Molecular Spectra and Molecular Structure - 1987 PS1

Problem Set #1 5. J. I. Steinfeld (2nd Ed.), p. 36, #2 (a) Given the matrix elements of the coordinate x for a harmonic oscillator � � ψ�v xψ�v dx = 0 unless v� = v ± 1 �v|x|v � = and �v + 1|x|v� = (2β)−1/2 (v + 1)1/2 , �v − 1|x|v� = (2β)−1/2 (v)1/2 , where β = 4π2 mν/h where ν = 21π (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 � r � ψv xr ψv� dx �v|x |v � = 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 � �2 state. Is it true that x2 = (x)2 , or that x4 = x2 ? 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 (52 P3/2 → 52 S1/2 and 52 P1/2 → 52 S1/2 ) 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? 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 2 S]) and the (n = 2, � = 1, m = 1[2 2 P]) states of atomic hydrogen. The wave functions are ψ100 = ψ211 = 1 π1/2 a3/2 0 e−r/a0 Y00 (θ, φ), 1 4(2π)1/2 a5/2 0 re−r/2a0 Y11 (θ, φ), neglecting electron and nuclear spin. Remember that the dipole operator is a 3–vector, µ = e0 r = e0 (î sin θ cos φ + ĵ sin θ sin φ + k̂ cos θ). Transition Amplitudes for np2 ← np n’s Transitions in the L–S–J Limit � ∞ −1/2 Rnp rRn� s dr µ ≡ −e 3 Condon & Shortley, p. 245. 0 Condon & Shortley, p. 247 give all non–zero transition amplitudes: � � p2 1 S|µ|sp 1 P1 = −(20)1/2 µ � � p2 1 D|µ|sp 1 P1 = +10µ � � p2 3 P0 |µ|sp 3 P1 = −(20)1/2 µ � � p2 3 P1 |µ|sp 3 P0 = −(20)1/2 µ � � p2 3 P1 |µ|sp 3 P1 = +(15)1/2 µ � � p2 3 P1 |µ|sp 3 P2 = −5µ � � p2 3 P2 |µ|sp 3 P1 = −5µ � � p2 3 P2 |µ|sp 3 P2 = +(75)1/2 µ. Page 3 All other transition amplitudes are zero, most notably: � � p2 3 P0 |µ|sp 3 P0 = 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): S0 F0 + 10F2 P0 F0 − 5F2 ee 3 F0 − 5F2 H = P1 3 F0 − 5F2 P2 1 D2 F0 + F2 1 3 S0 0 −21/2 ζ 3 P0 −21/2 ζ −ζ 3 − 12 ζ = P1 1 3 P2 ζ 2−1/2 ζ 2 1 D2 2−1/2 ζ 0 1 HSO So we have three effectivce Hamiltonians for (np)2 . � �� � �� ��F0 + 10F2 ��Δ0 V0 ��� 5 1 −2−1/2 ζ ��� = F0 + F2 − ζ + � H(0) = � � −2−1/2 ζ F0 − 5F2 − ζ �� �V0 −Δ0 �� 2 2 15 1 F2 + ζ V0 = −2+1/2 ζ Δ0 = 2 2 1 H(1) = F0 − 5F2 − ζ 2 �� � � �� ��F0 − 5F2 + ζ/2 2−1/2 ζ ��� ��−Δ2 V2 ��� 1 = F0 − 2F2 + ζ + � H(2) = � � � V2 +Δ2 �� F0 + F2 �� 2−1/2 ζ 4 1 Δ2 = 3F2 − ζ V2 = 2−1/2 ζ. 4 Similarly, for the sp configuration P2 F0 − G1 + 12 ζ 3 F0 − G1 − 12 ζ 2−1/2 ζ P H= 1 1 P1 2−1/2 ζ F0 + G1 3 P0 F0 − G1 − ζ 3 Page 4 and there are three effective Hamiltonians for (n� s)(np) H(0) = F0 − G1 − ζ �� � ��−Δ1 V1 ��� 1 H(1) = F0 − ζ + � � V1 Δ1 �� 4 1 H(2) = F0 − G1 + ζ. 2 1 Δ1 = G1 + ζ, 4 V1 = 2−1/2 ζ Now 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 np2 and 4 L–S–J terms in np n’s, in principle giving rise to a “transition array” consisting of 5 × 4 transitions. The 5 parameters determine 20 frequencies and 20 intensities! We are not limited to the L–S–J or the j1 − j2 − J limit. 