Yuri Dunaev ([email protected])
DETERMINING MAIN PARAMETERS OF THE HELIUM ATOMS Hе I and Hе II
© Yuri Dunaev, 2011
THE SUMMARY
In the article there are determined main parameters of the atoms of He I and He II, particularly the
planetary satellites orbits radii, their angular velocities, and Kepler constants. There is determined that
the Kepler constants of atomic and molecular planetary systems relate between themselves as their
mean electric charges that is as the screening areas of their nuclei. The helium II spectrum is composed
with series analogues to those of the hydrogen molecule spectrum, the frequencies of which relate to
the respective hydrogen molecule spectrum frequencies as 4.00177:1. The helium I atom being ionized
and transformed to helium II atom, loses its external valent pair of electrons, while the pair of dyads
moves to an orbit approximately 10 times broader, on which they rotate with an approximately 33 times
slower angular velocity. As a result of such transformation the pair of dyads of helium II atom incurs the
action of the same braking torque as the pair of electrons together with the pair of dyads have incurred
in the helium I atom.
Preface
The article is a continuation of the previous one entitled “HELIUM ATOMIC STRUCTURE SEEN THROUGH
ITS RADIATION SPECTRUM)” http://www.wbabin.net/files/4361_dunaev11.pdf. In the previous article
there was disclosed the formation mechanism of the helium (helium I) atom radiation spectrum, and
based on the study of this mechanism there was disclosed the structure of the atom itself. There was
realized that the helium atom is a planetary system composed with a nucleus and planetary satellites
orbiting the nucleus in the same plane, in which rotates the nucleus itself. The planetary satellites form
two pairs in which they rotate around the nucleus at different distances and with different angular
velocities, in every pair one of the satellites rotating in one, and the other – in the opposite direction.
The satellites of one pair are electrons, while those of the other pair are groups of two electrons (dyads)
bonded with the strong interaction. The rotation frequency of the electrons around the helium atom
nucleus, which was found to be 3.262609 ∙ 10 17 s-1, relates to the rotation frequency of dyads as 1 to
1.35. The main attention of the previous work being focused on examining its radiation spectrum, some
questions related with main parameters of the helium atom have remained unrevealed. The task of the
article is to disclose these unrevealed questions. Furthermore in the previous article there have
remained passed over the radiation spectrum and structure of the helium II that is the helium I ion.
Therefore the task of the article is also to disclose questions related to the helium II.
Determining the main parameters of the Helium I atom
The main parameters of a helium I atom should be the electric charge of its nucleus, constant of Kepler,
orbits radii of electrons and dyads, and their rotation frequencies around the nucleus. The electric
charge of the helium atom nucleus together with those of some other elements was determined earlier
in my article “REAL SENSE OF ELECTRIC CHARGE” http://www.wbabin.net/physics/dunaev3.pdf by the
way of calculating its mean screening area. To better understand the employed methods let us reiterate
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them here at greater length, but only for the helium-4 nucleus, that as we know is composed with four
nucleons (according to modern views: with two protons and two neutrons of the nearly equal mass). As
I can see, the top view of this nucleus that is its view in the atom’s rotation plane has to be such as
represented below on fig.1.
A
B
fig.3
fig.4
fig.1
fig.2
If to look at this nucleus from a point of electron’s or dyad’s orbit, then depending on the point of view,
its profile would make a figure of changing form and dimensions from that of maximum area (fig.2 that
represents the view on fig.1 along the arrow A) to that of minimum area (fig.3 that represents the view
on fig.1 along the arrow B). The mean value of the atom’s nucleus charge has to be the mean value of
the fig.2 and 3 areas.
The value of this mean charge is calculated by the formula
𝑞𝑞𝐻𝐻𝐻𝐻 = 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 + 0.637(𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 − 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 ),
(1), where
𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 is the minimum charge of the nucleus equaling two proton charges that are meant to equal unitary
charges 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 = 2𝑞𝑞𝑝𝑝 =2𝑞𝑞𝑒𝑒 , and 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 is the maximum charge, which resulting from measuring the area of
the fig.4 is 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 =2.6488𝑞𝑞𝑝𝑝 = 2.6488𝑞𝑞𝑒𝑒 . Resulting from calculations by the above formula (1)
𝑞𝑞𝐻𝐻𝐻𝐻 = 2.4133𝑞𝑞𝑝𝑝 = 2.4133𝑞𝑞𝑒𝑒
Analogously to determining the radius of the electronic orbit of the hydrogen molecule, made in my
previous
article
“QUANTUM
MECHANICS’
FOUNDATION,
HOW
STRONG
IS
IT?”
