Electromagnetic Nature of Nuclear Forces and the Toroid Structure of the Helion and the Alpha Particle

Vol.4, No.7, 484-491 (2012) Natural Science http://dx.doi.org/10.4236/ns.2012.47065

Electromagnetic nature of nuclear forces and the toroid structure of the helion and the alpha particle

Kiril Kolikov

Plovdiv University “P. Hilendarski”, Plovdiv, Bulgaria; [email protected]

Received 20 May 2012; revised 18 June 2012; accepted 2 July 2012

ABSTRACT in the hydrogen atom, with which he obtained good results for its spectrum within the framework of the old In the present paper, we consider the nucleons quantum mechanics. in the helium-3 and helium-4 nuclei as tori. Additional reasoning to consider the nucleons as spa- These tori rotate with constant angular velocity tially extended objects is the fact that they are located around an axis, which is perpendicular to the close to each other in the atomic nuclei. Thus, we assume rotation plane and passes through the centre of that they are tori and their electric charge can be redis- mass of the nuclei. Based on exact analytical tributed. Essentially, this idea does not contradict to the expressions for the electrostatic interaction be- quark substructure model of nucleons, and it enables us tween two spheres with arbitrary radii and to determine the electrostatic interaction between them. charges derived by us recently, we find that the Based on the exact analytical expressions for the electro- well-known potential binding energy for the static interaction between two charged conducting spheres helion and for the alpha particle is of electro- with arbitrary radii and charges derived for the first time magnetic character. Using the above mentioned in [10], we conclude in [8] that strong interactions are formulae, we find the interaction force in the electromagnetic in nature. nuclei under consideration. Our toroid model In [9], we ascertain that the experimentally determined recovers the basic experimental results not only potential binding energy between nucleons in the deu- for the binding energy, but also for the stability, teron and triton can be obtained taking into account the radii, spins and the magnetic moments of the electrostatic interaction only. To do that, the values of the helion and the alpha particle. experimentally known radii, and masses of the nucleons

and the corresponding nuclei have been utilized. In [9], Keywords: Helion; Alpha Particle; Strong the volume and the mass density of nucleons, as well as Interaction; Potential Binding Energy; Electrostatic their interaction force in the deuteron and the triton have Interaction been calculated—important results in nuclear physics presented for the first time. 1. INTRODUCTION Applying the method developed in [8,9], we calculate in the present paper the potential energy and the interac- There are two basic models in the theory of elemen- tion force between the nucleons in the helium-3 and he- tary particles: the standard model [1-3] and the helicon model [4-7]. lium-4 nuclei. Basic experimentally measured character- In [8] and [9] the toroid model of nucleons has been istics such as stability, spin, radius and magnetic moment proposed, which is in certain contradiction with the stan- of the helion and the alpha particle are also explained. According to R. Feynman’s conjecture at distances dard model, however in perfect agreement with the heli- −15 con model. less than 10 m either Coulomb’s law is not valid, or Every nucleon is modelled with an imaginary torus, electrons and protons cannot be considered as point-like which rotates with constant angular velocity around an particles [11]. In accordance with this conjecture, we assume that the distance within the corresponding proton- axis z passing through its mass centre (the geometric neutron pairs is less 10−15 m. centre) O and perpendicular to the rotation plane of its central circle. From quantum mechanical point of view, 2. TOROID MODEL OF NUCLEONS the nucleon is not a localized object in configuration space. Therefore, our model is valid in a formal heuristic In [8,9] the nucleons are modeled as tori. Moreover, sense a la Niels Bohr similar to his model for the electron we assume that they rotate with constant angular velocity

