COMPLETE UNDERSTANDING OF NEUTRON, DEUTERON, ALPHA PARTICLE AND NUCLEI - A NEW APPROACH

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1 1 COMPLETE UNERSTANING OF NEUTRON, EUTERON, ALPHA PARTICLE AN NUCLEI - A NEW APPROACH a) ) KUNWAR JAGISH NARAIN Regarding neutron, numerous questions arise, e.g.: 1. A neutron happens to e unstale in its free state and ecomes stale in stale nuclei and systems (e.g. deuterons and α particles), ut why and how?. What does happen or situation is created in stale nuclei/systems such that the neutrons ecome stale in them?. Why and how is that situation not eing created in nuclei/systems having, e.g. two-neutrons, three-neutrons and one-proton etc.?. Why and how does neutron have unstale and stale, oth the states, while the rest of all the elementary particles have only one state, either stale or unstale? 5. Why and how does neutron survive for time t = ± 0.8 seconds (mean life time of neutron) and then decays, while the rest of all the unstale elementary particles decay within fraction of a second? But the standard quark model and the other neutron models fail to give explanation of the aove questions. Therefore, presently a new model for neutron structure has een proposed that gives clear and complete explanation of all the aove questions along with very clear and complete explanation of the following numerous greatly important phenomena/events: 1. Why and how β particles, which are electrons, are emitted from the nuclei during β decay while it is elieved that the electrons do not reside inside the nuclei;. Why and how energy of the emitted β particles varies in the form of a continuous energy spectrum;. Why and how the neutrons have high penetrating power and distinguishale low and high-energy ranges;. How one-neutron and one-proton are arranged in a deuteron such that a inding force is generated etween them which persists and consequently deuteron exists in nature; 5. Why and how di-proton and di-neutron do not exist in nature; 6. Why and how inding energy per nucleon ( E ) of H and are increased to > E of deuteron, and E of H > E of ; 7. Why and how H is radioactive, decaying into through β decay; 8. How two-neutrons and two-protons are arranged in anα particle such that it persists and ehaves like a particle and eams ofα particles are otained despite having repulsive Coulom force etween them; 9. Why and how E ofα particle is increased to > 6 E of deuteron, instead of increasing to E of deuteron; 10. How are nucleons arranged in nuclei having mass numer (A) integer multiple of such that the nuclei are most strongly stale; 11. Why and how E 8 of Be < E of, while E of nuclei increases as their A increases; 1. Why and how other nuclei are not strongly stale; 1. Why and how near A = 6, E is maximum and then it gradually decreases as A increases and ultimately for A > 00, the nuclei ecome radioactive andα, β,γ,ν are emitted from them; 1. Howγ andν otain particle like physical existence and so high energy and penetrating power. These explanations give almost a complete understanding of neutron, deuteron,α particle and nuclei (structure, staility and decay). Finally: 1. An important conclusion has een drawn that the strength of staility of a nucleus does not depend only upon its E ut also upon the strength of staility of its neutrons ecause the later one too varies;. The existence of antineutrino has een question marked. a) kjnarain@yahoo.co.in ; kjnarain@hotmail.com ) Former address: Physics epartmentovt. Bilasa Girls P.G. (Autonomous) College, Bilaspur (C.G.) 95001, INIA

2 1. INTROUCTION It is elieved that the neutrons are unstale in their free state, ut in stale nuclei/systems (e.g. deuteron, α particle), they ecome stale. But it raises question, do the neutrons ecome stale due to the staility of nuclei/ systems, or in stale nuclei/systems, since the neutrons ecome stale, consequently the nuclei/systems too ecome stale? If the former part of the aove question is true, the question arises, how due to staility of nuclei/systems, do the neutrons ecome stale? If the later part is true, the question arises, how do the neutrons ecome stale and how consequently do the nuclei/systems ecome stale? It is also elieved that the β particles are emitted from the unstale nuclei, not from the stale nuclei. But it raises several questions, e.g., do the nuclei ecome stale and hence the β particles stop emitting from them, or when the β particles stop emitting, is it assumed that the nuclei ecome stale? Secondly, how and from where are the β particles emitted? If, according to quark model [1], the β particles are emitted from the neutrons as the consequence of decay of their down quarks into up quarks and the + β particles are emitted from the protons as the consequence of decay of their up quarks into down quarks, then the protons too should have unstale and stale oth the sates as the neutrons have. But, on the contrary, the protons are always stale. There are numerous such questions ut we find their no explanation. The quark model [, ] is supposed to e the standard model of particle physics, ut that too fails to give explanation of these questions. Further, we find almost no explanation of: 1. Why and how β particles, which are electrons, are emitted from the nuclei during β decay while it is elieved that the electrons do not reside inside the nuclei;. Why and how energy of the emitted β particles varies in the form of a continuous energy spectrum;. Why and how the neutrons have high penetrating power and distinguishale low