1. Construct level diagrams for p2 and sp at the L–S–J limit (ζ = 0), the j − j limit (F2 = 0 for p2 , G1 = 0 for sp), and at several intermediate values of ζ/F2 or ζ/G1 . This sort of diagram is called a “correlation diagram”. For graphical purposes it is convenient to keep constant the quantity, which determines the splitting between highest and lowest levels of p2 , 225 2 15 9 F2 + (F2 )(ζ) + ζ 2 ≡ ΔE(p2 ), 4 2 4 and a similar quantity for sp, G21 + 9 2 1 ζ + (G1 )(ζ) ≡ ΔE(sp). 16 2 2. Use the first order non-degenerate perturbation theory correction to the wavefunctions to compute the intensities for p2 ← sp transitions near the L–S–J limit (ζ � F2 for p2 , ζ � G1 for sp). For Page 5 example, the “nominal” sp 1 P1 level becomes H1 P1 3 P1 �� 3 � � sp P1 E10P1 − E30P1 � � 2−1/2 ζ �� 3 � � sp P1 . = �� sp 1 P1 + 2G1 + 12 ζ � � � �� �‘sp 1 P1 ’ = �� sp 1 P1 + The transition probability is the square of the transition amplitude, so the “nominally forbidden” transition p2 3 P1 ← sp 1 P1 has a transition probability ��� ���2 2ζ 2 2 P(3 P1 ← 1 P1 ) = �� ‘sp 1 P1 ’ |µ| ‘p2 3 P1 ’ �� = � �2 µ (15). 2G1 + 12 ζ Note that, for the transitions between either of the two sp J = 1 levels and either of the two p2 J = 2 or J = 0 levels, the transition probability includes two amplitudes which must be summed before squaring. This gives rise to quantum mechanical interference effects. In fact, it is because of these interference effects that, in the j − j limit, (3/2, 3/2)2 ← (1/2, 1/2)1 and (1/2, 1/2)0 ← (1/2, 1/2)1 transitions become rigorously forbidden. 3. Condon and Shortley (p. 294) give the transformations from the L–S–J to the j1 − j2 − J basis set. These transformed functions correspond to the functions that diagonalize HSO . � � � �1/2 � �1/2 ��3 � ��1 � 33 2 1 2 � P2 − � D2 p = 3 3 22 2 � � � �1/2 � �1/2 ��3 � ��1 � 31 1 2 � P2 + � D2 = 22 2 3 3 � � � � 31 = ��3 P1 22 1 � � � �1/2 � �1/2 ��1 � ��3 � 33 2 1 � S0 − � P0 = 3 3 22 0 � � � �1/2 � �1/2 ��1 � ��3 � 11 1 2 � S0 + � P0 = 3 3 22 0 � � � � 13 sp = ��3 P2 22 2 � � � �1/2 � �1/2 ��1 � ��3 � 13 2 1 � P1 − � P1 = 3 3 22 1 � � � �1/2 � �1/2 ��3 � ��3 � 11 1 2 � P1 + � P1 = 22 1 3 3 � � � � 11 = ��3 P0 . 22 0 Page 6 Construct the new p2 H(0), H(1), H(2) and sp H(0), H(1), H(2) matrices in the j − j basis using the above transformations. 4. Use perturbation theory as in Problem 2 to compute the transition intensities near the j− j limit (F2 � ζ or G1 � ζ). You should discover that destructive interference starts to turn off the transitions that will become the forbidden 3 P2 ← 1 P1 and 3 P0 ← 1 P1 transitions in the L–S–J limit.