around an axis z passing through their mass (geometric) 2 22 hkk np  RR n p . (2) centre O and perpendicular to the rotation plane of their central circle (Figure 1). Obviously, 0  hkpn k  . Next, we study the system consisting of a proton and a We assume that the mass and number densities of the neutron and determine the electrostatic interaction be- proton and the neutron are equal, that is p  n . If mp tween them. and mn are the masses of Tp and Tn , then The tori corresponding to the proton and the neutron mp mn are denoted by Tp and Tn , while their centres are de-  p  and n  . (3) Vp Vn noted by Op and On , respectively. We assume that the central circles of Tp and Tn lie in parallel or merging According to [12], the volumes of the tori Tp and planes and rotate in the same or opposite direction with T are Vk 2π22R and V 2π22kR, respectively. the same angular velocity ω around the line z, which n p pp nnn Therefore, from p  n and from Eq.3 it follows that passes through Op and On being perpendicular to the rotation plane. Then, if OO  h, it follows that h  0 mRnp pn kk . (4) (Figure 2). np mRp n Let K p and Kn be the centres of the forming cir- Eqs.3 and 4 contain the proton and the neutron mass cles of the tori Tp and Tn , respectively, and whose values are experimentally known and can be read- ROp  pKp, ROnnn K be the radii of the central cir- ily substituted. cles of Tp and Tn . We assume that the distance be- 15 Let q  0 be the radius of the of the empty part of tween Tp and Tn is 0 10 m. According to the available experimental data the pro- the circle with radius equal to that of the proton rp . Then ton radius rp is not greater than the neutron radius rn . That is why we must have RR . p n qr p  2 kp . (5) Let k p and kn be the radii of the forming circles In order to apply the results from [10] derived for the K p and Kn , respectively. It is clear that kRp  p and case of spheres, we remodel the tori as in [8,9]. kn Rn. Moreover, Due to the spherical symmetry of the proton charge Rk r and Rk r, (1) p pp nnn [13], we can assume that all its charge p is concentrated in the geometric centre O of the torus T . where rp and rn are the radii of the proton and the p p neutron in the configuration shown in Figure 2. It is We remodel the proton with a sphere S p having the same centre O lying on the axis z , such that its area worthwhile to note that in some nuclei rp , k p , Rp p is equal to that of the torus T . Moreover, the charge p and rn , kn , Rn may take different values. p of the sphere is spherically symmetric and it can be re- Since OOpn h, simple geometric consideration yields distributed. According to [12] the area Lp of the torus Tp is

2 Lkp  4π ppR. (6)

Since the areas of the torus Tp and the sphere S p are equal, from Eq.6 it follows that the radius rP of S p is

rkP  π ppR. (7) Figure 1. Toroid model of a nucleon.

Next, we remodel the neutron Tn whose torus TN has equivalent area and the same centre On . The distance  between its surface and that of S p is the same as the distance between Tn and Tp (Figure 3). Let SN be the sphere, whose central circle is forming the torus TN . We denote the centre of SN with KN so that OKnN RN, and let rN be the radius of SN . If

OKpN R, then Rr PN r  and OOpn h im- 222 plies RRN   h, that is

Figure 2. Cross section of the system 2 2 containing a proton and a neutron. RrrNPN  h. (8)

Eq.11 hold because the forming spheres of the torus

TN are located symmetrically with respect to the centre of the sphere SP . 3. VOLUME MASS DENSITY OF NUCLEONS Let us consider the proton in a free state as the torus 0 01 5 Tp . According to [14] its radius is rp 0.84184 10 m. Let K 0 is the centre of forming circle of the torus p T 0 and let k 0 is the radius of this circle. Moreover, Figure 3. Cross section of the reduced p p RO00 K. model of the proton-neutron system. p p Utilizing expressions (1) and (5) for different values of Clearly, max Rrr and min Rr . the radius q of the empty part of the circle with radius NpN N N r , we have calculated in [9] the quantities k 0 , R0 The equality of the areas L and L of the tori T p p p N n N 2 0200 and Tn , together with Eq.6 yields and the volume of the proton torus Vkp  2π ppR R [12]. The mass of the proton in a free state is rk n . (9) Nn m01.672621638 1027 kg [15]. This can be used to RN p 0 calculate  p . For the number l of the spheres SN whose total These results are depicted in Table 1 for various val- area is equal to that of the torus TN , the following rela- 22 ues of q . tion lr.4π N  4π RNN r holds. Therefore Since, the mass of the neutron 027 RN mn 1.674927212 10 kg [15] is also known, its vol- l  π . (10) 0 r ume Vn in a free state can be determined, where N 00 p  n has been taken into account as already men- Let us assume that the centre Op is fixed with re- tioned in Section 2. spect to some inertial reference frame J . Next, we in- Due to the mass defect in the atomic nuclei, the vol- troduce a rigid inertial coordinate system G , which ro- umes of the nucleons change. We conjecture however, tates with the constant angular velocity ω of the torus that the mass densities of the proton and the neutron in T rotation with respect to J . The point O is the 00 N p all nuclei do not change pnconst . Based on centre of the coordinate system Oxyp z , which is firmly the values presented in Table 1, we ascertain the poten- fixed with G and the spheres SP and SN are at rest tial binding energy and calculate the interaction force in one against the other (Figure 3). the deuteron and the triton in [9]. Thus, taking into account the experimentally found We model the deuteron as two immersed concentric value for rr (r is the radius of the considering knk tori. The inner one Tp corresponds to the proton, while nucleus), assigning different values for q and varying the outer Tn corresponds to the neutron, such that the 01 015 m and 0 hk  k  , from Eqs.1-5 pn distance between Tp and Tn is of the order of 15 we can find Rn , kn and Rp , k p , rp . Combining with 01 0 m. In addition, we assume that the two tori

