◦ ′ ◦ ψ1 |H |ψ4 = 2 π 225 2k 48 ◦ ′ ◦ ψ4 |H |ψ3 = 2 π 49 2k 80 ◦ ′ ◦ ψ4 |H |ψ3 = 2 π 81 2k 112 ◦ ′ ◦ ψ4 |H |ψ7 = 2 π 1089 Δn ≥ 3 perturbations contribute extraordinarily little relative to Δn = 1 ◦ E 1 = E 1 + ψ◦ | ′| ◦ 1 H ψ1 1 h2 = α + k ; α = 2 8mL2 ◦ ′ ◦ ◦ E 4 = E 4 + ψ4 | | 4 Hψ 1 = 16α + k 2 To first order | 1 ψ | 4 ψ 2k 8 ◦ = − 2 ψ π α 27 2 2k 48 80 ◦ ◦ ◦ = ψ4 + 2 ψ − ψ π α 343 3 729 5 1 ψ◦ We can rewrite the ψ’s in terms of a single parameter, β 1 | 1 = ψ◦ − 0.296β ψ◦ ψ 2 4 | 4 = ψ◦ + 0.140β ψ◦ − 0.110β ψ◦ ψ 3 5 2k |β | ≪ 1 ; β= 2 πα Establishing the qualitative effect on |ψ1 is simple 1 ψ◦ = 0 , L 1 ψ◦ = 0 L Mixing in −0.296β ψ◦ makes |ψ1 asymmetric, with a slightly increased probability of finding the particle 2 on “L–side” of the well. Problem Set #1 ANSWERS Spring, 1994 Answer: #11, continued n ± m = odd, 1 | 1 = ψ◦ − 0.296β ψ◦ ψ 2 β = 0, 0.05, 0.1, 0. 2 4 | 4 = ψ◦ + 0.140β ψ◦ − 0.110β ψ◦ ψ 3 5 β = 0, 0.2, 0.5, 1 Page 15 ...
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This note was uploaded on 11/28/2011 for the course CHEM 5.74 taught by Professor Robertfield during the Spring '04 term at MIT.

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