3 and high-energy ranges;. How one-neutron and one-proton are arranged in a deuteron such that it persists and hence exists in nature; 5. Why and how di-proton and di-neutron do not exist in nature; 6. Why and how inding energy per nucleon ( E ) of H and are increased to > E of deuteron, and E of H > E of ; 7. Why and how H is radioactive, decaying into through β decay; 8. How two-neutrons and two-protons are arranged in anα particle such that it persists and ehaves like a particle and eams ofα particles are otained despite having repulsive Coulom force etween them; 9. Why and how E ofα particle is increased to > 6 E of deuteron, instead of increasing to E of deuteron; 10. How are nucleons arranged in nuclei having mass numer (A) integer multiple of such that the nuclei are most strongly stale; 11. Why and how E 8 of Be < E of, while E of nuclei increases as their A increases; 1. Why and how other nuclei are not strongly stale; 1. Why and how near A = 6, E is maximum and then it gradually decreases as A increases and ultimately for A > 00, the nuclei ecome radioactive andα, β,γ,ν are emitted from them; 1. Howγ andν otain particle like physical existence (as the electrons have, otherwise, the collisions etween electrons and γ photons and hence the Compton s effect cannot take place) and so high energy and penetrating power. Few phenomena have een tried to explain using the quark model, ut there arise numerous such questions which raise serious question mark over the truth of this model (see section-.9). The present model gives very clear and complete explanation of all the questions mentioned aove and in the astract, and of all the phenomena/events mentioned aove (see sections- to 9). [The rest of other properties of neutron, deuteron,α particle and nuclei, and the related phenomena/ events too can e explained.] These explanations give almost a complete understanding of neutron, deuteron,α particle and nuclei (structure, staility and decay).

4 . THE PRESENT MOEL The electron possesses magnetism y virtue of nature similarly as it possesses its charge -e y virtue of nature (see reference-). This magnetism occurs in the form of a circular ring, shown y a dark solid line circle, Fig. 1, around the charge of electron, where the charge has een shown y a spherical all, Fig. 1(a), and y a thick dark circle, Fig. 1(). Around the magnetic ring, there occurs magnetic field. The magnetic ring and the all of charge of the electron oth spin ut in directions opposite to each other. (How and from where the magnetic ring and the all of charge of the electron otain their spin motion and how their spin motion persists, see reference-.) The spin (intrinsic) magnetic moment ( µ ), which the electron possesses, arises due to the spin motion of its magnetic s ring and occurs in the direction of spin angular momentum of the magnetic ring L sm (see reference-). The existing concept that µ of electron arises due to spin motion of its charge -e is not true []. s The frequencies of spin motions of the all of charge and of the magnetic ring of the electron happen to e such that the spin angular momentum of the all of charge of electron L, is greater than the spin angular momentum of the magnetic ring of electron L sm. The spin angular momentum L, which the electron (as whole) possesses, happens to e the resultant of these two, i.e. s L s = L sc - L sm. Consequently the electron possesses its linear motion in the direction of spin angular momentum of its charge, i.e. in the direction of L sc (ecause the spinning particles, e.g. electron, proton etc. possess linear motion in the directions of their respective spin angular momentum, see reference-5). The proton too possesses magnetism y virtue of nature in the form of a circular ring around the charge of proton similarly as the electron possesses, Fig. 1, and this magnetism and the charge of proton oth spin in directions opposite to each other. The magnetic moment ( µ ), the proton s sc