Eq.7 we determine the radius rkRP  π pp of the sphere rotate in the same direction with constant angular velocity ω around a straight line z passing through their SP with equivalent area. From Eqs.8 and 9 we determine the radius r of the forming sphere S . common centre and perpendicular to the rotation plane. N N We model the triton with three tori whose central circles From RrPN r  and taking into account ([10], Eqs.10 and 11 therein), we can determine the interaction Table 1. Dimensions, volume and mass density of the proton. force F SSP , N  and the binding energy WS P , SN  . q 0 0 0  0 According to Eq.10 the binding energy WTp , Tn and k p Rp V p p 15 15 34 5 318 the interaction force F TTp , n  between the proton and m m10 m10 m 10 kg m 10 the neutron become 0.3 rp 0.294644 0.547196 0.93771 1.78373 R N 0.4 r WTp , TnP π WS, SN p 0.252552 0.589288 0.74192 2.25444 rN 0.5 r 0.21046 0.63138 0.55203 3.02997 and (11) p 0.6 r 0.168368 0.673472 0.37685 4.43843 R p F TT,  π N FSS,. pn P N 0.7 r 0.126276 0.715564 0.22523 7.42639 rN p

1 2 lie in three parallel planes. The tori Tn and Tn representing the neutrons are located symmetrically on both sides of the torus Tp representing the proton. The dis- 1 2 tance between the neutron tori Tn , Tn and the proton 15 one Tp is the same and of the order of 010 m. i The tori Tp and Tn i 1, 2  rotate around an axis z , which passes through their centres and is perpendicular 1 to the rotation plane. The tori Tp and Tn rotate with constant angular velocity ω in the same direction 2 (clockwise, for example), while Tn rotates with the same angular velocity in the opposite direction. Figure 4. Cross section of the helion In the subsequent exposition we will model the he- model. lium-3 and the helium-4 nuclei. The nucleon disposition must comply with the principle of the minimum of po- spin of the helion is s3 12, which is an experimen- tential energy. Taking into account the mass defect in 2 He atomic nuclei, the potential energy of interaction can be tally established fact [16]. calculated according to the following formula [16] The proton rotating in an opposite direction decreases the centrifugal force, which is originated by the two nu- 2 WNmNmmcKppnnK. (12) cleons rotating in the same direction. That is why the 15 helion radius r3 1.9506 10 m [17] is smaller than 2 He Here N p is the number of protons, Nn is the num- 27 15 ber of neutrons, while mp 1.672621638 10 kg and that of the deuteron r2 2.140210 m [15]. m 1.6749 27212 1027 kg are the proton and the neu- 1 H n The magnetic moments due to the proton charges tron mass, respectively [15], mK is the mass of the nucleus and c 2.99792458 108 ms−1 is the speed of light compensate each other. Thus, the magnetic moment of in vacuum [15]. the helion is generated by the charge of the neutron only. As a consequence of the enlarged neutron radius as Based on the values for mass densities in Table 1, we compared to that in a free state, the magnetic moment of calculate the corresponding values of W for the helion K 26 −1 the helion  3 1.0746  10 JT is greater in ab- and the alpha particle, we follow the procedure described 2 He in Section 2. Finally, our results will be compared with solute value than the magnetic moment of the neutron the values obtained by virtue of Eq.12. 26 −1 n 0.9662  10 JT , which is in agreement with the experimental data [15]. 4. MODEL OF THE HELION i Let Kn and K p be the centres of the forming cir- i The helion is a mirror nucleus of the triton and there- cles of Tn and Tp , respectively with radii kn and i fore its structure is analogous to that of the latter [9]. It kkp  p i 1, 2  . We introduce the notations Rnn OK ii 1 consists of one neutron T and two protons T and and for i 1, 2 , ROKp  pp (Figure 4). n p i 2 1 2 This means that hOO p . Since the neutron radius is Tp . Since the proton tori Tp and Tp repel due to the equal to the radius of the helion, it follows that rrn  3 electrostatic interaction between them, their central cir- 2 He and Rrnn 3  k. cles will lie in two parallel planes symmetrically located 2 He on both sides of the plane in which the central circle of Due to the spherical symmetry of the proton charges, the neutron torus Tn is located. In addition, the distance we can assume that their charges 1 2 19 between the surfaces of Tp , Tp and that of Tn will be p 1.602176487 10 C are concentrated in their geo- the same 0 1 015 m (Figure 4). i metric centres Op i 1, 2  . On these grounds we can This configuration ensures symmetry with respect to i model