English version   Atommag szimmetria

Rohán János

Az atommag felépítése proton és elektron tóruszokból

Az izotópok rugalmas gyűrűs atommag szerkezete, atommag szimmetria, bomlás mechanizmusa


A színek jelentése:   1  ,   1  ,   1  ,   1  ,   1  
Sor-
szám
Rend-
szám
Neutron
szám
Vegyjel m t1/2 rendszám = pep hamburger + proton szimmetria
  1   1   0n H
.ooOOoo.
1p0nH199.985%stable1/2+0 = p - (1n)Proton  
1stable 0 = pp
  2     1n n
1nn613.9 s1/2+Neutron  
1613.9 s nn = pe
  3   1   1n D
1p1nH20.015%stable1+1 = 1 + (0n)Deutérium
2stable1 = 1D = pep
  4   2   1n He
2p1nHe30.000137 %stable1/2+2 = p+1 - (1n)
3stable2 = p+1He3 = pnp
  5   1   2n T
1p2nH3e em12.32 y1/2+1 = 1 + (1n)Trícium
312.32 y1 = 1T = pe p ep
  6   3   1n Li
3p1nLip em91 yoctos2-3 = 2p+1 - (2n)
491 ys3 = 2p+1Li4 = p D p
  7   2   2n He
2p2nHe499.999863%stable0+2 = 2
4stable2 = 2He4 = 2 D
  8   1   3n H
1p3nH4n em139 ys2-1 = 1 + (2n)
4139 ys1 = 1H4 = pep e pep
  9   3   2n Li
3p2nLi5p+He em370 ys3/2-3 = 1p+2 - (1n)
5370 ys3 = 1p+2Li5  = D p D
10   2   3n He
2p3nHe5n em7x10-22 s3/2-2 = 1+1 + (1n)
50.7 zs2 = 1+1He5 = D n D
11   1   4n H
1p4nH52n em>910 ys1/2+1 = 1 + (3n)
5>910 ys1 = 1H5 = pepe p epep
12   4   2n Be
4p2nBe6He em5x10-21 s0+4 = 2p+2 - (2n)
65 zs4 = 2p+2Be6 = 2 He3
13   3   3n Li
3p3nLi67.59%stable1+3 = 1+2
6stable3 = 1+2Li6  = 3 D
14   2   4n He
2p4nHe6e em806.7 ms0+2 = 1+1 + (2n)
6806.7 ms2 = 1+1He6 = D 2n D = (T T)
15   1   5n H
1p5nH63n em290 yoctos2-#1 = 1 + (4n)
6290 ys1 = 1H6 = pepep e pepep
16   5   2n B
5p2nB7p em0.32 zeps3/2-5 = 3p+2 - (3n)
70.32 zs5 = 3p+2B7 = He3 p He3
17   4   3n Be
4p3nBe7Ke53.22 d3/2-4 = 1p+3 - (1n)
753.22 d4 = 1p+3Be7 = D He3 D
18   3   4n Li
3p4nLi792.41%stable3/2-3 = 1+2 + (1n)
7stable3 = 1+2Li7  = D T D
19   2   5n He
2p5nHe7n em2.9x10-21 s3/2-2 = 1+1 + (3n)
72.9 zs2 = 1+1He7 = D 3n D = (T n T)
20   6   2n C
6p2nC82p em1.98x10-21 s0+6 = 4p+2 - (4n)
82 zs6 = 4p+2C8  = He3 2p He3
21   5   3n B
5p3nB8b+770 ms2+5 = 2p+3 - (2n)
8770 ms5 = 2p+3B8  = He3 D He3
22   4   4n Be
4p4nBe82He em6.7x10-17 s0+4 = 2+2
867 as4 = 2+2Be8 = 2 He4 = 4 D
23   3   5n Li
3p5nLi8e em840.3 ms2+3 = 2+1 + (2n)
8840.3 ms3 = 2+1Li8 = D n D n D = T D T
24   2   6n He
2p6nHe8e em119.0 ms0+2 = 2 + (4n)
8119 ms2 = 1+1He8 = T 2n T = (D 4n D)
25   6   3n C
6p3nC9b+126.5 ms3/2-6 = 3p+3 - (3n)
