5.80 Small-Molecule Spectroscopy and Dynamics

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1 MIT OpenCourseWare Small-Molecule Spectroscopy and Dynamics Fall 008 For information about citing these materials or our Terms of Use, visit:

2 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Chemistry 5.76 Spring 1976 Closed Book Slide Rules and Calculators Permitted Examination #1 ANSWERS March 1, 1976 Answer any THREE of the four questions. You may work a fourth problem for extra credit. All work will be graded but no total grade will exceed 80 points. 1. A. (10 points) Give a concise statement of Hund s three rules. First rule: The lowest energy state belonging to a configuration has maximum S. Second Rule: Of the states of maximum spin, the lowest term has maximum L. Third Rule: The lowest J state of the lowest term is the one with maximum J for a more than half-filled shell and minimum J for less than half-filled shell. None of the Hund s rules apply to any but the lowest energy term belonging to a configuration. B. (10 points) State the definition of a vector operator. B is a vector with respect to A if [A i, B j ] = ǫ i jk iħb k C. (10 points) If B and C are vector operators with respect to A, then what do you know about matrix elements of B C in the AM A basis? B C is scalar with respect to A, therefore A M A B C AM A = δ A Aδ MA M A A B C A independent of M A. D. ( 5 points) The atomic spin-orbit Hamiltonian has the form H SO = ξ(r i )l i s i i Classify H SO as vector or scalar with respect to J, L, and S. State whether H SO is diagonal in the JM J LS or LM L S M S basis. H SO is scalar, vector, vector with respect to J, L, S. H SO is diagonal with respect to J and M J but not L and S in the JM J LS basis but diagonal in nothing in the LM L S M S basis.

3 5.76 Exam #1 ANSWERS March 1, 1976 page. Consider the following multiplet transition array: J? Lower State (L, L )??? (180) (16) (0) Upper State (L,S )?? (40) (16) (310) Intensities are in parentheses above transition frequencies in cm 1 ; line separations in cm 1 are given between relevant transition frequencies. A. (10 points) Use the Landé interval rule E(L, S, J) E(L, S, J 1) = ζ(nls )J to determine J and J values. Rather than list the J and J assignments of each line, only list J and J for the line observed to be most intense and for the line observed to be least intense. Consider upper state term separations first Upper state J ranges 5 3 Lower state: E(J MAX ) E(J MAX 1) J MAX = = = E(J MAX 1) E(J MAX ) J MAX J MAX = = J MAX lower state J ranges 4 Most intense line cm 1 in J = 4 J = 5 Least intense line cm 1 in J = 4 J = 3 B. (10 points) Use the range of J and J and the intensity distribution (i.e., that the most intense transition is not ΔJ = 0) to determine the term symbols ( S +1 L) for the upper and lower states. Assume ΔS = 0. Range of J implies either S = 1, L = 4, L = 3 or L = 1, S = 4, S = 3 The second possibility is ΔS 0 forbidden. Upper state is 3 G, Lower state is 3 F.

4 5.76 Exam #1 ANSWERS March 1, 1976 page 3 C. ( 5 points) Is the upper state regular (highest J at highest term energy) or inverted (highest J at lowest term energy)? Is the lower state regular or inverted? [Partial energy level diagrams might be helpful here.] 3 G J = 5 3 G J = J = 3 3 F upper state regular J = 4 3 F lower state inverted J = J = 4 3. A. ( 5 points) List the L-S terms that arise from the (ns)(np) and (ns) (np) configurations. [HINT: (np) gives 1 S, 3 P, 1 D; to get sp couple an s electron to these three states.] (ns)(np) S, P, 4 P, D (ns) (np) P B. ( 5 points) Which configuration gives rise to odd terms and which to even? (ns)(np) is even because Σl i = (ns) (np) is odd because Σl i = 1 C. ( 5 points) List the electric dipole allowed transitions between terms of the sp and s p configurations. (Ignore fine-structure splitting of L-S terms into J-states.) S P P P D P are the allowed transitions. D. (10 points) Construct qualitative energy level diagrams on which you display all allowed J J components of P S, P P, and P D transitions. Indicate which J J line you would expect to be strongest for each of these three transitions. J J J 5/ 3/ D P 3/ 1/ S 1/ P 3/ 3/ 3/ 1/ 1/ 1/ Strongest Strongest Strongest 3/ - 5/ 3/ - 3/ 3/ - 1/

5 5.76 Exam #1 ANSWERS March 1, 1976 page 4 4. (5 points) Calculate transition probabilities for the two transitions given the following information: nsnp 1 P 10 (np) 1 S 00 nsnp 1 P 10 (np) 1 D 0 1 P 10 = J = 1, M J = 0, L = 1, S = = s0 p0 + s0 + p0 1 S 00 J = 0, M J = 0, L = 0, S = = 3 p1 p p1 + p p0 + p0 1 D p1+ p p1 p p0+ p0 The electric dipole transition moment operator, µ, does not operate on spin coordinates, is a one-electron operator, and is a vector with respect to l i. nsnp (np) transitions are Δl =+1 processes. The relevant Δl =+1 matrix elements, as given by the Wigner-Eckart theorem for vector operators are 1 1 n,l = 1, m l = 1 (µ + + µ ) n,l = 0, m l = 0 = µ + (ns) n,l = 1, m l = 0 µ z n,l = 0, m l = 0 = µ + (ns) 1 1 n,l = 1, m l = 1 (µ ++ µ ) n,l = 0, m l = 0 =+ µ + (ns) where µ + (ns) is the reduced matrix element np µ ns.

6 5.76 Exam #1 ANSWERS March 1, 1976 page 5 Since µ is a one electron operator, the two-electron Slaters must match for one spin-orbital and must have identical spin in the other. This means we need only consider part of 1 S 00 and 1 D 0. 1 S p0+ p0 1 D 0 6 p0+ p0 because the p1 + p 1 and p1 p 1 + Slaters differ from the s0 p0 + and s0 + p0 Slaters by two spin-orbitals. So we do not even need to evaluate matrix elements to get the ratio of transition probabilities Actually evaluating matrix elements gives ( ) 1 1 P 1 10 S = ( = P 1 ) D [ ] 1 [ ] 1 P s0 p0 + p0 + s0 + p µ 1 S 00 = µ p0 p0 µ p0 6 1 [ = µ + (ns) µ + (ns) ] = µ + (ns) 6 6 Probability is 1 µ = 3 µ +(ns) = 4 3 µ +(ns) for P S for P D Show all your work including false starts. If you are unable to express the transition probabilities in terms of µ + (ns), lavish partial credit will be given for the ratio of transition probabilities 1 P 1 S P 10 1 D 00