5 5 possesses, arises due to the spin motion of its magnetic ring and occurs in the direction of spin angular momentum of the magnetic ring ( L sm ). The frequencies of spin motions of the all and of the magnetic ring of the proton happen to e such that the spin angular momentum of magnetic ring ( L sm ), is lesser than the spin angular momentum of the all of proton ( L sc ). Consequently the spin angular momentum L s, which the proton (as whole) possesses, happens to e = L sc - L sm, and the proton possesses its linear motion in the direction of spin angular momentum of its all ( L sc ). The frequency, say ω ', corresponding to resultant spin angular momentum L of proton, happens to e the actual frequency, which the proton possesses. The magnetic moments ( µ ) of electron ( µ se ) and proton ( µ sp ) depend upon: 1- Amount of s magnetism in their magnetic rings; - Frequencies of spin motion of their rings. The frequency of spin motion of the magnetic ring of the electron (and similarly of proton) depends upon the magnitude of interaction etween the magnetic and electric fields respectively of magnetic ring and of the all of charge of the electron (and proton). And the magnitude of interaction depends upon: i- Amount of charge in the all; ii- Frequency of spin motion of the all; iii- Amount of magnetism in the ring. As we know, lighter the all, it spins with greater frequency as compared to a heavy all if they find same amount of energy to spin, therefore, the all of charge of the electron spins with much greater frequency as compared to that with which the all of charge of the proton spins ecause the electron is aout 10 less massive than the proton. And consequently, µ se is aout 7 10 greater than µ sp. The amount of magnetism the proton possesses too may e greater than the amount of magnetism the electron possesses and it may contriute in increasing µ, ut its contriution cannot increase µ aout 7 10 times. sp s sp

6 6 In the structure of neutron, the electron and proton are set such that the direction of Lsm of the proton lies opposite to the direction of Lsm of the electron, and similarly the direction of Lsc of the proton lies opposite to the direction of Lsc of the electron. Since the electron and proton oth possess their linear motion along the directions of their respective L, in the neutron, they travel in directions opposite to each other. Then, it is ovious that they front on collide with each other. When they collide, since the proton is much heavier (aout 10 times), oviously it shall e larger too in size than the electron, the proton is neither eing pushed ehind nor stops moving. It goes on moving forward in the direction of its linear motion ut its velocity is of course eing reduced (according to the law of conservation of momentum). The electron stops moving forward and its velocity is eing reduced to zero. Then oviously it is dragged along with the proton in the direction of linear motion of the proton. After collision with the proton, since the velocity of electron is reduced to zero, the frequency of spin motion of its all (of charge), say sc ω c, starts decreasing according to expression [5] m v =hω (where h is Planck s constant, m, v, and ω are respectively the mass, linear velocity and frequency of spin motion of the particle), ecause, when v of a particle starts decreasing, its ω starts decreasing according to this expression. The frequency of spin motion of the all ( ω c ) of electron goes on decreasing till a situation arises when Lsc of the electron is reduced as much that Lsm of the electron starts ecoming greater than its (electron) L sc. And when the difference ( L sc ) is increased as much that the electron can move in the direction of its L - sm L sm with velocity greater than the velocity of proton and against the attractive Coulom force y the proton on it, it is separated from the proton and starts moving in the direction of its L sm. And, in this way, the neutron decays into a proton and an electron. (When the neutron ecomes stale, the separation of its electron from its proton is stopped, see section-..)

7 7 After, or may e efore the separation of electron from the proton, the frequency of spin motion of the charge of electron starts increasing again, and during the course of motion of electron in the direction of its L sm, a situation comes when its L sc starts ecoming greater than its L sm. When the difference ( Lsc - L sm ) is increased as much that the electron can move towards it L sc, it starts moving again towards it L, i.e. towards the proton, and comines with the proton. Thus the whole sc process (i.e. the comination of electron with the proton, their separation, and their comination again) goes on continuously. uring this process, since the whole system (i.e. electron and proton of the neutron) continues moving in the direction of motion of the proton, the neutron possesses its linear motion in the direction of linear motion of the proton. (When the electron and proton collide, how their energy and momentum etc. conserve, see section-9...).1 Note Currently it is elieved that the electrons do not reside inside the nuclei. They are generated inside the nuclei at the time of β decay. This elief has een developed due to the following reasons: i- Finite size; ii- Spin angular momentum; iii- Statistics; iv- Magnetic moment; v- Principle of uncertainty; vi- Compton wave length etc. of electron. But the present model resolves all the aove first five reasons (see section-). Sixth reason is also resolved ecause the electrons (and other matter particles too) do not have wave nature (for detail, see reference-5).. EXPLANATION OF THE PROPERTIES OF NEUTRON AN OF SOME RELATE PHENOMENA/EVENTS.1 Mean life time (t) of neutron The time elapsed in etween when the proton and electron comine (collide) with each other and when they are separated from each other, happens to e the mean life time (t) of neutron.. Why and how t happens to e = ± 0.8 seconds