9126.5 ms6 = 3p+3C9  = 3 He3
26   5   4n B
5p4nB9p+2He em8x10-19 s3/2-5 = 1p+4 - (1n)
9800 zs5 = 1p+4B9 = He4 p He4
27   4   5n Be
4p5nBe9100%stable3/2-4 = 1+3 + (1n)
9stable4 = 1+3Be9 = He4 n He4
28   3   6n Li
3p6nLi9e em178.3 ms3/2-3 = 1+2 + (3n)
9178.3 ms3 = 1+2Li9  = T T T
29   2   7n He
2p7nHe9n em7 zs1/2-#2 = 1+1 + (6n)
97 zs2 = 1+1He9 = T 3n T
30   7   3n N
7p3nN10p em200 yoctos2-7 = 4p+3 - (4n)
10200 ys7 = 4p+3N10 = He3 p D p He3
31   6   4n C
6p4nC10b+19.290 s0+6 = 2p+4 - (2n)
1019.290 s6 = 2p+4C10 = He3 He4 He3
32   5   5n B
5p5nB1019.8 %stable3+5 = 3+2
10stable5 = 2+1+2B10 = 5 D = He4 D He4
33   4   6n Be
4p6nBe10e em1.51x10+6 y0+4 = 2+2 + (2n)
101.51x10+6 y4 = 2+2Be10 = T He4 T
34   3   7n Li
3p7nLi10n em2.0 zs1+3 = 1+2 + (4n)
102 zs3 = 1+2Li10 = T H4 T
35   2   8n He
2p8nHe102n em2.68 zeps0+2 = 2 + (6n)
102.68 zs2 = 2He10 = H5 H5 = T 4n T
36   7   4n N
7p4nN11p em590 ys1/2+7 = 3p+4 - (3n)
11590 ys7 = 3p+4N11 = D p D p D p D = He3 Li5 He3
37   6   5n C
6p5nC11b+20.334 min3/2-6 = 1p+5 - (1n)
1120.334 min6 = 1p+5C11 = He4 He3 He4 = (D D p n p D D)
38   5   6n B
5p6nB1180.2 %stable3/2-5 = 1+4 + (1n)
11stable5 = 1+4B11 = He4 T He4
39   4   7n Be
4p7nBe11e em13.81 s1/2+4 = 3+1 + (3n)
1113.81 s4 = 3+1Be11 = T D n D T = D n D n D n D
40   3   8n Li
3p8nLi11e em8.59 ms3/2-3 = 2+1 + (5n)
118.59 ms3 = 2+1Li11 = H4 T H4 = T n T n T
41   8   4n O
8p4nO122p em580x10-24 s0+8 = 4p+4 - (4n)
12580 ys8 = 4p+4O12 = 4 He3
42   7   5n N
7p5nN12b+11.000 ms1+7 = 2p+5 - (2n)
1211 ms7 = 2p+5N12 = D D p D p D D
43   6   6n C
6p6nC1298.89%stable0+6 = 3+3
12stable6 = 3+3C12 = 3 He4 = 6 D = 2 Li6
44   5   7n B
5p7nB12e em, 3He em20.20 ms1+5 = 2+3 + (2n)
1220.20 ms5 = 2+3B12 = D T D T D
45   4   8n Be
4p8nBe12e em21.49 ms0+4 = 2+2 + (4n)
1221.49 ms4 = 2+2Be12 = T T T T
46   3   9n Li
3p9nLi12n em<10 ns3 = 2+1 + (6n)
12<10 ns3 = 2+1Li12 = H4 H4 H4
47   8   5n O
8p5nO13b+8.58 ms3/2-8 = 3p+5 - (3n)
138.58 ms8 = 3p+5O13 = He3 D He3 D He3
48   7   6n N
7p6nN13b+9.965 min1/2-7 = 1p+6 - (1n)
139.965 min7 = 1p+6N13 = D D D p D D D = Li6 p Li6
49   6   7n C
6p7nC131.11%stable1/2-6 = 1+5 + (1n)
13stable6 = 1+5C13 = 3 D n 3 D = Li6 n Li6
50   5   8n B
5p8nB13e em17.33 ms3/2-5 = 3+2 + (3n)
1317.33 ms5 = 3+2B13 = T D T D T
51   4   9n Be
4p9nBe13n em0.5 ns1/2+4 = 1+3 + (5n)
