Sor- szám	Rend- szám	Neutron szám	Vegyjel	m	t1/2	rendszám = pep hamburger + (n többlet)	szimmetria
1	1	0n	H .ooOOoo. 1p 0n H1 99.985% stable 1/2+ 0 = p - (1n) Proton	1	stable	0 = p	p
2		1n	n 1n n 613.9 s 1/2+ Neutron	1	613.9 s	nNeutron = proton (kék) + elektron (piros)	n = pe
3	1	1n	D 1p 1n H2 0.015% stable 1+ 1 = 1 + (0n) Deutérium	2	stable	• 1 = 1	D
4	2	1n	He 2p 1n He3 0.000137 % stable 1/2+ 2 = p+1 - (1n)	3	stable	• 2 = p+1	He3 = pnp
5	1	2n	T 1p 2n H3 e em 12.32 y 1/2+ 1 = 1 + (1n) Trícium	3	12.32 y	• 1 = 1 + (1n)	T
6	3	1n	Li 3p 1n Li p em 91 yoctos 2- 3 = 2p+1 - (2n)	4	91 ys	• 3 = 2p+1 - (2n)	Li4 = p D p
7	2	2n	He 2p 2n He4 99.999863% stable 0+ 2 = 2	4	stable	• 2 = 2	He4 = 2 D
8	1	3n	H 1p 3n H4 n em 139 ys 2- 1 = 1 + (2n)	4	139 ys	• 1 = 1 + (2n)
9	3	2n	Li 3p 2n Li5 p+He em 370 ys 3/2- 3 = 1p+2 - (1n)	5	370 ys	• 3 = 1p+2 - (1n)	Li5  = D p D = (pnpnp)
10	2	3n	He 2p 3n He5 n em 7x10-22 s 3/2- 2 = 1+1 + (1n)	5	0.7 zs	2 = 1+1 + (1n)	He5 = D n D
11	1	4n	H 1p 4n H5 2n em >910 ys 1/2+ 1 = 1 + (3n)	5	>910 ys	• 1 = 1 + (3n)
12	4	2n	Be 4p 2n Be6 He em 5x10-21 s 0+ 4 = 2p+2 - (2n)	6	5 zs	• 4 = 2p+2 - (2n)	Be6 = 2 He3
13	3	3n	Li 3p 3n Li6 7.59% stable 1+ 3 = 1+2	6	stable	• 3 = 1+2	Li6  = 3 D
14	2	4n	He 2p 4n He6 e em 806.7 ms 0+ 2 = 1+1 + (2n)	6	806.7 ms	• 2 = 1+1 + (2n)	He6 = D 2n D = (T T)
15	1	5n	H 1p 5n H6 3n em 290 yoctos 2-# 1 = 1 + (4n)	6	290 ys	• 1 = 1 + (4n)
16	5	2n	B 5p 2n B7 p em 0.32 zeps 3/2- 5 = 3p+2 - (3n)	7	0.32 zs	• 5 = 3p+2 - (3n)	B7 = He3 p He3
17	4	3n	Be 4p 3n Be7 Ke 53.22 d 3/2- 4 = 1p+3 - (1n)	7	53.22 d	4 = 1p+3 - (1n)	Be7 = D He3 D = (pnpnpnp)
18	3	4n	Li 3p 4n Li7 92.41% stable 3/2- 3 = 1+2 + (1n)	7	stable	• 3 = 1+2 + (1n)	Li7  = D T D
19	2	5n	He 2p 5n He7 n em 2.9x10-21 s 3/2- 2 = 1+1 + (3n)	7	2.9 zs	2 = 1+1 + (3n)	He7 = D 3n D = (T n T)
20	6	2n	C 6p 2n C8 2p em 1.98x10-21 s 0+ 6 = 4p+2 - (4n)	8	2 zs	• 6 = 4p+2 - (4n)	C8  = He3 2p He3
21	5	3n	B 5p 3n B8 b+ 770 ms 2+ 5 = 2p+3 - (2n)	8	770 ms	• 5 = 2p+3 - (2n)	B8  = He3 D He3
22	4	4n	Be 4p 4n Be8 2He em 6.7x10-17 s 0+ 4 = 2+2	8	67 as	• 4 = 2+2	Be8 = 2 He4 = 4 D
23	3	5n	Li 3p 5n Li8 e em 840.3 ms 2+ 3 = 2+1 + (2n)	8	840.3 ms	• 3 = 2+1 + (2n)	Li8 = D n D n D = T D T
24	2	6n	He 2p 6n He8 e em 119.0 ms 0+ 2 = 2 + (4n)	8	119 ms	• 2 = 1+1 + (4n)	He8 = T 2n T = (D 4n D)
25	6	3n	C 6p 3n C9 b+ 126.5 ms 3/2- 6 = 3p+3 - (3n)	9	126.5 ms	6 = 3p+3 - (3n)	C9  = 3 He3
