02psetsp94

02psetsp94 - MIT OpenCourseWare http://ocw.mit.edu 5.80...

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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 5.76 Modern Topics in Physical Chemistry Spring, 1994 Problem Set #2 Reading Assignment: Bernath, Chapter 5 The following handouts also contain useful information: C & S, page 117, radial expectation values of r k for 1-e – atoms LS → (j,j ′ ) J Coupling Patterns Herzberg pp. 177-181, The Interval Rule: Analysis of Multiplets Problems 1-4 deal with material from my 2/11/94 lecture (Lecture 7). 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. j 1 – j 2 – J limiting cases. C & S references are to Condon and Shortley “The Theory of Atomic Spectra.’ Problems 5-9 are standard textbook problems, more basic, and much easier than 1-4 and 10. BACKGROUND MATERIAL FOR PROBLEMS 1-4 (i) Transition Amplitudes for np 2 ← np n ′ s Transitions in the L- S- J Limit µ ≡ − e3 − 1/2 ∫ ∞ R np r R n s ′ dr C&S, p. 245. C&S, p. 247 gives all nonzero transition amplitudes: p 2 1 S µ sp 1 P 1 = − (20) 1/2 µ p 2 1 D µ sp 1 P 1 = + 10µ p 2 3 P µ sp 3 P 1 = − (20) 1/2 µ p 2 3 P 1 µ sp 3 P = − (20) 1/2 µ p 2 3 P 1 µ sp 3 P 1 = + (15) 1/2 µ p 2 3 P 1 µ sp 3 P 2 = − 5µ p 2 3 P 2 µ sp 3 P 1 = − 5µ p 2 3 P 2 µ sp 3 P 2 = + (75) 1/2 µ. All other transition amplitudes are zero, most notably: p 2 3 P µ sp 3 P = because there is no way to add one unit of photon angular momentum to an initial state with J = 0 to make a final state with J = 0. (ii) Energy levels for np 2 and np n ′ s in the L–S–J Basis Set In the L–S–J limit, for p 2 (see C&S, pp. 198, 268): 1 S F + 10F 2 3 P F − 5F 2 H ee = 3 P 1 F − 5F 2 3 P 2 F − 5F 2 1 D 2 F + F 2 1 S − 2 1/2 ζ 3 P − 2 1/2 ζ −ζ H SO = 3 P 1 − 1 2 ζ 3 P 2 1 2 ζ 2 − 1/2 ζ 1 D 2 2 1/2 ζ So we have three effective Hamiltonians for (np) 2 F + 10F 2 –2 + 1/2 ζ H (0) = –2 + 1/2 ζ F − 5F 2 − ζ = F + 5 2 F 2 − 1 2 ζ + ∆ V V − ∆ ∆ = 15 1 = − 2 + 1/2 ζ 2 F 2 + 2 ζ V H 1 2 1 ζ ( ) = F − 5F 2 − F − 5F 2 + ζ / 2 2 − 1/2 ζ H (2) = 2 − 1/2 ζ = F − 2F 2 + 4 1 ζ + − ∆ 2 V 2 F + F 2 V 2 + ∆ 2 ∆ 2 = 3F 2 − 4 1 ζ V = − 2 − 1/2 ζ Similarly, for the sp configuration: 3 P 2 F − G 1 + 2 1 ζ H = 3 P 1 F − G 1 − 2 1 ζ 2 − 1/2 ζ 1 P 2 − 1/2 ζ F + G 1 3 P F − G 1 − ζ and there are three effective Hamiltonians for (n ′ s)(np) H (0) = F – G 1 – ζ Problem Set #2 Spring, 1994 Page 2 H 1 1 4 ζ + − V ∆ 1 1 ∆ V 1 1 ∆ 1 = G 1 + 1 4 ζ V 1 = 2 − 1/2 ζ ( ) = F − H (2) = F – G 1 + 1 2 ζ (iii) Now we are ready to discuss the energy level diagram and relative intensities of all spectral lines...
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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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02psetsp94 - MIT OpenCourseWare http://ocw.mit.edu 5.80...

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