8 8 uring the course of time in etween when the proton and electron comine (collide) with each other and when they are separated from each other, the following two events take place: i- The velocity of the electron is reduced to zero; ii- Consequently the frequency of spin motion of the charge of electron and hence L starts decreasing, and when L is reduced as much that the electron sc can move in the direction of its L sm with velocity greater than the velocity of proton, the electron is separated from the proton (see section-). The first event of course takes no time ut the second event takes time and that happens to e = ± 0.8 seconds [6].. Why and how neutron has stale and unstale oth the states In the free state of neutron, after time t the electron and proton are separated from each other, i.e. the neutron decays, and therefore the neutron happens to e unstale in its free state. In stale nuclei and systems, the separation of electron from the proton is stopped (why and how the separation is stopped, see sections-, 5, and 6), consequently the neutron ecomes stale. Therefore, the neutron has stale and unstale oth the states.. Magnetic moment of neutron According to existing expression µ s = ( q / m) Ls for the spin magnetic moment ( µ s ) of a spinning particle (having charge q, mass m, and spin angular momentum L s ), since the neutron has no charge, its spin magnetic moment ( µ ) should e zero. But on the contrary, the neutron has spin sn magnetic moment. For that, it is argued that neutron is not a charge less particle ut has net charge = 0. It means, neutron is constituted y two or more than two particles, each having charge and µ s such that the resultants of their charge and of their µ s give respectively the zero charge and µ sn (= J/T ) of neutron. sc

9 9 Suppose if it is argued that the constituent particles of neutron have charge ut do not have spin magnetic moment, and when they constitute the neutron, due to spin motion of its net charge, the neutron otains its µ, this argument cannot e accepted. Because: 1- When the net charge on sn the neutron ecomes zero and it spins, the zero charge spins, and hence µ should e zero; - The questions arise, how and from where does the net charge otain spin motion and how does that (spin motion) persist? Because for persistent spin motion, the net charge must have infinite energy or some source of infinite energy, how and from where does the net charge otain that? If it is argued that, efore constituting the neutron, the constituent particles possess spin motion, then they must have their spin magnetic moments too. Because when they have charge and possess spin motion, they must have their spin magnetic moments. The particles which constitute a neutron are a proton and an electron (see section-). Electron and proton have charge e and + e, and spin magnetic moment µ se and µ sp respectively (see section- ), and when comining with each other they constitute a neutron, the net charge of neutron ecomes = 0 and the net spin magnetic moment (µ sn ) = µ sp ± µ se. In the structure of neutron, since the directions of Lsm of electron and proton are in opposite directions (see section-), and the directions of µ se and µ sp of electron and proton lie respectively in the directions of their L sm, µ se and µ sp lie in directions opposite to each other. And hence µ sn = µ se (= J / T ) - µ sp (= J/T) = J/T, and it should occur in the direction of µ se ecause µ se > µ sp. The experimental value of µ = J/T has same sign as the theoretical (presently sn otained) value of µ has, ut is much lesser (aout 7 10 ) than the theoretical value ( sn 10 - J/T). The reason is proaly as follows: sn

10 10 The proton has same amount of charge (+e) as the electron has (-e), ut µ is aout 10 times lesser than µ se. The decrease of aout 10 times in the value of µ sp is due to having aout 10 times more mass y the proton in comparison to that of the electron. Since the neutron too is aout 10 times more massive than the electron, µ sn is reduced y aout 7 10 times. This conclusion cannot e overruled ecause, as the net charge of neutron is zero, it means, when electron and proton comine with each other, though they do not merge into a single particle ut comine such that the resultant comination (neutron) ecomes just like a single particle. Further, we find that µ sn is a little < µ sp while m n (mass of neutron = Kg) is a little > m (mass of proton = Kg), it confirms that due to increase in mass of the resultant system (i.e. neutron) y aout 10 times, µ sn is reduced y aout 7 10 times..5 While it is elieved that the electrons do not reside inside the nuclei, then why and how the electrons are emitted from the nuclei during β decay In the structure of a neutron (say N 1 ), after collision of its electron (say E 1 ) with its proton (say P 1 ), since the velocity of proton P 1 is decreased, hence as soon as the tendency in the electron E 1 to move along the direction of its L sm is developed, the velocity of proton P 1 starts increasing again, and when the electron is separated from the proton, the rate of increase in the velocity of proton is increased and the proton tries to regain the original value (say V) it had efore its collision with the electron. nce, the distance gap, which is created in etween the proton and the electron after their separation and when the electron is just to start moving again towards the proton, happens to e very short. Therefore, though the process of separation of electron from the proton and their recomination goes on continuously, ut together they ehave just like a single particle. sp p