26	5	4n	B 5p 4n B9 p+2He em 8x10-19 s 3/2- 5 = 1p+4 - (1n)	9	800 zs	• 5 = 1p+4 - (1n)	B9 = He4 p He4 = pnpnpnpnp
27	4	5n	Be 4p 5n Be9 100% stable 3/2- 4 = 1+3 + (1n)	9	stable	4 = 1+3 + (1n)	Be9 = He4 n He4
28	3	6n	Li 3p 6n Li9 e em 178.3 ms 3/2- 3 = 1+2 + (3n)	9	178.3 ms	• 3 = 1+2 + (3n)	Li9  = T T T
29	2	7n	He 2p 7n He9 n em 7 zs 1/2-# 2 = 1+1 + (6n)	9	7 zs	2 = 1+1 + (6n)	He9 = T 3n T
30	7	3n	N 7p 3n N10 p em 200 yoctos 2- 7 = 4p+3 - (4n)	10	200 ys	• 7 = 4p+3 - (4n)	N10 = He3 p D p He3
31	6	4n	C 6p 4n C10 b+ 19.290 s 0+ 6 = 2p+4 - (2n)	10	19.290 s	• 6 = 2p+4 - (2n)	C10 = D p He4 p D
32	5	5n	B 5p 5n B10 19.8 % stable 3+ 5 = 3+2	10	stable	• 5 = 3+2	B10 = 5 D = He4 D He4
33	4	6n	Be 4p 6n Be10 e em 1.51x10+6 y 0+ 4 = 2+2 + (2n)	10	1.51x10+6 y	• 4 = 2+2 + (2n)	Be10 = T He4 T
34	3	7n	Li 3p 7n Li10 n em 2.0 zs 1+ 3 = 1+2 + (4n)	10	2 zs	• 3 = 1+2 + (4n)	Li10 = T H4 T
35	2	8n	He 2p 8n He10 2n em 2.68 zeps 0+ 2 = 2 + (6n)	10	2.68 zs	• 2 = 2 + (6n)	He10 = H5 H5 = T 4n T
36	7	4n	N 7p 4n N11 p em 590 ys 1/2+ 7 = 3p+4 - (3n)	11	590 ys	• 7 = 3p+4 - (3n)	N11 = D p D p D p D
37	6	5n	C 6p 5n C11 b+ 20.334 min 3/2- 6 = 1p+5 - (1n)	11	20.334 min	6 = 1p+5 - (1n)	C11 = He4 He3 He4 = (D D p n p D D)
38	5	6n	B 5p 6n B11 80.2 % stable 3/2- 5 = 1+4 + (1n)	11	stable	• 5 = 1+4 + (1n)	B11 = He4 T He4
39	4	7n	Be 4p 7n Be11 e em 13.81 s 1/2+ 4 = 3+1 + (3n)	11	13.81 s	4 = 3+1 + (3n)	Be11 = T D n D T
40	3	8n	Li 3p 8n Li11 e em 8.59 ms 3/2- 3 = 2+1 + (5n)	11	8.59 ms	• 3 = 2+1 + (5n)	Li11 = H4 T H4 = T n T n T
41	8	4n	O 8p 4n O12 2p em 580x10-24 s 0+ 8 = 4p+4 - (4n)	12	580 ys	• 8 = 4p+4 - (4n)	O12 = 4 He3
42	7	5n	N 7p 5n N12 b+ 11.000 ms 1+ 7 = 2p+5 - (2n)	12	11 ms	• 7 = 2p+5 - (2n)	N12 = D D p D p D D
43	6	6n	C 6p 6n C12 98.89% stable 0+ 6 = 3+3	12	stable	• 6 = 3+3	C12 = 3 He4 = 6 D = 2 Li6
44	5	7n	B 5p 7n B12 e em, 3He em 20.20 ms 1+ 5 = 2+3 + (2n)	12	20.20 ms	• 5 = 2+3 + (2n)	B12 = D T D T D
45	4	8n	Be 4p 8n Be12 e em 21.49 ms 0+ 4 = 2+2 + (4n)	12	21.49 ms	• 4 = 2+2 + (4n)	Be12 = T T T T
46	3	9n	Li 3p 9n Li12 n em <10 ns 3 = 2+1 + (6n)	12	<10 ns	• 3 = 2+1 + (6n)	Li12 = H4 H4 H4
47	8	5n	O 8p 5n O13 b+ 8.58 ms 3/2- 8 = 3p+5 - (3n)	13	8.58 ms	8 = 3p+5 - (3n)	O13 = He3 D He3 D He3
48	7	6n	N 7p 6n N13 b+ 9.965 min 1/2- 7 = 1p+6 - (1n)	13	9.965 min	• 7 = 1p+6 - (1n)	N13 = D D D p D D D = Li6 p Li6
49	6	7n	C 6p 7n C13 1.11% stable 1/2- 6 = 1+5 + (1n)	13	stable	6 = 1+5 + (1n)	C13 = 3 D n 3 D = Li6 n Li6