http://www.wbabin.net/physics/dunaev1.pdf, let us work out a balance equation of forces applied to
the electron, particularly that of Coulomb and centrifugal one, and in order to determine the force of
Coulomb active on the helium atom electron let us employ the just found value of the nucleus’ electric
charge. The Coulomb’s force (the left part of the balance equation) would make
𝐹𝐹𝑐𝑐 = 𝑘𝑘𝑒𝑒
𝑞𝑞 𝑒𝑒 𝑞𝑞 𝑛𝑛
𝑟𝑟 2
= 𝑘𝑘𝑒𝑒
2.4133𝑞𝑞 𝑒𝑒2
,
𝑟𝑟 2
where
𝑘𝑘𝑒𝑒 is for the proportionality coefficient, 𝑞𝑞𝑒𝑒 − is the charge of electron, 𝑞𝑞𝑛𝑛 − the above charge of the
atomic nucleus, and 𝑟𝑟 − the distance between the charges.
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According to Wikipedia en.wikipedia.org/wiki/Coulomb's_law 𝑘𝑘𝑒𝑒 =8.9875517873681764·109 NmC-2, and
according to NIST physics.nist.gov/cuu/Constants/index.html the unitary charge of electron makes
1.602 176 487(40) · 10-19 C.
The right part of the balance equation is represented as the centrifugal force
𝐹𝐹𝑐𝑐𝑐𝑐 = 𝑚𝑚𝑒𝑒 𝑟𝑟𝜔𝜔𝑒𝑒 2 ,
where
𝑚𝑚𝑒𝑒 is the mass of electron, that is 9.109 382 15(45) · 10-31 kg physics.nist.gov/cuu/Constants/index.html
and 𝜔𝜔𝑒𝑒 is the angular velocity of the electron equaling 2𝜋𝜋𝑅𝑅, where 𝑅𝑅 is the above electron rotation
frequency determined in my previous article “HELIUM ATOMIC STRUCTURE SEEN THROUGH ITS
RADIATION SPECTRUM” as 𝑅𝑅 = 3.262609 ∙ 10 17 s-1. Then the balance equation will be as follows
𝐹𝐹𝑐𝑐 = 𝑘𝑘𝑒𝑒
2.4133 𝑞𝑞 𝑒𝑒2
𝑟𝑟𝑒𝑒 2
= 𝐹𝐹𝑐𝑐𝑓𝑓 = 𝑚𝑚𝑒𝑒 𝑟𝑟𝑒𝑒 (2𝜋𝜋𝜋𝜋)2 .
Here from one can find the electron orbit radius
3
2.4133 𝑘𝑘 𝑒𝑒 𝑞𝑞 𝑒𝑒2
.
4𝜋𝜋 2 𝑚𝑚 𝑒𝑒 𝑅𝑅 2
𝑟𝑟𝑒𝑒 = �
Employing the above numerical data one can obtain: 𝑟𝑟𝑒𝑒 = 5.258939·10-12 m = 0.05258939 nm, that makes
it possible to determine the overall size of the helium atom as 𝐷𝐷𝐻𝐻𝐻𝐻 = 2𝑟𝑟𝑒𝑒 =0.010517878 nm.
The constant of Kepler for the helium atom has to be: 𝐾𝐾 = 𝑟𝑟𝑒𝑒 3 𝜔𝜔𝑒𝑒 2 = (5.258939 · 10−12 )3 ∙
(2𝜋𝜋 ∙ 3.262609 ∙ 1017 )2 = 611.202 m3/s2 (for comparison – according to “REAL SENSE OF ELECTRIC
CHARGE“ http://www.wbabin.net/physics/dunaev3.pdf, p.8, the constant of Kepler for the hydrogen
molecule makes 506.527 m3/s2). It is easy to notice that the constants of Kepler relate between
themselves in the same way as the charges of the atomic nuclei that is as their mean screening areas.
The angular rotation velocity of the dyad will make
𝜔𝜔𝑑𝑑 = 1.35 𝜔𝜔𝑒𝑒 = 2.767443 ∙ 1018 s-1.
Then the dyad’s orbit radius will be
3
𝐾𝐾
𝑟𝑟𝑑𝑑 = � 2 = 4.30536·10-12 m, and the relation between these radii will be
𝜔𝜔
𝑟𝑟𝑒𝑒
𝑟𝑟 𝑑𝑑
𝑑𝑑
5.25894·10 −12
= 4.30536 ·10 −12 = 1.221487.
Determining main parameters of the He IІ atom
As it is known from modern literary sources He II is the helium atom ion deprived of one of its electrons.
In the light of the already exposed views on the helium atom structure it would be logical to imagine the
above ion as deprived of the external pair of electrons but still in possession of the pair of dyads. Then
the radiation spectrum of this ion would have the same structure as has an atom or molecule with only
one pair of photon generative factors, and as it is known one of such molecules is that of hydrogen.