11 11 In the stale nuclei and systems (e.g. deuterons and α particles), the neutrons and protons are so arranged and ound with each other such that the continuous process of separation and recomination of electrons with protons in the structures of their respective neutrons are eing stopped (for detail, see sections- to 6). nce, inside the nuclei and systems, the electrons are not found in states as, e.g., protons are found. Therefore, it can e said or concluded that the electrons do not reside inside the nuclei. They reside as a part of the structure of neutrons. In nuclei when A ecomes > 00, the stailities of neutrons start getting weak (see section- 9..1), i.e. in the structures of neutrons, the stale cominations of their electrons with their respective protons start getting loose. Then the separation and recomination of their electrons with their protons start again And when in any neutron, the comination of its electron with its proton is roken and the electron is separated from the proton, if this electron is found in position, as shown in Fig. (), with a group G (see section-7), the electron is ejected from its neutron and also from the nucleus, and thus a β decay is otained (for detail, see section-9..1). The ejection of electron is caused due to the repulsive force [] etween the electron and the group G..6 Why and how β particles emitted from the radioactive sources have continuous energy spectrum When a neutron (say N 1 ) ecomes unstale and its electron E 1, after getting separated from its proton P 1 starts moving away from the proton, the energy of the electron varies continuously and similarly when the electron starts moving again towards the proton to comine with that, then too its energy varies continuously. [The course of time t (mean life time of neutron), during which the electron remains in contact with the proton, then too the energy of the electron varies continuously due to the variation in frequency of spin motion of the charge of electron, ecause it varies the spin energy of the electron.] The situation, Fig. (), etween the electron and the group G can come at

12 1 any instant during the course of time, say t (when the electron is moving away from the proton), or during the course of time, say t (when the electron is moving towards the proton). Therefore, the electron can e ejected at any instant during the course of time t + t and hence accordingly the ejected electron possesses its energy. (The emitted electron possesses that energy too, it otains y the repulsive force.) The energy of the electron since varies continuously throughout the full course of time t + t, the energy of the emitted electron (i.e. β particle) too varies accordingly and hence the energy of the emitted β particle is otained in the form of a continuous energy spectrum..7 Why and how the neutron has high penetrating power In order to explain why neutrons have high penetrating power, let us first take an example. We take two ullets (spherical/cylindrical) of same mass, size and sustance, and to one ullet we give a conical shape at its front side. If these ullets are fired with the same energy on the same target from the same distance one after the other, we shall find that the depth of penetration of the ullet, having conical shape at its front side, is more as compared to the depth of penetration of the other ullet. In the structure of neutron, since the electron lies always in front of the proton during its (proton) motion (whether the electron is moving away from the proton getting separated from that or moving towards that), and the electron is much lighter than the proton (then oviously the electron shall e smaller too), the electron produces almost the same effect as the conical shape at the front of the ullet produces. Consequently the neutrons possess high penetrating power. Further, the neutron possesses motional energy [5] M.E. {= K.E. (kinetic energy) + S.E. (spin energy [5])} = M.E. of proton + M.E. of electron, and motional momentum [5] M.M. {= L.M. (linear momentum) + S.M. (spin momentum [5])} = M.M. of proton + M.M. of electron. The M.E. and M.M. of the electron too increase the penetrating power of the neutron. Thus the penetrating power of a neutron is increased quite a lot.

