HBr, angular distribution analysis E(0) Updated,

Peak “A”

E0,J´=3

E0,J´=4

Rydb. H + Br/Br* r(HX) J´´ v´´= 0 HX HX** J´ v´ E H + X - /Ion-pair/V v´,J´ H* + Br H* + Br* H + + Br H + + Br* 2 h resonance excitation 3 h dissociation / H* formation 4 h ionization / H + formation

2 h resonance excitation 3 h dissociation / H* formation According to :

where Unknown (variable in a fit) Unknown (variables in a fit)

BUT simpler for “non Q branches” (O, S) (i.e. for J´´  J´): 2 h resonance excitation 3 h dissociation / H* formation …. i.e. independent of the R´s

An alternative way to analyse the angular distribution data is by the procedure given by Chichinin et al.: -which involves the use of relative intensities of spectral lines I Q /I S and I Q /I O which could be derived from our REMPI spectra: i f ph HBr(Ji) HBr** H* + Br/Br*

1. Determine „b“ factor, via mass resolved REMPI spectra (see p: 9 ) from, 2. Determine alignment parameter A 20 (<=„b“ ) (P: 9, ) (43)(for J f = 1) (24) (25)

4. Determine the angular distribution for one-photon photodissociation of the unpolarized „f“ state (i.e.  (n,n ph )) from for k = 1 (one photon), where  (n) is the photofragment angular distribution produced by a multiphoton excitation via the intermediate state (i.e. angular distribution derived from our experiments) (see p: 6 in ) (1) ph 3. Determine angular distribution (  f (n)) via  –factor (  (f) ) determination for„f“(see slide 10): (P:6 (and 9), ) (30) See: ;J f = 1

1. Determine „b“ factor, via mass resolved REMPI spectra (see p: 9 ) from, Detailed analysis, see : agust,heima, …./PPT pptx a´ c´ d´ a´´ c´´ d´´

HBr; E(0) angular distribution analysis; Areas a´c´d´a"c"d" b**2 IO IQIS IQ/IOIQ/IS IQ/IO SO J´´LorentzGaussLorentz GaussLorentz J´(S)LorentzJ´(O) , , , ,735833, ,50,21, ,183454, , , ,666671,50, , , ,777844, ,250,1211,666671, , , , , , ,20, ,750, , , , ,168845, , , , ,80, , , , ,739515, , , , , , , , , ,13895, ,1250, , , , , , , , , , , ,8750, , , , ,10, , , , , , , , , ,90, , , , , , ,096 OVERLAP av b**21, , values with1, values V(m+4)av(av) 3 values1, ,19 av1,278447S(J´=2-8) ATH Gaussian fit to Q linesav1,159578O(J´=1-3) ERGO, (b**2) ca.:1,2 b: 1,095445or-1,

2. Determine alignment parameter A 20 (<=„b“ ) (P: 9, ) (43)(for J f = 1) (24) (25)

Alignment parameter A 20 : p:319 b A 20 (J f =1) A 20 (min) = -1 A 20 (max) =+0.5 b=1 Perpend. domin. b=-0.5 parallel. domin. Gr0,Lay0 <=

Alignment parameter A 20 : p:319 b factor A 20 (J f ) A 20 (min;J f =1) = -1 A 20 (max;J f =1) =+0.5 b = 1 Perpendicular dominant b=-0.5 parallel dominant J f = J f = Gr0,Lay0 <= Our A 20 values = A 20 (b = 1.1; b 2 = 1.2)

f <-<- i transition dominantly perpendicular transition, i.e.  3. Determine angular distribution (  f (n)) via  –factor (  (f) ) determination for„f“(see slide 10): (P:6 (and 9), ) (30) See:  (f) = -0, ;J f = 1

Gr1,Lay1<=   f (n; J f =1)

J´> 1: i.e.: ( -but to a first approximation (?)  L = 0 for L = 4,6,…. ERGO: the angular distribution shown on slide 18 holds for all J f ´s ! -in which case the alteration in angular distribution vs. J observed (Slide 2) is due to the photofragmentation step(?!)

Now what?! How do we derive  (n,n ph ) ? (1) ph How about a „fitting procedure“:

i.e. something like:  f i ph f Experiment  (ph) (1) + A ~ -for red coloured parameters unknown => derive  (ph)