PERIODIC PROGRESSION OF WAVE LENGTH ENERGY
Copyright 1993, 1997, 2001, 2002
by Gerald E. Mowery
No claim for government or public domain information
007***Li*****.0543****6.0000000
*************************5.988960*********6.9100*Li
009***Be****.1085****11.988950
*************************8.265194*********9.0122*Be
011***B*****.1833****20.254144
************************10.353591*******10.8110**B
013***C*****.2770***30.607726
************************12.784530*******12.0107**C
015***N*****.0393***43.392265
************************14.607730*******14.0067**N
017***O*****.5249***58.000000
************************16.784530*******
15.9994**O
019***F*****.6768***74.784530
************************18.983426*******18.9984**F
021***Ne****.8486***93.767956
************************21.259668*******20.1830*Ne
023***Na***1.0410***115.027624
************************23.491713*******22.9898*Na
025***Mg***1.1253***138.519337
************************25.709392*******24.3120*Mg
027***Al***1.4863***164.2287292
************************27.967950*******26.9815*Al
029***Si***1.7394***192.196685

************************30.201105*******
28.0860**Si
031***P****2.0127***222.397790

************************32.479558********
30.9738**P
033***S****2.30664***254.877348
************************34.711602********32.0640**S
035***Cl****2.6208***289.588950
************************37.000000********35.4530*Cl
037***Ar****2.9556***326.58895
************************41.407304********39.9480*Ar
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Kev Value of K wave Length
-
--------------------------------x 6 = Atomic Mass
Divided by .0543
OBTAINING A DIFFERENCE BY APPLYING THE EQUATION TO   TWO ADJACENT ELEMENTS AND SUBTRACTING "ONE FRM THE OTHER" PREDICTS THE ATOMIC MASS OF THE HEAVIER ELEMENT.
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