the protons with spheres SP having the same the mass centre (the geometric centre) of the helion. This i i centres Op and radii rrP  P i 1, 2  with areas equal also implies the helion stability T . i i 3 He to the areas of Tp . Moreover, the charge of each SP is 12 2 1 2 The centres OOp , p and On of the tori Tp , Tp p , which is spherically symmetric and can be redistrib- and Tn , respectively lie on one axis z perpendicular to uted. their plane of rotation with On  O being the mass cen- With TN we denote the torus which is equivalent in 1 tre of the helion. The tori Tp and Tn rotate with con- area to Tn with the same centre O . The distance be- i stant angular velocity ω in a certain direction, while tween SP and TN is equal to the distance  between 2 i Tp rotates with the same angular velocity but in the Tp and Tn . Let us denote: with SN the sphere whose opposite direction. From here it follows directly that the central circle is the forming circle of the torus TN ; with

K the center of the sphere and with r the radius of 13 N N W3 12.352  10 J. Using the corresponding Eq. the sphere (Figure 5). 2 He i 11, we confirm this value for We assume that the points Op are at rest with respect i to an inertial frame J . Let us also introduce two rigid WWTTWp3 2, pn  ,p , where i 1 or i  2 . By 2 He non inertial reference frames G and G rotating with 1 2 virtue of the same Eq.11 we find the magnitude of the respect to J with angular velocity equal to that of the interaction force for the helion rotation of the torus T . N i i F3 2,FTpn T F pp , , where i 1 or i  2 . The point Op is the centre of the coordinate system 2 He   Oxii y z and is fixed firmly with the reference frame Gi In Table 2, k and k denote the radii of the form- i n p i 1, 2  relative to which the spheres SP and SN are ing circles, while Rn and Rp are the radii of the cen- at rest one against the other (Figure 5). tral circles of the neutron and the proton tori in the helion, The torus T is located symmetrically with respect to N respectively. In addition, F3 is the interaction force i 2 He the two spheres SP . It suffices to analyse the electro- between the nucleons in the helion. static interaction between TN and any of the spheres. It is important to note that in order to obtain the value Based on the calculations presented in [8], we assume of W3 , we can vary h as well for different values of 1 2 He that the electrostatic interaction between the spheres SP the distance  , such that khkk   holds. This 2 ppn and SP can be approximated with interaction between implies that the distance  p between the surfaces of the point-like charges p , p concentrated in their centres, two protons will vary according to 02 k  2. 12 12 pn i.e. WSpp,,S  Wpp and F SSpp,, Fp p . For the sake of simplicity, we assume that the distance 5. MODEL OF THE ALPHA PARTICLE 1 2  p between the surfaces of the protons Tp and Tp is i Structurally, the alpha particle is obtained by adding the same as the distance between Tp and Tn , i.e.  one neutron to the helion. We assume that the centres of i 1 2 1 2  p  . Therefore, we must have OOPp h k  (Fig- the protons T , T and the neutrons T , T are 2 p p n n ure 4). symmetrically located with respect to the mass centre of the nucleus. The central circles of the tori lie in parallel Using Eqs.1-5 from Section 2, we calculate kn , Rn i i planes. The distance between the tori Tp , Tn (i 1, 2 ) and k p , Rp , rp for different values of the mass den- 15 sity  taken from Table 1 of Section 3 and different is the same 01 0 m, while the elongation between T 1 , T 2 is some small distance (Figure 6). values of the distance  . The mass defect of the two n n protons and the neutron has been taken into account in From this configuration it follows the stability of the alpha particle is T4   . Eqs.3 and 4 accordingly to the mass proportions of the 2 He nucleons. According to Eqs.7-9 we calculate r , r i i i P N The centres Op and On (i 1, 2 ) of the tori Tp and the distance Rr r between the centers of i PN and Tn lie on one axis z perpendicular to the plane of the spheres S i and S . 1 1 P N rotation. The tori Tp and Tn rotate in the same direc- The experimentally established mass of the helion is tion around z with constant angular velocity ω , while m 5.00641192 1027 kg [15]. According to Eq.12 T 2 and T 2 rotate with the same velocity in the oppo- 3 He p n 2 site direction. It follows that both the spin and the mag- the binding energy of the helion is netic moment of the alpha particle are zero, which is an