130.5 ns4 = 1+3Be13 = T T n T T
52   9   5n F
9p5nF14p emnodata2-9 = 4p+5 - (4n)
14nodata9 = 4p+5F14 = He3 D Li4 D He3
53   8   6n O
8p6nO14b+70.606 s0+8 = 2p+6 - (2n)
1470.606 s8 = 2p+6O14 = D He3 D D He3 D = Be7 Be7
54   7   7n N
7p7nN1499.634%stable1+7 = 4+3
14stable7 = 4+3N14 = 7 D = Li6 D Li6
55   6   8n C
6p8nC14e em5700 y0+6 = 2+4 + (2n)
145700 y6 = 2+4C14 = T D D D D T = D T D D T D
56   5   9n B
5p9nB14e em12.5 ms2-5 = 4+1 + (4n)
1412.5 ms5 = 4+1B14 = T T D T T
57   4   10n Be
4p10nBe14e + n em4.84 ms0+4 = 2+2 + (6n)
144.84 ms4 = 2+2Be14 = H4 T T H4
58   9   6n F
9p6nF15p em410 ys1/2+9 = 3p+6 - (3n)
15410 ys9 = 3p+6F15 = He3 D p n p n p D He3
59   8   7n O
8p7nO15b+122.24 s1/2-8 = 1p+7 - (1n)
15122.24 s8 = 1p+7O15 = D D D He3 D D D = Be7 n Be7
60   7   8n N
7p8nN150.366%stable1/2-7 = 1+6 + (1n)
15stable7 = 1+6N15 = D D D T D D D
61   6   9n C
6p9nC15e em2.449 s1/2+6 = 3+3 + (3n)
152.449 s6 = 3+3C15 = He5 He5 He5
62   5   10n B
5p10nB15e em9.87 ms3/2-5 = 3+2 + (5n)
159.87 ms5 = 3+2B15 = T T T T T
63   4   11n Be
4p11nBe15n em<200 ns4 = 3+1 + (7n)
15<200 ns4 = 3+1Be15 = H4 T n T H4
64   10   6n Ne
10p6nNe162p em3.74 zeps0+10 = 4p+6 - (4n)
163.74 zs10 = 4p+6Ne16 = He3 D He3 He3 D He3 = 2 B8
65   9   7n F
9p7nF16p em11.42 zeps0-9 = 2p+7 - (2n)
1611.42 zs9 = 2p+7F16 = He4 He3 D He3 He4
66   8   8n O
8p8nO1699.762%stable0+8 = 4+4
16stable8 = 4+4O16 = 4 He4 = 8 D
67   7   9n N
7p9nN16e em7.13 s2-7 = 2+5 + (2n)
167.13 s7 = 2+5N16 = T D D D D D T = T B10 T
68   6   10n C
6p10nC16e em0.747 s0+6 = 4+2 + (4n)
160.747 s6 = 4+2C16 = T T He4 T T
69   5   11n B
5p11nB16n em<190 pics0-5 = 1+4 + (6n)
16<190 pics5 = 1+4B16 = T T H4 T T
70   4   12n Be
4p12nBe162n em<200 ns0+4 = 2+2 + (8n)
16<200 ns4 = 2+2Be16 = H4 H4 H4 H4
71   10   7n Ne
10p7nNe17b+109.2 ms1/2-10 = 3p+7 - (3n)
17109.2 ms10 = 3p+7Ne17 = He3 He4 He3 He4 He3
72   9   8n F
9p8nF17b+64.49 s5/2+9 = 1p+8 - (1n)
1764.49 s9 = 1p+8F17 = D D D D p D D D D
73   8   9n O
8p9nO17  0.038%stable5/2+8 = 1+7 + (1n)
17stable8 = 1+7O17 = He4 He4 n He4 He4
74   7   10n N
7p10nN17e em4.173 s1/2-7 = 3+4 + (3n)
174.173 s7 = 3+4N17 = T He4 T He4 T
75   6   11n C
6p11nC17e em193 ms3/2+6 = 5+1 + (5n)
17193 ms6 = 5+1C17 = T D T n T D T
76   5   12n B
5p12nB17e em5.08 ms3/2-5 = 2+3 + (7n)
175.08 ms5 = 2+3B17 = T H4 T H4 T
77   11   7n Na
