Element

Mass defect in number of Δm = δ

Nber δs of Δm remaining after subtraction of δ from previous even précédent & of the new helions

neutrons in addition to helions

probable

role of

δs

N˚ of the last helion shell given by the type of bond

N˚ shell formula

(18) (19) (20)

comments

¹²₆C

68.81

1

²⁴₁₂Mg

148.38

148 − 69 − 63 = 16

2 double bonds (L2) per helion (or 12Δm)

2

2.66

³⁶₁₈Ar

229.77

230 − 148 − 63 = 19

2 L3/helion (or18Δm)

3

3.17

⁵⁰₂₄Cr

326.53

327 − 230 − 63 = 34

2

2 L4/helion (24Δm) 10 left either 1L4 per n or at least 2L2 per n

4

4.86

The 2 n stable would be more at the 2nd shell level

⁶⁴₃₀Zn

419.96

420 − 327 − 63 = 30

4

2 L4/helion or 24Δm, 6 left for 2n (2L3)

4

4.28

shell 5 requires 30Δm and there is 2n more. So rather shell 4 (24). les 2 n are on a lower shell

⁸⁰₃₆Kr

522.89

523 − 420 − 63 = 40

8

2L5/helion or 30Δm, 10 left for 4n

5

5.00

The last 4 n are on a lower shell with several bonds

⁹²₄₂Mo

598.58

599 − 523 − 63 = 13

8

6L2 for 3 helions

2

2.17

Shell 2 fills up

¹⁰⁶₄₈Cd

680.11

680 − 599 − 63 = 18

10

6L3 for 3 helions

3

3.00

Shell 3 fills up. The 2 n take bonds to the other n

¹²⁴₅₄Xe

786.73

787 − 680 − 63 = 44

16

2L5/helion or 30Δm, 14 left for 6n

5

5.50

Shell 5 fills

¹⁴²₆₀Nd

891.63

892 − 787 − 63 = 42

22

2L5/helion or 30Δm,

8 left for 6n

5

5.25

Shell 5 fills

¹⁵⁴₆₆Dy*

948.43

948 − 892 − 63 = −7

22

The next 3 elements are unstable

¹⁷⁴₇₂Hf*

1055.87

1056 − 948 − 63 = 45

30

¹⁹⁰₇₈Pt*

1135.31

1135 − 1056 − 63 =

34