experimentally established fact [15,16]. Moreover, the decrease of the centrifugal force as compared to the helion implies that the radius of the

Figure 5. Cross section of the reduced Figure 6. Cross section of the model model of the triton. of the alpha particle.

Table 2. Dimensions of nucleons and interaction force between them in the helion.

  k R k R F3 n n p p 2 He 318 17 kg m 10 m10 m10 15 m10 15 m10 15 m10 15 N 1.78373 0.728856 0.16290 1.7877 0.17828 1.4905 −13415.87

2.25444 0.695851 0.14415 1.8065 0.15580 1.5441 −15321.68

3.02997 0.648141 0.12364 1.8270 0.13192 1.6025 −16853.54

4.43843 0.588400 0.10154 1.8491 0.10693 1.6650 −19439.99

7.42639 0.508946 0.078006 1.8726 0.08107 1.7313 −23415.29

Table 3. Dimensions of nucleons and the interaction force in the alpha particle.

  k R k R F4 n n p p 2 He 318 17 kg m 10 m10 m10 15 m10 15 m10 15 m10 15 N 1.78373 1.582281 0.17769 1.4953 0.20746 1.0954 –60565.5

2.25444 1.421146 0.15697 1.5160 0.17881 1.1667 –72347.6

3.02997 0.793140 0.13440 1.5386 0.14916 1.2475 –84474.6

4.43843 0.313777 0.11018 1.5628 0.11934 1.3303 –100313.4

7.42639 0.246313 0.08449 1.5885 0.089552 1.4121 –139022.6

15 i 1, 2 are at rest with respect to an inertial reference alpha particle r4 1.673 10 m [17] is smaller   2 He frame J . In addition, four firm noninertial coordinate 15 than the radius of the helion r 1.9506 10 m. j 3 He systems G ij1, 2 ; 1, 2 are introduced rotating at i i 2 i   Let K p and Kn be the centres of the forming cir- constant angular velocity ω (or ω ) equal to the an- i i i i 1 2 cles of T and Tn with radii kk and kknn p p p gular velocity of rotation of the neutrons TN and TN ii i i 1, 2 . For i 1, 2 , we denote ROp  ppK and with respect to J . Every point O is a centre of two ii p j jj ROn nKn (Figure 6). coordinate systems Oxii y z j 1, 2  fixed firmly The radius of the neutron is equal to the radius of the j with the reference frames Gi with respect to which the i j alpha particle, that is rrn  4 and Rrnn4 k. From 2 He 2 He spheres SP and SN are at rest. 1 2 1221 Since the total charges of the tori TN and TN are the isosceles trapezoid KnnKKK p p it follows that the 1 zero it can be assumed with good approximation that distance   between the surfaces of the tori Tp and T 2 is there is no electrostatic interaction between them. n Due to the symmetry, it suffices to find the interaction 2 2 i  2khRRkk. (13) between any of the spheres SP  SP and the spheres nnpnp j SN  j 1, 2  . We assume as we noted in Section 4 that 1 2 We assume again that the proton charges are concen- the interaction between the spheres SP and SP is trated in their geometric centres Oi . Similar to Section 12 p taken as point charges pp, , i.e. WSpp,,S  W pp i i 12 4, we model the protons as spheres SP with centres Op and F SSpp,, Fpp  . For the sake of calculative i   and radii rrP  P i 1, 2 , whose areas are equal to simplicity it is assumed that the distance  p between i i 1 2 that of Tp . Moreover, the charge of SP is p , which is the surfaces of the proton tori Tp and Tp , as well as spherically symmetric and can be redistributed. the distance  n between the surfaces of the neutron tori i 1 2 i With TN i 1, 2 , we denote the tori, whose areas Tn and Tn , is the same as the one between Tp and i i i are equal to the areas of Tn with the same centres On . Tn i 1, 2  , i.e.  pn . i j 12 12 