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 80 Small Molecule Spectroscopy and Dynamics Fall 2008 Problem Set 2 ANSWERS Reading Assignment Bernath Chapter 5 The following handouts also contain useful information C S page 117 radial expectation values of rk 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 j1 j2 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 np2 np n s Transitions in the L S J Limit e3 1 2 R np 0 r R n s dr C S p 245 C S p 247 gives all nonzero transition amplitudes p 2 1 S sp 1 P1 20 1 2 p 2 1 D sp 1 P1 10 p 2 3 P0 sp 3 P1 20 1 2 p 2 3 P1 sp 3 P0 20 1 2 p 2 3 P1 sp 3 P1 15 1 2 p 2 3 P1 sp 3 P2 5 p 2 3 P2 sp 3 P1 5 p 2 3 P2 sp 3 P2 75 1 2 All other transition amplitudes are zero most notably p 2 3 P0 sp 3 P0 0 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 Energy levels for np2 and np n s in the L S J Basis Set ii In the L S J limit for p2 see C S pp 198 268 1 3 F0 10F2 S0 P0 F0 5F2 H P1 ee F0 5F2 3 3 1 F0 5F2 P2 F0 F2 D2 1 S0 3 H SO P0 3 P1 3 1 0 21 2 21 2 12 2 1 2 2 0 1 2 1 2 P2 D2 So we have three effective Hamiltonians for np 2 0 2 1 2 5 1 F0 F2 V0 2 2 F0 5F2 F 10F H 0 0 1 2 2 2 V0 0 15 1 F2 V0 2 1 2 2 2 1 H 1 F0 5F2 2 2 F0 5F2 2 2 1 2 1 H 2 F 2F 0 2 V2 4 2 1 2 F0 F2 0 2 3F2 1 4 V2 2 V0 2 1 2 Similarly for the sp configuration 3 P2 1 F0 G1 2 3 H P1 1 3 P0 P0 1 F0 G1 2 1 2 2 1 2 2 F0 G1 F0 G1 and there are three effective Hamiltonians for n s np H 0 F0 G1 Problem Set 2 ANSWERS Spring 1994 Page 2 H 1 F0 1 1 V1 4 V1 1 1 4 1 G1 V1 2 1 2 1 H 2 F0 G1 2 iii 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 F0 difference in repulsion energy for np by np vs np by n s F0 0 if n n np spin orbit parameter for np same for both configurations 0 by definition F2 np np quadrupolar repulsion between two np electrons F2 0 G1 n s np exchange integral G1 0 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 the p2 and sp configurations at the L S J limit 0 the j j limit F2 0 for p2 G1 0 for sp and at several intermediate values of F2 or 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 2 E p 2 4 2 4 and a similar quantity for sp 9 2 1 G1 E sp 16 2 Correlation diagrams for sp and p2 configurations can be found on pages 272 and 275 of Condon and Shortley G12 3 For sp 4G 1 Problem Set 2 ANSWERS Spring 1994 Page 3 G1 3 4 1 2 1 2 1 1 2 1 2 For p2 5F 2 F2 3 1 2 1 2 1 1 2 1 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 F2 for p2 G1 for sp For example the nominal sp 1P1 level becomes 2 1 2 sp P1 sp P1 0 sp P1 sp P1 sp 3 P1 0 1 E1 P E 3 P 2G1 1 1 2 The transition probability is the square of the transition amplitude so the nominally forbidden transition p2 3P1 sp 1P1 has a transition probability 1 Problem Set 2 ANSWERS 1 H 1 P 3P 1 1 3 Spring 1994 1 Page 4 P 3 P1 1 P1 sp 1P1 p 2 3P1 2 2 2 2 2 15 1 2G1 2 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 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 s corrected to 1st order p2 sp 2 1 2 3 D2 D2 P2 1 6F2 2 1 2 2 1 3 P2 3 P2 D2 1 6F2 2 3 3 P1 P1 1 1 3 P0 3 P0 2 1 S 15F2 0 1S0 1 S0 2 3 P 15F2 0 3 P2 3 P2 2 1 2 1 P1 P1 P 2G1 12 1 3 3 2 1 2 3 P1 P1 P 2G1 12 1 1 1 3 P0 3 P0 Calculate intensities of selected forbidden transitions 0 Problem Set …