50	5	8n	B 5p 8n B13 e em 17.33 ms 3/2- 5 = 3+2 + (3n)	13	17.33 ms	• 5 = 3+2 + (3n)	B13 = T D T D T
51	4	9n	Be 4p 9n Be13 n em 0.5 ns 1/2+ 4 = 1+3 + (5n)	13	0.5 ns	4 = 1+3 + (5n)	Be13 = T T n T T
52	9	5n	F 9p 5n F14 p em nodata 2- 9 = 4p+5 - (4n)	14	nodata	• 9 = 4p+5 - (4n)	F14 = He3 D Li4 D He3
53	8	6n	O 8p 6n O14 b+ 70.606 s 0+ 8 = 2p+6 - (2n)	14	70.606 s	• 8 = 2p+6 - (2n)	O14 = D He3 D D He3 D = Be7 Be7
54	7	7n	N 7p 7n N14 99.634% stable 1+ 7 = 4+3	14	stable	• 7 = 4+3	N14 = 7 D = Li6 D Li6
55	6	8n	C 6p 8n C14 e em 5700 y 0+ 6 = 2+4 + (2n)	14	5700 y	• 6 = 2+4 + (2n)	C14 = T D D D D T = D T D D T D
56	5	9n	B 5p 9n B14 e em 12.5 ms 2- 5 = 4+1 + (4n)	14	12.5 ms	• 5 = 4+1 + (4n)	B14 = T T D T T
57	4	10n	Be 4p 10n Be14 e + n em 4.84 ms 0+ 4 = 2+2 + (6n)	14	4.84 ms	• 4 = 2+2 + (6n)	Be14 = H4 T T H4
58	9	6n	F 9p 6n F15 p em 410 ys 1/2+ 9 = 3p+6 - (3n)	15	410 ys	• 9 = 3p+6 - (3n)	F15 = He3 D p n p n p D He3
59	8	7n	O 8p 7n O15 b+ 122.24 s 1/2- 8 = 1p+7 - (1n)	15	122.24 s	8 = 1p+7 - (1n)	O15 = D D D He3 D D D = Be7 n Be7
60	7	8n	N 7p 8n N15 0.366% stable 1/2- 7 = 1+6 + (1n)	15	stable	• 7 = 1+6 + (1n)	N15 = D D D T D D D
61	6	9n	C 6p 9n C15 e em 2.449 s 1/2+ 6 = 3+3 + (3n)	15	2.449 s	6 = 3+3 + (3n)	C15 = He5 He5 He5
62	5	10n	B 5p 10n B15 e em 9.87 ms 3/2- 5 = 3+2 + (5n)	15	9.87 ms	• 5 = 3+2 + (5n)	B15 = T T T T T
63	4	11n	Be 4p 11n Be15 n em <200 ns 4 = 3+1 + (7n)	15	<200 ns	4 = 3+1 + (7n)	Be15 = H4 T n T H4
64	10	6n	Ne 10p 6n Ne16 2p em 3.74 zeps 0+ 10 = 4p+6 - (4n)	16	3.74 zs	• 10 = 4p+6 - (4n)	Ne16 = He3 D He3 He3 D He3 = 2 B8
65	9	7n	F 9p 7n F16 p em 11.42 zeps 0- 9 = 2p+7 - (2n)	16	11.42 zs	• 9 = 2p+7 - (2n)	F16 = He4 He3 D He3 He4
66	8	8n	O 8p 8n O16 99.762% stable 0+ 8 = 4+4	16	stable	• 8 = 4+4	O16 = 4 He4 = 8 D
67	7	9n	N 7p 9n N16 e em 7.13 s 2- 7 = 2+5 + (2n)	16	7.13 s	• 7 = 2+5 + (2n)	N16 = T D D D D D T = T B10 T
68	6	10n	C 6p 10n C16 e em 0.747 s 0+ 6 = 4+2 + (4n)	16	0.747 s	• 6 = 4+2 + (4n)	C16 = T T He4 T T
69	5	11n	B 5p 11n B16 n em <190 pics 0- 5 = 1+4 + (6n)	16	<190 pics	• 5 = 1+4 + (6n)	B16 = T T H4 T T
70	4	12n	Be 4p 12n Be16 2n em <200 ns 0+ 4 = 2+2 + (8n)	16	<200 ns	• 4 = 2+2 + (8n)	Be16 = H4 H4 H4 H4
71	10	7n	Ne 10p 7n Ne17 b+ 109.2 ms 1/2- 10 = 3p+7 - (3n)	17	109.2 ms	10 = 3p+7 - (3n)	Ne17 = He3 He4 He3 He4 He3
72	9	8n	F 9p 8n F17 b+ 64.49 s 5/2+ 9 = 1p+8 - (1n)	17	64.49 s	• 9 = 1p+8 - (1n)	F17 = D D D D p D D D D
