Dany jest (z "gory") 50-cio elementowy , uporzadkowany zbior C-"czastek fundamentalnych":
C = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 30 , 31 , 32 , 33 , 34 , 35 , 36 , 37 , 38 , 39 , 40 , 41 , 42 , 43 , 44 , 45 , 46 , 47 , 48 , 49 , 50}
Z elementow zbioru C (z "czastek fund.") kreujemy 5-cio "czastkowe" "wzory-atomy" w ten sposob ze, w "atomie" kazda "czastka fund." jest rozna od pozostalych.
Oto uporzadkowany "narastajaco" zbior A takowych "wzorow-atomow".
A={ [1,2,3,4,5] ; [1,2,3,4,6] ; [1,2,3,4,7] ;
; ........ pomiedzy reszta "wzorow-atomow" ..... ;
[44,47,48,49,50] ; [45,47,48,49,50] ; [46,47,48,49,50] }
"Najlzejszy" i "najmniejszy"-"wzor-atom" zbioru A czyli :
a(1)=[1,2,3,4,5] ma "mase"= sumie swoich "czastek fund.":
Ma(1)=1+2+3+4+5=15
oraz "objetosc" = iloczynowi" czastek fund.":
Oa(1)=1x2x3x4x5=120
"Najciezszy" i "najwiekszy"-ostatni "wzor-atom" zbioru A czyli a(N)=[46,47,48,49,50] ma "mase":
Ma(N)=46+47+48+49+50=240
oraz "objetosc":
Oa(N =46x47x48x49x50=254 251 200
-gdzie N jest nieznana narazie liczba okreslajaca numer ostaniego "wzoru-atomu" zbioru A czyli liczba N jest rowna ilosci wszystkich wykreowanych "wzorow-atomow" 5-cio "czastkowych".
Okazuje sie ze, liczba N jest rowna ilorazowi objetosci "najwiekszego" i "najmniejszego" "wzoru-atomu" czyli:
N= Oa(N) / Oa(1) = 254 251 200 / 120 =2 118 760
wiec ilosc wszystkich "wzorow-atomow" N=2 118 760
"Masa" najwiekszego "wzoru-atomu" jest 16 razy wieksza od "masy" najmniejszego-
Ma(N) / Ma(1) = 240/15 = 16
Majac wykreowane "wzory-atomy" mozemy przystapic do kreacji "czasteczek".
Do tego niezbedne beda "wiazania".Oto zalozony zbior "wiazan":
W={1, 2, 3, 4, 5, 6, 7, 8, 9}
Kazdy "atom-pierwiastek" dysponuje dwoma roznymi "wiazaniami" czyli ma do wyboru zbior par "wiazan" Wp={
(1,2);(1,3);(1,4);(1,5);(1,6);(1,7);(1,8);(1,9);
(2,3);(2,4);(2,5);(2,6);(2,7);(2,8);(2,9);
(3,4);(3,5);(3,6);(3,7);(3,8);(3,9);
(4,5);(4,6);(4,7);(4,8);(4,9);
(5,6);(5,7);(5,8);(5,9);
(6,7);(6,8);(6,9);
(7,8);(7,9);
(8,9) }
skladajacy sie z 36 elementow-par "wiazan".
Kazdy "wzor-atom" istnieje w 36 odmianach ze wzgledu na wybrane "wiazanie".Tak wiec ilosc wszystkich mozliwych "wzorow-atomow" wraz z "wiazaniami" wynosi:
36xN=36 x 2 118 760 =76 275 360
Majac "atomy-pierwiastki" oraz "wiazania" mozemy przystapic do kreacji "czasteczek".
"Czasteczek"11-sto "atomowych"-(zalozenie).
Ponizej pokazany jest schemat "wiazan miedzyatomowych".Oto przykladowo "wzor-atom"-
a(1)=[1,2,3,4,5] wraz z pierwsza para "wiazan"-
w(1)=(1,2)
a(1)w(1) = ~(1)~[1,2,3,4,5]~(2)~
Oto przykladowo najprostsza "czasteczka" C1-zlozona z 11 identycznych "atomow"- a(1)w(1) przedstawiona w konfiguracji liniowej-
11C1<a(1)w(1)>
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Jest to 11-sto "atomowa czasteczka najprostszego pierwiastka"- w najmniej zlozonej konfiguracji -lancuchowej.
Ponizej ta sama 11 "atomowa czasteczka"
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w konfiguracji zlozonej-cyklicznej.
I ta sama 11 "atomowa czasteczka"
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w konfiguracji zlozonej-krystalicznej.
Ilosc L roznych 11-sto "atomowych czasteczek pierwiastkow i zwiazkow" zgodnie ze wzorem:
k=11 , n=76 275 360
wynosi:
L=1,4010841082480147276444939760468e+80
Ta liczba odpowiada szacunkowej liczbie atomow w obserwowanym Wszechswiecie!
c.d.n.
Inne tematy w dziale Technologie