11p7nNa18b+1.3 zeps1-11 = 4p+7 - (4n)
181.3 zs11 = 4p+7Na18 = He3 D He3 D He3 D He3
78   10   8n Ne
10p8nNe18b+1672 ms0+10 = 2p+8 - (2n)
181672 ms10 = 2p+8Ne18 = He4 He3 He4 He3 He4
79   9   9n F
9p9nF18b+109.77 min1+9 = 5+4
18109.77 min9 = 5+4F18 = 9 D
80   8   10n O
8p10nO18  0.200%stable0+8 = 2+6 + (2n)
18stable8 = 2+6O18 = 2 Be9
81   7   11n N
7p11nN18e em622 ms1-7 = 4+3 + (4n)
18622 ms7 = 4+3N18 = T D T D T D T = Li8 D Li8
82   6   12n C
6p12nC18e em92 ms0+6 = 3+3 + (6n)
1892 ms6 = 3+3C18 = 6 T = 2 Li9
83   5   13n B
5p13nB18n em<26 ns4-#5 = 3+2 + (8n)
18<26 ns5 = 3+2B18 = T n T H4 T n T = He7 H4 He7
84   12   7n Mg
12p7nMg192p emnodata1/2-#12 = 5p+7 - (5n)
19nodata12 = 5p+7Mg19 = He3 D He3 He3 He3 D He3 = B8 He3 B8
86   10   9n Ne
10p9nNe19b+17.22 s1/2+10 = 1p+3 + 6 - (1n)
1917.22 s10 = 1p+3 + 6Ne19 = Be7 6 D
87   9   10n F
9p10nF19100%stable1/2+9 = 1+8 + (1n)
19stable9 = 1+2 + 6F19 = Li7 6D (Be9 p Be9 ??)
88   8   11n O
8p11nO19e em26.464 s5/2+8 = 3+5 + (3n)
1926.464 s8 = 3+5O19 = Be9 n Be9 = He4 n He4 n He4 n He4
89   7   12n N
7p12nN19e em271 ms1/2-7 = 5+2 + (5n)
19271 ms7 = 5+2N19 = T T Li7 T T
90   6   13n C
6p13nC19e em46.2 ms1/2+6 = 1+5 + (7n)
1946.2 ms6 = 1+5C19 = T T T n T T T = Li9 n Li9
91   5   14n B
5p14nB19e em2.92 ms3/2-5 = 4+1 + (9n)
192.92 ms5 = 4+1B19 = T n T n T n T n T = T n Li11 n T

A szimmetria nem abszolút, mert a szerkezet rugalmas. A magelektronok ragasztó szerepet töltenek be, gluonok nem léteznek.

A magelektronok egymást taszitó hatása nagyobb mint a protonokat összeragasztó hatás, mert az elektronok közelebb kerülhetnek egymáshoz mint a protonokhoz, a protonok kisebb átmérője miatt. Ezért nem is létezhet negativ atommag ahol protonok ragasztanának össze elektronokat. Vagy nem.

J.Lucas, Atommag modell         Stabil izotópok,       Nuclear properties         Neutronok nem léteznek
Foton és anyagmodell
 

Az atommag szerkezete

Problémák a relativitáselmélet körül

 

Vita 1,   Vita 2,   Vita 3,   Vita 4,         Alias rigó fórum:   OF   OF2   OF3     és       Tudomány cikkek archiv       SG   SG2       Szkept       Hypog       Csivar

Az oxigénnél kisebb atommagok hengeres csöves szerkezete felépítése

Irj a vendégkönyvembe! János Irj a vendégkönyvembe!
janos@biochem.szote.u-szeged.hu


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