13 1.8 Why and how the neutron has distinguishale low and high energy ranges In the structure of neutron, the course of time t, during which the electron moves away from the proton after separation from that, the neutron possesses M.E. = continuously increasing M.E. of proton (ecause, after collision, the proton immediately starts trying to regain its original velocity V, which it had efore its collision with the electron, see section-.5) + varying K.E. of electron [which varies accordingly as its difference ( Lsm - L sc ) and the Coulom attractive force acting on it due to positive charge on the proton, vary] + S.E. of the magnetic ring of electron S.E. of the all of charge of electron. Therefore, accordingly the effective M.E. of neutron varies and it (effective M.E. of neutron) happens to e continuously increasing. And the course of time t, during which the electron of the neutron moves towards its (neutron) proton, the neutron possesses M.E. = increasing M.E. of proton continuously increasing K.E. of electron [which varies accordingly as its difference ( Lsc - L sm ) and the Coulom attractive force on it vary] + S.E. of the magnetic ring of electron - S.E. of the charge of electron. Therefore, accordingly M.E. of neutron varies and it (M.E.) happens to e continuously decreasing. The course of time t, during which the electron of the neutron remains in contact with its (neutron) proton, then too the effective M.E. of neutron varies, ecause, after collision of the electron with the proton, since the proton immediately starts trying to regain its original velocity V, the M.E. of proton goes on increasing. And also (after collision) since ɷ of the all of charge of electron starts decreasing, which goes on decreasing till Lsm ecomes greater than Lsc and the difference ( L sm - L sc ) is increased as much that the electron starts moving in the direction of it s S.E. of the all of charge of electron goes on decreasing. Therefore, M.E. of neutron varies, as the increasing M.E. of proton the decreasing S.E. of the all of electron varies. This variation occurs in the form of a gradual increasing order. L sm,

14 1 Thus, during the course of time t + t, M.E. of neutron goes on increasing, and during the course of time t, it goes on decreasing. The moment when the disintegration takes place, the neutron may e at any instant of the courses of time t + t + t. Consequently, accordingly the neutron possesses its energy. If the otained neutron is of the course of time t + t, it possesses an energy range which happens to e of increasing order, i.e. high-energy range. And if the otained neutron is of the course of time t, it possesses an energy range ut happens to e of decreasing order, i.e. low-energy range..9 iscussion According to Quark model [], a neutron is composed of two down quarks ( d 1 and d ), each having charge [] e/ and mass [].1 to 5.8 MeV/c, and one up quark (u ), having charge [] +e/ and mass [] 1.7 to. MeV/c, and thus has zero net charge. The up and down quarks in the neutron are arranged as ( u d 1 d ), Fig. []. A proton is composed of one down quark ( d ) and two up quarks ( u 1 andu ) and thus has +e net charge and the quarks are arranged as ( u 1 d u ), Fig.. This model explains the eta decay successfully, i.e., a down quark d decays into a lighter up quark u emitting a virtualw oson [] having charge [7] -e and mass [7] ± 0.0 GeV/c, and W oson decays into an electron ( β ) and an antineutrino, Fig.. But this model gives rise to numerous very serious such questions of which no explanation can e given. For example: 1- How (i.e. mechanism) does the down quark d decays into the lighter up quark u emitting a virtual oson? If it is argued that it occurs due to weak interaction [], then the questions arise, does the weak interaction take ± 0.8 s? Or especially in this case does it take ± 0.8 s? If especially in this case it takes ± 0.8 s, then the question arises why and how? Further, does the weak interaction take place within the quark d itself or etween the quarks d 1 and d or among the W

15 15 quarks u, d1 and d? If it takes place etween the quarks d 1 and d or among the quarks u, d1 and d, then what does happen to the electrostatic Coulom interaction? Since the electrostatic Coulom interaction is times stronger than the weak interaction, the role of electrostatic Coulom interaction cannot e ignored or overruled. Then how does the weak interaction come into play? If the weak interaction takes place within the quark d itself, how, why and what does happen within the quark d that it decays into a lighter up quark u emitting a virtual W oson? Further, the assumptions, e.g.: 1. The decay of quark d having charge e/, into a quark u having charge +e/ emitting a virtual possession of mass y W oson;. Emission of a virtual W oson;. Even eing a virtual particle, W oson, that too aout 10 times more than that of the mother quark d ;. Even eing a virtual particle, possession of charge y d has only charge e/; 5. ecay of virtual electron W oson, that too e while the mother quark W oson, which physically does not exist, into a real β, which physically exists; are puzzling. These assumptions are unelievale and hence cannot e accepted. How, y which mechanism and/or according to which scientific (physical/ chemical) law, do the aove events take place? As the consequence of decay of a quark d into a quark u, a real particle of mass (.1 to to.) MeV/c can e emitted, or, according to mass-energy equivalence principle (theory of relativity), an energy (.1 to to.) MeV can e emitted, ut the occurrence of the aove events is not possile to elieve. What are the physical interpretations of: 1. Virtual charge y virtual W oson;. Possession of mass and W oson;. Emission of virtual W oson as the consequence of decay of a real quark d ;. ecay of that virtual W oson into a real electron ( β ) etc.? As far as the author s knowledge is concerned, it is elieved that there exist only matter and energy in the universe and they are inter-convertile. In which category does the virtual W oson lie?

