HBr, angular distribution analysis

E(0)

Updated, 24.12.2013

2.0

1.5

1.0

0.5

0.0

-0.5

-1.0

8642

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 hn resonance excitation

3 hn dissociation / H* formation

4 hn ionization / H+ formation

2 hn resonance excitation

3 hn dissociation / H* formation

According tohttps://notendur.hi.is/~agust/rannsoknir/papers/jcp121-11802-04.pdf :

where

Unknown(variable in a fit)

Unknown(variables in a fit)

BUT simpler for “non Q branches” (O, S) (i.e. for J´´ ¹ J´):

2 hn resonance excitation

3 hn dissociation / H* formation

https://notendur.hi.is/~agust/rannsoknir/papers/jcp121-11802-04.pdf

…. i.e. independent of the R´s

An alternative way to analyse the angular distribution data is bythe procedure given by Chichinin et al.:https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf -which involves the use of relative intensities of spectral lines IQ/IS and IQ/IO

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 https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf ) from,

2. Determine alignment parameter A20 (<=„b“ )(P: 9, https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf )

(43)(for Jf = 1)

(24)

(25)

4. Determine the angular distribution for one-photon photodissociation of the unpolarized„f“ state (i.e. w (n,nph)) from

for k = 1 (one photon), where w (n) is the photofragment angular distribution produced by amultiphoton excitation via the intermediate state (i.e. angular distribution derived from our experiments) (see p: 6 in https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf )

(1)ph

3. Determine angular distribution (wf(n)) via b –factor (b (f)) determination for„f“(see slide 10): (P:6 (and 9), https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf )

(30)

See: http://mathworld.wolfram.com/LegendrePolynomial.html

;Jf = 1

1. Determine „b“ factor, via mass resolved REMPI spectra (see p: 9 https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf ) from,

Detailed analysis, see : agust,heima, …./PPT-131219.pptx

a´ c´ d´

a´´ c´´ d´´

HBr; E(0) angular distribution analysis; Areas a´ c´ d´ a" c" d" b**2 b**2

IO IO IQ IS IS IQ/IO IQ/IS IQ/IS IQ/IS IQ/IS IQ/IO IQ/IO IQ/IO S O

J´´ Lorentz Gauss Lorentz Lorentz Gauss Lorentz Lorentz Lorentz Lorentz Lorentz Lorentz Lorentz Lorentz Lorentz J´(S) Lorentz J´(O)

0 73217 10297 7294 7,110518 1,666667 3 0 1,422104 2

1 263790 24571 18460 10,73583 3,333333 2,5 0,2 1,208299 3

2 427040 38185 28692 11,18345 4,166667 2,333333 0,133333 16,66667 1,5 0,085714 1,093155 4

3 23416 16781 547620 42857 33308 23,38657 12,77784 4,666667 2,25 0,12 11,66667 1,666667 0,088889 1,163604 5 1,149405 1

4 27174 17033 623810 44897 34919 22,95613 13,89425 5 2,2 0,114286 10 1,75 0,090909 1,211165 6 1,259831 2

5 32831 21523 607140 37550 27657 18,49289 16,16884 5,238095 2,166667 0,111111 9,166667 1,8 0,092308 1,373385 7 1,069499 3

6 43837 28994 558280 31471 16560 12,73536 17,73951 5,416667 2,142857 0,109091 8,666667 1,833333 0,093333 1,477418 8 0,750617 4

7 58895 41854 447480 2555 1797 7,597929 175,1389 5,555556 2,125 0,107692 8,333333 1,857143 0,094118 14,78462 9 0,440264 5

8 53011 41966 281260 5,305691 5,666667 2,111111 0,106667 8,095238 1,875 0,094737 0,299025 6

9 37166 30645 69388 1,866975 5,757576 2,1 0,105882 7,916667 1,888889 0,095238 0,07443 7

10 13743 11193 5,833333 2,090909 0,105263 7,777778 1,9 0,095652

11 2882 2019 5,897436 2,083333 0,104762 7,666667 1,909091 0,096

OVERLAP

OVERLAP av b**2 1,210327 1,057338 4 values

with 1,159578 3 values

V(m+4) av(av) 3 values 1,188578

1,19

av 1,278447 S(J´=2-8)

ATH Gaussian fit to Q lines av 1,159578 O(J´=1-3)

ERGO, (b**2) ca.: 1,2 b: 1,095445or -1,09545

https://notendur.hi.is/~agust/rannsoknir/Crete/XLS-131221.xlsx

2. Determine alignment parameter A20 (<=„b“ )(P: 9, https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf )

(43)(for Jf = 1)

(24)

(25)

Alignment parameter A20:http://www.ejournal.unam.mx/rmf/no503/RMF50315.pdf , p:319

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

2.01.51.00.50.0-0.5 b

A20(Jf=1)

A20(min) = -1

A20(max) =+0.5

b=1Perpend. domin.

b=-0.5parallel. domin.

https://notendur.hi.is/~agust/rannsoknir/Crete/PXP-131222.pxp; Gr0,Lay0 <= https://notendur.hi.is/~agust/rannsoknir/Crete/XLS-131221.xlsx

-1.0

-0.5

0.0

0.5

2.01.51.00.50.0-0.5

Alignment parameter A20:http://www.ejournal.unam.mx/rmf/no503/RMF50315.pdf , p:319

b factor

A20(Jf)

A20(min;Jf=1) = -1

A20(max;Jf=1) =+0.5

b = 1Perpendicular dominant

b=-0.5parallel dominant

Jf= 1

2 11

Jf =

11 2 1

https://notendur.hi.is/~agust/rannsoknir/Crete/PXP-131222.pxp; Gr0,Lay0 <= https://notendur.hi.is/~agust/rannsoknir/Crete/XLS-131221.xlsx

Our A20 values =A20 (b = 1.1; b2 = 1.2)

f <-<- i transition dominantly perpendicular transition,i.e. S <- P <- S

3. Determine angular distribution (wf(n)) via b –factor (b (f)) determination for„f“(see slide 10): (P:6 (and 9), https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf )

(30)

See: http://mathworld.wolfram.com/LegendrePolynomial.html

b (f) = -0,62215

https://notendur.hi.is/~agust/rannsoknir/Crete/XLS-131221.xlsx

;Jf = 1

0.10

0.09

0.08

0.07

0.06

0.05

0.04

0.03

150100500

https://notendur.hi.is/~agust/rannsoknir/Crete/PXP-131222.pxp; Gr1,Lay1<= https://notendur.hi.is/~agust/rannsoknir/Crete/XLS-131221.xlsx

q

wf(n; Jf =1)

J´> 1:

https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf:

i.e.:

(https://notendur.hi.is/~agust/rannsoknir/papers/jcp121-11802-04.pdf)

-but to a first approximation (?) bL = 0 for L = 4,6,….

ERGO: the angular distribution shown on slide 18 holds for all Jf´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 w (n,nph) ?

(1)ph

How about a „fitting procedure“:

i.e. something like:

q0 180

f i

ph f

Experiment

b(ph)

(1) + A~

-for red coloured parameters unknown => derive b(ph)