Experimental data borrowed from the site www.nist.gov/pml/data/asd.cfm allowed working out the
Table 1, which columns 1 and 5 contain data about the intensity of the helium II spectral lines, which
lengths in Å are shown in the columns 2 and 6. The said lines may be grouped in 6 series, in which the
3
wave lengths relate between themselves as do those of Lyman, Balmer, Paschen, Brackett, Pfund and
Humfrey series in the hydrogen molecule spectrum.
Table 1
Intensity Wave
lengths (Å)
Interrela
tions of
wave
lengths
1
15 c
20 c
30 c
50 c
100 c
300 P
150 P
1000 P
500 P
2
231.4541
232.5842
234.3472
237.3307
243.0266
256.3166
256.3177
303.7804
303.7858
3
0.76190
0.76562
0.77142
0.78124
0.79999
0.84374
0,84374
1
1
5c
6c
8c
15 c
30 c
35
50 c
120 P
50 P
7P
25 P
180 P
25 P
15 P
958.70
972.11
992.36
1025.27
1084.94
1215.09
1215.17
1640.3321
1640.3447
1640.3750
1640.3914
1640.4742
1640.4897
1640.5326
0.58443
0.59261
0.60495
0.62502
0.66139
0.74073
0.74078
1
1
1
1
1
1
1
Wave
lengths
interrelati
ons in Н2
spectrum
4
0.76191
0.76562
0.77143
0.78125
0.80000
0.84375
1 Lyman
0.58442
0.59259
0.60494
0.625
0.66138
0.74074
Wave
lengths
interrelatio
ns in Н2
spectrum
8
0.50909
0.53594
0.58333
0.68362
Intensity
Wave
lengths
(Å)
Interrelat
ions of
wave
lengths
5
7c
9c
12 c
15 c
4 P,c
3 P,c
15* P,c
15* P,c
12 P,c
6
2385.40
2511.20
2733.30
3203.10
4685.3769
4685.4072
4685.7038
4685.7044
4685.8041
7
0.50908
0.53593
0.58333
0.68359
1
1
1
1
1
5c
8c
15 c
5411.52
6560.10
10123.6
0.53455
0.64800
1
0.53455
0.648
1 Brackett
4c
6c
11626.4
18636.8
0.62384
1
0.62384
1 Pfund
3c
30908.5
1 Paschen
1 Humfrey
1 Balmer
The data of the Table testify that the helium II spectrum has its own series of Lyman, Balmer, Paschen,
Brackett, Pfund and Humfrey.
The comparison of the wave lengths of the H2 molecule and helium II spectra is made in the Table 2.
Table 2
H2
926.226
930.748
937.803
949.743
972.537
1 025.722
1 215.(668)
4
He II
231.454
232.584
234.347
237.331
243.027
256.317
303.78(6)
H2/ He II
4.00177
4.00177
4.00177
4.00177
4.00177
4.00177
4.00172
The wave lengths being inversely proportional to the angular velocities or rotation frequencies of the
planetary satellites, the angular velocity of the dyads’ rotation in the helium ion will be𝜔𝜔𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 =
4.00177𝜔𝜔𝐻𝐻2 , where 𝜔𝜔𝐻𝐻2 is the angular velocity of the electrons’ rotation in the hydrogen molecule. The
last being 𝜔𝜔𝐻𝐻2 =2𝜋𝜋𝑐𝑐𝑐𝑐𝐻𝐻 =2𝜋𝜋 ∙ 2.9979 ∙ 108 ∙ 1.0973732 ∙ 107 = 2.0670518 ∙ 1016 s-1, the angular velocity
of the dyads’ rotation in the ion of helium will make
𝜔𝜔𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 = 4.00177𝜔𝜔𝐻𝐻2 = 8.2718659 ∙ 1016 s-1.
The last allows compiling the Table 3 that makes comparison between the main parameters of the
hydrogen molecule and those of the helium I and helium II atoms.
Table 3
H2
Не I - electron
Не I - dyad
He II
Orbit radius
Angular velocity
1.058354·10-10 m
5.25894 · 10−12 m
4.30536 · 10−12 m
4.470189· 10−11 m
2.0670518 ∙ 1016 s-1
2.0499577∙ 1018 s-1
2.767443 ∙ 1018 s-1
8.2718659 ∙ 1016 s-1
Nucleus’
screening
area
2
2.4133
Constant of
Kepler
Relation
506.527 m3/s2
611.202 m3/s2
1
1.20665
Quite appropriate would be a question: why the helium atom being deprived of its pair of valent
electrons, the dyads pass to a tenfold greater orbit and slow down their angular velocity nearly 33 times.
