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 1987 Problem Set #1 Due February 23, 1987 Problems 1-4 deal with material from the February 11, 1987 Lecture. A lot of background material is provided. These problems illustrate non-text material dealing with 2 × 2 secular equations, perturbation theory, transition probabilities, quantum mechanical interference effects, and atomic L–S–J vs. ji − j2 − J limiting cases. Problems 5-8 are standard textbook problems, more basic, and much easier than 1-4. 5. J. I. Steinfeld (2nd Ed.), p. 36, #2 (a) Given the matrix elements of the coordinate x for a harmonic oscillator �v|x|v�� = ψv � xψv�dx = 0 unless v� = v ± 1 and �v + 1|x|v� = (2β)−1/2(v + 1)1/2 , �v − 1|x|v� = (2β)−1/2(v)1/2 , where β = 4π2mν/h where ν = 21 π (k/m)1/2 and v is the vibrational quantum number. Evaluate the nonzero matrix elements of x2, x3, and x4; that is, evaluate the integrals �v|xr|v�� = ψvxrψv� dx for r = 2, 3, and 4 (without actually doing the explicit integrals, of course!). (b) From the results of (a), evaluate the average values of x, x2, x3, and x4 in the vth vibrational state. Is it true that x2 = (x)2, or that x4 = � x2�2? What conclusions can you draw about the results of a measurement of x in the vth vibrational state? 6. J. I. Steinfeld (2nd Ed.), p. 74, #1 In the spectrum of rubidium, an alkali metal, the short-wavelength limit of the diffuse series is 4,775 Å. The lines of the first doublet in the principal series (52P3/2 → 52S1/2 and 52P1/2 → 52S1/2) have wavelengths of 7,800 Å and 7,947 Å, respectively. (a) By means of term symbols, write a general expression for the doublets of the sharp series, giving explicitly the possible values for n, the principal quantum number. (b) What is the spacing in cm−1 of the first doublet in the sharp series?� � � Problem Set #1 Spring, 1987 Page 2 (c) Compute the first ionization potential of rubidium in cm−1 and electron volts (eV). 7. J. I. Steinfeld (2nd Ed.), p. 74, #2 In the first transition row of the periodic table there is a regular trend in ground state multiplicities from calcium (singlet) to manganese (sextet) to zinc (singlet), with one exception. (a) Why does the multiplicity rise to a maximum and then fall? (b) Explain the discontinuity shown by chromium (atomic number 24). (c) Niobium, the element under vanadium in the second transition row, also shows a discontinuity in multiplicity, though vanadium does not. Explain. 8. J. I. Steinfeld (2nd Ed.), p. 75, #7 Evaluate the transition dipole moment matrix element between the (n = 1, � = 0, m = 0[1 2S]) and the (n = 2, � = 1, m = 1[22P]) states of atomic hydrogen. The wave functions are 1 ψ100 = π1/2a3/2e−r/a0Y00(θ, φ), 0 1 ψ211 = 5/2re−r/2a0Y11(θ, φ), 4(2π)1/2a0 neglecting electron and nuclear spin. Remember that the dipole operator is a 3–vector, µ = e0r = e0(ˆi sin θ cos φ + ˆj sin θ sin φ + kˆ cos θ). Transition Amplitudes for np2 np n’s Transitions in the L–S–J Limit ← RnprRn� sdr Condon & Shortley, p. 245. µ ≡ −e 3−1/20 ∞ Condon & Shortley, p. 247 give all non–zero transition amplitudes: � p2 1S|µ|sp 1P1 � = −(20)1/2 µ � p2 1D|µ|sp 1P1 � = +10µ � p2 3P0|µ|sp 3P1 � = −(20)1/2 µ � p2 3P1|µ|sp 3P0 � = −(20)1/2 µ � p2 3P1|µ|sp 3P1 � = +(15)1/2 µ � p2 3P1|µ|sp 3P2 � = −5µ � p2 3P2|µ|sp 3P1 � = −5µ p2 3P2|µ|sp 3P2 = +(75)1/2 µ.� � Problem Set #1 Spring, 1987 Page 3 All other transition amplitudes are zero, most notably: p2 3P0|µ|sp 3P0 = 0 because there is no way to add one unit of photon angular momentum to initial state J = 0 to make a final state J = 0. Energy Levels for np2 and npn�s in the L–S–J Basis Set In the L–S–J limit, for p2 (see Condon & Shortley, pp. 198, 268): 1S0 3P0 Hee = 3P1 3P2 1D2 1S0 3P0 HSO = 3P1 3P2 1D2 F0 + 10F2 F0 − 5F2 F0 − 5F2 F0 − 5F2 F0 + F2 0 −21/2ζ −21/2ζ −ζ 1 2− ζ 1ζ 2−1/2ζ22−1/2ζ 0 So we have three effectivce Hamiltonians for (np)2. ������ F0 + 10F2 −2−1/2ζ ������ 5 1 ������ Δ0 V0 ������H(0) = −2−1/2ζ F0 − 5F2 − ζ = F0 + 2F2 − 2ζ + V0 −Δ0 15 1 Δ0 = F2 + ζ V0 = −2+1/2ζ2 21H(1) = F0 − 5F2 − ζ ������ F0 − 5F2 + 2ζ/2 2−1/2ζ ������ 1 ������−Δ2 V2 ������H(2) = 2−1/2ζ F0 + F2 = F0 − 2F2 + 4ζ + V2 +Δ2 1 Δ2 = 3F2 − ζ V2 = 2−1/2ζ. 4Similarly, for the sp configuration 3P2 F0 − G1 + 1ζ2F0 − G1 − 21ζ 2−1/2ζH = 13PP11 2−1/2ζ F0 + G1 3P0 F0 − G1 − ζProblem Set #1 Spring, 1987 Page 4 and there are three effective Hamiltonians for (n� s)(np) H(0) = F0 − G1 − ζ 1 1H(1) = F0 − ζ + ������−Δ1 V1 ������ Δ1 = G1 + ζ, V1 = 2−1/2ζ4V1 Δ141H(2) = F0 − G1 + ζ. 2Now we are ready to discuss the energy level diagram and relative intensities of all spectral lines for transitions between (np)2 (n� s)(np) configurations. The relevant parameters are ← F0(np np) - F0(n� s np) ≡ ΔF0 (difference in repulsion energy for np by np vs. np by n� s; ΔF0 > 0 if n� = n). ζ(np) (spin-orbit parameter for np; same for both configurations) ζ > 0 by definition. F2(np np) (quadrupolar repulsion between two np electrons) F2 > 0. G1(n� s np) (exchange integral) G1 > 0. µ (np n� s transition moment integral). ← All spectral line frequencies and intensities may be derived from these 5 fundamental electronic constants. Note that there are 5 L–S–J terms in np2and 4 L–S–J terms in np n’s, in principle giving rise to a “transition array” consisting of 5 × 4 transitions. The 5