" You need to work out additional material and add or change this form in order to complete this work!!!!

number of degenrate states;

"1eVis 1,60219 10^-19 J is 96.458 kJ mol-1  is 8065.5 cm-1;;;;;;";

" k is  1.38066 10^-23 J K-1;;;;";

"R is k Na  is  8.31441 K-1 mol-1";

Clear[k,R,m]

s

K

k=1.38066  10^-23  kg m^2  s^-2     K^-1

m

hh=6.626 10^-34   kg m^2  s^-1

cc=2.99799 10^8  m/s

"This is conversion factor   in J *m  or if energy is in m-1,it is J";

hh  cc

"so if you have cm-1 \!\(\*FractionBox[\(1.9864681740000004`*^-23\\ kg\\ \*SuperscriptBox[\(m\), \(2\)]\\ \\ \\ cm\), SuperscriptBox[\(s\), \(2\)]]\) "

"convert energy kT <--> hcB  from  J<-->cm-1  "

T=298 K

k T/(hh cc)

or 207 cm-1

"SO CONVERSION FACTOR /K is  "

k T/(hh cc)

or  0.695033 cm-1

T=1  K

1/.695033

298 K *0.6950

"END OF SI UNITS";

ROTATIONAL BOLTZMANN

g=Table[2  j +1,{j,0,15}]

use  B== hcB in cm-1 and T==kT/hc in cm-1;

for T=298.15 stand room temp T=207.22cm-1;

298.15 * .695033

tt=Table[ i*10*.695033,{i,0,20}]

"HCl B=10 check;";

B=10.59

bt[j_]:=2 j

bt[2]

------------------------------------------------------------------

B=10.59

Table[B  j ( j+1),{j,0,20}]

Plot[{0.`,21.18`,63.54`,127.08`,211.8`,317.70000000000005`,444.78`,593.04`,762.48`,953.1`,1164.9`,1397.8799999999999`,1652.04`,1927.3799999999999`,2223.8999999999996`,2541.6`,2880.48`,3240.54`,3621.78`,4024.2000000000003`,4447.8`},{x,0,2}]

energy

---------------------------------------------------------------------------

Bol[j_,t_]:=(2  j +1) *Exp[-B  j ( j+1) / (t*0.6950)]

Bol[2.,290]

Plot[Evaluate[Table[Bol[j,t],{j,1,12}]],{t,290,300}]

"note energy of j=0 is zero, start with 1";

Plot[Evaluate[Table[Bol[j,t],{j,1,12}]],{t,1,300},AxesLabel->Automatic]

Tabley=Table[Bol[j,300.],{j,1,20}]

ListPlot[Tabley,PlotJoined->True]

Plot[Evaluate[Table[Bol[j,t],{t,1,300,50}]],{j,1,20},AxesLabel->Automatic,PlotRange->Full,AxesOrigin->{0,0}]

How to normalize?

Question - how would the normalized distribution look like - show for a range of temperatures.
What is the conlculsion aboyt the distributions

Plot[Evaluate[Table[Bol[j,t]/Bol[1,t],{t,1,300,50}]],{j,1,20},AxesLabel->Automatic,AxesOrigin->{0,0},PlotRange->All]

"IF MOLECULE SURVIVES AND IS NOT coupled with other degries";

"experimnetal data";

"assigned m now goes to j";

ListPlot[{{2571.04`,63.025`},{2598.04`,60.7376`},{2624.66`,56.5307`},{2651.08`,50.6777`},{2676.72`,42.7661`},{2701.99`,33.696`},{2726.67`,24.567`},{2750.97`,17.7626`},{2774.88`,12.4204`},{2798.03`,9.4842`},{2820.59`,8.4309`},{2842.57`,8.6026`},{2863.79`,11.4713`},{2905.25`,11.4448`},{2924.92`,7.5972`},{2943.82`,6.0691`},{2962.14`,6.1261`},{2979.69`,7.4544`},{2996.66`,10.3977`},{3012.86`,15.1432`},{3028.67`,21.5899`},{3043.52`,28.7497`},{3057.98`,35.8124`},{3071.48`,41.6841`},{3084.21`,46.1068`},{3095.98`,49.3055`},{3131.85`,53.1381`},{3133.77`,53.2278`},{3135.7`,53.2859`},{3137.63`,53.3374`}}]

"put it into "absorbance shape";

"concentratikons here are low - assume one layer";

"correct P";

hclP={{13,64-63.025`},{12,64-60.7376`},{11,64-56.5307`},{10,64-50.6777`},{9,64-42.7661`},{8,64-33.696`},{7,64-24.567`},{6,64-17.7626`},{5,64-12.4204`},{4,64-9.4842`},{3,64-8.4309`},{2,64-8.6026`},{1,64-11.4713`}}

ListPlot[hclP]

"correct R";

54-11.4448

hclR={{0,(54-11.4448`)},{1,(54-7.5972`)},{2,(54-6.0691`)},{3,(54-6.1261`)},{4,(54-7.4544`)},{5,(54-10.3977`)},{6,(54-15.1432`)},{7,(54-21.5899`)},{8,(54-28.7497`)},{9,(54-35.8124`)},{10,(54-41.6841`)},{11,(54-46.1068`)},{12,(54-49.3055`)},{13,(54-53.1381`)},{14,(54-53.2278`)},{15,(54-53.2859`)}}

gPex=ListPlot[{hclP},PlotJoined->True]

ListPlot[{hclP,hclR}]

(2  j +1) *Exp[-B  j ( j+1) / (t*0.6950)]

hclR

"FINF FIT";

hclP

"you can normalize ,Boltzman is the ratio between pop[ulations that will not change th efit";

hclP0={{13,0.9750000000000014`/52.5287`},{12,3.2623999999999995`/52.5287`},{11,7.469299999999997`/52.5287`},{10,13.322299999999998`/52.5287`},{9,21.2339`/52.5287`},{8,30.304000000000002`/52.5287`},{7,39.433`/52.5287`},{6,46.2374`/52.5287`},{5,51.5796`/52.5287`},{4,54.5158`/52.5287`},{3,55.5691`/52.5287`},{2,55.3974`/52.5287`},{1,52.5287`/52.5287`}}

"norm with correct T:in order to get the fit closer to the experimental numerical domain and make it easier to search,  you can go back and open that one two, not affect result";

if yoiu use j=1 norm;

(2  1 + 1) *Exp[-B  1 ( 1 + 1) / (289*0.6950)]

Find the temperature  t from th efit function here:

FindFit[DATAFILE, {a + (1/NORM)* Function(j, t)},{a,t}, j]