The distance between the objects SP and TN is the We have OOpp 2 k p and OOnn2kn . There- i j ii same as the distance between Tp and Tn fore, OOp npn h k k . Taking into account the mass i ij1, 2 ; 1, 2 . Let us denote: with SN the sphere density  of the nucleons from Table 1 of Section 3 i whose central circle is the forming circle of the torus TN ; and using Eqs.1-5 from Section 2 for different values  i i with KN the center of the sphere and with rN  rN the all parameters kn , Rn and k p , Rp , rp can be found. radius of the sphere (Figure 6). In Eqs.3 and 4 the corresponding mass defect for the two i i We further assume that the points Op and On protons and the two neutrons has been taken into account.

According to Eqs.7-9 we calculate rP , rN and the dis- the volumes and the mass densities of the nuclei, as well 11 1 tances ROKrr1 PN P N  and as the force of interaction in the considered nuclei. 12 2 ROKrr2 PN P N   between the centres of the 1 1 1 2 spheres SP , SN and SP , SN , where  is given by 7. CONCLUSIONS Eq.13. The presented here model can be applied for the more The experimentally established mass of the alpha par- complicated atomic nuclei. The electrostatic interaction 27 ticle is m4 6.64465620 10 kg [15]. According to 2 He between nucleons changes their electric structure. Thus, Eq.12 the binding energy for the alpha particle all nucleons enter to various extents in interactions, 13 which compensate each other. W4 32.923210 J. Using the corresponding Eq. 2 He Let us also note that the electromagnetic forces be- 11 we confirm this value for tween nucleons depend on whether they rotate in the ii12 same or different directions that is they depend on the WW4 2,TpnT2,WTpnTWpp,, where 2 He orientation of their spins. i 1 or i  2 . By virtue of the same Eq.11 we find the Based on our research, we are confident that all avail- magnitude of the interaction force for the alpha particle able experimental data about atomic nuclei can be ex- F 2,FTiiT12,FT T2F p,p, where i 1 plained by the model proposed. Other new properties of 4 He pn pn  2 atomic nuclei can also be found. or i  2 . It is essential that we obtain the basic nuclear charac- The quantities k and k in Table 3 denote the ra- n p teristic—the binding energy in all of the considered nu- dii of the forming circles of the tori, R and R are n p clei, using only electromagnetic interactions. The most the radii of the central circles of the neutron and proton significant conclusion from our studies is that nuclear tori, respectively and F4 is the interaction force be- 2 He forces are electromagnetic in origin. tween the nucleons in the alpha particle. We would like to note that in order to obtain the value 8. ACKNOWLEDGEMENTS of W4 , the quantity h for different values of the dis- 2 He tance  can be varied as well, such that The author would like to thank Stefan Bozhkov, for performing the

0 hkpnk . It follows that the distance  p be- calculations in the present paper on Wolfram Mathematica 7.0. tween the proton surfaces will vary 04 pnk 3 . The results of the present studies are published with the financial According to Eq.13 the distance   will vary, too. support from the Fund for Scientific Research at the Ministry of Educa- tion and Science of Bulgaria under contract DTC No. 02/35. 6. DISCUSSION

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