Element

Mass defect in number

of Δm = δ

δ’s

N˚ shell given by Table A4

Probable role of this remaining Δm

Neutrons in add. to hélions

Comments

¹⁶₈O

95.41

5

95.41 − 68.81 − 21.07 = 5.53

2

2 double bonds for the last hellion (2 × 2.76d)

²⁰₁₀Ne

120.12

4

120.12 − 95.41 − 21.07 = 3.64

2

Idem (2 × 1.82d)

²⁴₁₂Mg

148.38

7

148.4 − 120.12 − 21.07 = 7.19

2

idem

Calculation/C (shell 1) gives: 148 − 69 − 63 = 16 or 16/6 = 2.66Δm per bond thus shell 2

²⁸₁₄Si

177

8

3

2 triple bonds

³²₁₆S

204

6

3

2 triple bonds

³⁶₁₈Ar

229.77

230 − 204 − 21 = 5

3

2 triple bonds

230 − 148 − 63 = 19

or 19/6 = 3.17 (shell 3)

⁴⁰₂₀Ca

256.28

256 − 230 − 21 = 5

4

⁴⁶₂₂Ti

298.85

299 − 256 − 21 = 22

4

2

.δs increases strongly when 2 add. n are needed.

⁵⁰₂₄Cr

326.53

327 − 299 − 21 = 7

4

2

⁵⁴₂₆Fe

354.09

354 − 327 − 21 = 6

4

2

⁵⁸₂₈Ni

380.10

380 − 354 − 21 = 5

4

2

⁶⁴₃₀Zn

419.96

420 − 380 − 21 = 19

4

4

⁷⁰₃₂Ge

458.88

459 − 420 − 21 = 18

5

6

⁷⁴₃₄Se

483

483 − 459 − 21 = 3

5

6

⁸⁰₃₆Kr

522.89

523 − 483 − 21 = 19

5

8

⁸⁴₃₈Sr

548

548 − 523 − 21 = 4

2

8

⁹⁰₄₀Zr

589.63

590 − 548 − 21 = 21

2

10

⁹²₄₂Mo

598.58

599 − 548 − 42 = 9**

2

8

⁹⁶₄₄Ru

620.95

621 − 599 − 21 = 1

3

8

¹⁰²₄₆Pd

657.79

658 − 621 − 21 = 16

3

10

¹⁰⁶₄₈Cd

680.11

680 − 599 − 63 = 18

3

10

¹¹²₅₀Sn

716.70

717 − 680 − 21 = 16

5

12

¹²⁰₅₂Te

765.14

765 − 717 − 21 = 27

5

16

¹²⁴₅₄Xe

786.73

787 − 680 − 63 = 44

5

16

¹³⁰₅₆Ba*

821.83

822 − 787 − 21 = 14

5

The new He go to

shell 5 with 2n

18

¹³⁶₅₈Ce

856.62

857 − 822 − 21 = 14

5

The new He go to

shell 5 with 2n

20

¹⁴²₆₀Nd

891.63

892 − 857 − 21 = 14

5

The new He go to

shell 5 with 2n

22

¹⁴⁴₆₂Sm

899.02

899 − 892 − 21 = −14

20

From Sm to Er, we see 2 decreases of −14 followed by 2 strong increases of 21 and 24

¹⁵²₆₄Gd

941.29

941 − 899 − 21 = 21

24

It is interpreted as saying that the last helion of the Sm is placed at a nivel 4 and drives at least one other helion previously arrived in 5 at a level 4

¹⁵⁴₆₆Dy*

948.43

948 − 941 − 21 = −14

22

2 n are no longer necessary The helion brought by Gd is then placed at level 5 and requires 4n. The same goes for Dy and Er

¹⁶²₆₈Er

993.17

993 − 948 − 21 = 24

26

¹⁶⁸₇₀Yb

1024.89

1025 − 993 − 21 = 11

The new He go to

shell 5 with 2n

28

Yb, Hf, W. The 3 He occupy first 3 places in level 5 (filling sequence idem Kr Cd)

¹⁷⁴₇₂Hf*

1055.87

1056 − 1025 − 21 = 10

The new He go to

shell 5 with 2n

30

¹⁸⁰₇₄W

1086.48

1087 − 1056 − 21 = 10

The new He go to

shell 5 with 2n

32

Os sees the end of the filling of level 4 with another He going

¹⁸⁴₇₆Os

1105.26

1105 − 1087 − 21 = −3

32

in 4. No new n. n decrease in level

¹⁹⁰₇₈Pt

1135.31

1135 − 1105 − 21 = 5

34

¹⁹⁶₈₀Hg

1166.47

1167 − 1135 − 21 = 11

36

Pt to Pb end of filling shell 5

²⁰⁴₈₂Pb

1209.15

1209 − 1167 − 21 = 21

40

²⁰⁹₈₄Po*

1231.57

1232 − 1209 − 21 = 1

41

²²²₈₆Rn*

1285.32

1285 − 1232 − 21 = 32

50

²²⁶₈₈Ra*

1302.62

1303 − 1285 − 21 = −3

50

²³²₉₀Th*

1328.93

1329 − 1303 − 21 = 5

52

²³⁴₉₂U*

1337.32

1337 − 1329 − 21 = −13

50