73	8	9n	O 8p 9n O17   0.038% stable 5/2+ 8 = 1+7 + (1n)	17	stable	8 = 1+7 + (1n)	O17 = He4 He4 n He4 He4
74	7	10n	N 7p 10n N17 e em 4.173 s 1/2- 7 = 3+4 + (3n)	17	4.173 s	• 7 = 3+4 + (3n)	N17 = T He4 T He4 T
75	6	11n	C 6p 11n C17 e em 193 ms 3/2+ 6 = 5+1 + (5n)	17	193 ms	6 = 5+1 + (5n)	C17 = T D T n T D T
76	5	12n	B 5p 12n B17 e em 5.08 ms 3/2- 5 = 2+3 + (7n)	17	5.08 ms	• 5 = 2+3 + (7n)	B17 = T H4 T H4 T
77	11	7n	Na 11p 7n Na18 b+ 1.3 zeps 1- 11 = 4p+7 - (4n)	18	1.3 zs	• 11 = 4p+7 - (4n)	Na18 = He3 D He3 D He3 D He3
78	10	8n	Ne 10p 8n Ne18 b+ 1672 ms 0+ 10 = 2p+8 - (2n)	18	1672 ms	• 10 = 2p+8 - (2n)	Ne18 = He4 He3 He4 He3 He4
79	9	9n	F 9p 9n F18 b+ 109.77 min 1+ 9 = 5+4	18	109.77 min	• 9 = 5+4	F18 = 9 D
80	8	10n	O 8p 10n O18   0.200% stable 0+ 8 = 2+6 + (2n)	18	stable	• 8 = 2+6 + (2n)	O18 = 2 Be9
81	7	11n	N 7p 11n N18 e em 622 ms 1- 7 = 4+3 + (4n)	18	622 ms	• 7 = 4+3 + (4n)	N18 = T D T D T D T = Li8 D Li8
82	6	12n	C 6p 12n C18 e em 92 ms 0+ 6 = 3+3 + (6n)	18	92 ms	• 6 = 3+3 + (6n)	C18 = 6 T = 2 Li9
83	5	13n	B 5p 13n B18 n em <26 ns 4-# 5 = 3+2 + (8n)	18	<26 ns	• 5 = 3+2 + (8n)	B18 = T n T H4 T n T = He7 H4 He7
84	12	7n	Mg 12p 7n Mg19 2p em nodata 1/2-# 12 = 5p+7 - (5n)	19	nodata	12 = 5p+7 - (5n)	Mg19 = He3 D He3 He3 He3 D He3 = B8 He3 B8
86	10	9n	Ne 10p 9n Ne19 b+ 17.22 s 1/2+ 10 = 1p+3 + 6 - (1n)	19	17.22 s	10 = 1p+3 + 6 - (1n)	Ne19 = Be7 6 D
87	9	10n	F 9p 10n F19 100% stable 1/2+ 9 = 1+8 + (1n)	19	stable	• 9 = 1+8 + (1n)	F19 = Be9 p Be9 (Li7 6D ?)
88	8	11n	O 8p 11n O19 e em 26.464 s 5/2+ 8 = 3+5 + (3n)	19	26.464 s	8 = 3+5 + (3n)	O19 = Be9 n Be9 = He4 n He4 n He4 n He4
89	7	12n	N 7p 12n N19 e em 271 ms 1/2- 7 = 5+2 + (5n)	19	271 ms	• 7 = 5+2 + (5n)	N19 = T T Li7 T T
90	6	13n	C 6p 13n C19 e em 46.2 ms 1/2+ 6 = 1+5 + (7n)	19	46.2 ms	6 = 1+5 + (7n)	C19 = T T T n T T T = Li9 n Li9
91	5	14n	B 5p 14n B19 e em 2.92 ms 3/2- 5 = 4+1 + (9n)	19	2.92 ms	• 5 = 4+1 + (9n)	B19 = 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.

J.Lucas, Atommag modell         Nuclear properties
Foton és anyagmodell

Az atommag szerkezete

Problémák a relativitáselmélet körül

További origó fórum kérdések   OF   OF2   OF3     és     Tudomány cikkek archiv     Szkept     Hypog     Csivar

AstrojaN az oxigénnél kisebb atommagok hengeres csöves szerkezete felépítése

	janos@biochem.szote.u-szeged.hu

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