To reply this question one first of all need to understand why the atoms and molecules on the whole
need electronic satellites. The situation analysis convinces that without these satellites the atom or
molecule nucleus, owing to its minuscule dimensions, would not practically meet any resistance to its
rotation from the part of the ether, and its state would be very unstable. The planetary satellites
interacting with ether while rotating around the nucleus create some kind of “planetary”
electromagnetic field, which while interacting with an electromagnetic field induced by the rotation of
the nucleus (“nuclear” electromagnetic field) influence the rotation of the last. The response to the
question why do rotate the planetary satellites and why they rotate with such and no other velocities
was already disclosed in my previous article “REAL SENCE OF ELECTRIC CHARGE”
http://www.wbabin.net/physics/dunaev3.pdf, and further to the expressed there ideas one can assert
that the mutual rotation is an intrinsic property of any pair of nearly placed in ether elementary
particles, while the velocity of this rotation and distance between the particles adjust to the resistance
acting on them from the environmental ether. Returning now to the helium atom, one can assume that
its planetary pairs play the role of some kind of anchor that through the interaction of the above
electromagnetic fields retain stable the nucleus rotation velocity. Then having lost the pair of electrons,
the helium II atom has to provide the pair of dyads with such parameters of movement, by which its
braking capability might equal the integral braking capability of the pair of electrons and the pair of
dyads of the helium I atom.
The ethereal resistance to the motion of a particle has to be proportional to its frontal area and some
degree of its velocity relative to the ether. If to designate the unitary resistance, that is the resistance
produced by the motion through the ether of a particle with the unitary frontal area and unitary velocity
as𝑦𝑦, then the produced by the ether resistant torque could be determined as
5
𝑀𝑀 = 𝑦𝑦𝑦𝑦𝑦𝑦(𝑟𝑟𝑟𝑟)𝑥𝑥
(2), where
𝑠𝑠 is the particle’s frontal area, 𝑟𝑟 – its orbital radius, 𝜔𝜔 – the angular velocity of its rotation, and 𝑥𝑥 – a
certain degree, which the value is to be determined. If to assume the frontal area of electron as 1, the
mean frontal area of dyad, as rotating in the plane of motion would be 1.637. By an empirical way there
was found that by 𝑥𝑥 = 1.578 the resistant torque opposed to the He I atom electron would be
1.578
𝑀𝑀е1 = 𝑦𝑦𝑠𝑠𝑒𝑒 𝑟𝑟𝑒𝑒 (𝑟𝑟𝑒𝑒 𝜔𝜔𝑒𝑒 )1.578 = 𝑦𝑦5.25894 · 10−12 (5.25894 · 10−12 ∙ 2.0499577 ∙ 1018 )
0.658277606𝑦𝑦.
=
While that opposed to the dyad would be
𝑀𝑀𝑑𝑑1 = 𝑦𝑦𝑠𝑠𝑑𝑑 (𝑟𝑟𝑑𝑑 𝜔𝜔𝑑𝑑 )1.578 = 𝑦𝑦1.637 · 4.30536 · 10−12 (4.30536 · 10−12 ∙ 2.767443 ∙ 1018 )1.578 =
1.033059269𝑦𝑦.
In sum the 2 torques would make 𝑀𝑀е1 + 𝑀𝑀𝑑𝑑1 =1.6913369 𝑦𝑦.
The torque resistant to the helium II atom dyad will be
𝑀𝑀𝑑𝑑2 = 𝑦𝑦𝑠𝑠𝑑𝑑 (𝑟𝑟𝑑𝑑 𝜔𝜔𝑑𝑑 )1.578 = 𝑦𝑦1.637 · 4.470189 · 10−11 (4.470189 · 10−11 ∙ 8.2718659 ∙ 1016 )1.578 =
1.692649505𝑦𝑦,
that practically equals the sum of the resistant torques of the electron and dyad in the He I atom.
Conclusions:
1) The previously proposed model of the helium atom together with an analysis of the He II
radiation spectrum allowed determining main parameters of the He I and He II atoms,
particularly the planetary satellites orbits radii, their angular velocities, and Kepler constants;
2) There is determined that the Kepler constants of atomic and molecular planetary systems relate
between themselves as their mean electric charges that is as the screening areas of their nuclei;
3) The helium II spectrum is composed with series analogues to those of the hydrogen molecule
spectrum, the frequencies of which relate to the respective hydrogen molecule spectrum
frequencies as 4.00177:1;
4) The helium I atom being ionized and transformed to helium II atom loses its external valent pair
of electrons, while the pair of dyads moves to an orbit approximately 10 times broader, on
which they rotate with an approximately 33 times slower angular velocity;
5) As a result of such transformation, the pair of dyads of the helium II atom incurs the action of
the same braking torque as the pair of electrons together with the pair of dyads have incurred in
the helium I atom.
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