16 16 Further, why and how does the decay of quark d into a quark u take place only in unstale nuclei, why and how not in stale nuclei/systems? Why, how and what does happen in stale nuclei/ systems such that the decay of quark d stops? Furthermore, how does the neutron otains its spin magnetic moment ( µ )? Suppose if it is argued that when the neutron spins, since its all the three quarks (one u quark and two d quarks) which have charge spin, the neutron acquires µ, this argument cannot e accepted. Because one u sn quark two d quarks of a neutron give its net charge = 0, and when they (one u and two d ) spin, zero charge spins and hence µ should e zero. Secondly, how and from where do the neutrons (or sn quarks) otain their spin motion and how does that persist? How and from where do they otain infinite energy for their persistent spin motion?. EXPLANATION OF WHY AN HOW NATURE HAS PROVIE US ONLY EUTERON (NP) WHILE THE BOUN STATES, I-PROTON (PP), I-NEUTRON (NN) ARE ALSO THEORETICALLY POSSIBLE BUT O NOT EXIST.1 Why and how nature has provided us deuteron (NP). When the difference ( L sm - L sc ) is increased as much that the electron E 1 [of neutron N 1, Fig. (a)] can move in the direction of its L sm with velocity greater than the velocity of proton P 1 (of neutron N 1 ) against the attractive Coulom force y the proton, the electron is separated from the proton and starts moving in the direction of its L sm (see section-). If y some means, the separation of electron from the proton is stopped, the electron shall remain with the proton, i.e. the neutron shall remain stale till the electron is separated again y some means or due to some reason, as, e.g., happens when A of the nuclei ecomes > 00 (see section-9..1). The separation of electron from the proton can e stopped if during the process of reducing L sc (see section-) ut efore the difference sn

17 17 ( Lsm - L sc ) is increased as much that the electron is separated from the proton, the effect of Coulom force of attraction on the electron y the proton and the velocity of the proton are increased as much that the electron may not e separated from that. The effect of Coulom force of attraction and the velocity of proton are increased if a proton P, moving parallel to a neutron N 1 and in the same direction in which the neutron N 1 is moving, comes in the plane of proton P 1 (of neutron N 1 ), adjacent and so close to P 1 that the magnetic fields around protons P and P 1 start interacting [], as shown in Fig. (). Because then, due to +e charge on proton P, the effect of attractive Coulom force y proton P 1 on electron E 1 (of neutron N 1 ) is increased, and due to interaction etween the magnetic fields of protons P 1 and P a inding force F is generated etween the protons P 1 P and they are ound together, Fig. () [for detail, see reference-]. When they are ound together, proton P increases the velocity of proton P 1 [which was earlier reduced due to its (proton P 1 ) collision with electron E 1, see section-] y dragging proton P 1 along with it. The increases in the velocity of proton P 1 and in the effect of Coulom force of attraction do not let electron E 1 to get separated from proton P 1. Consequently, electron E 1 and proton P 1 remain in contact with each other, i.e. neutron N 1 ecomes stale. Thus in this comination of a proton P and a neutron N 1 (i.e. in system N 1P ), Fig. (a), neutron N 1 ecomes stale, and with this stale neutron, a proton P is ound with a inding force F. The inding force F etween the proton P and neutron N 1 generates the total inding energy ( E t ) and the inding energy per nucleon ( E ) of deuteron. The magnetic field around proton P interacting with the magnetic field around proton P 1 (of neutron N 1 ) creates a magnetic field around them, having direction as shown in Fig. (). The outer

18 18 portion of the magnetic field otained around them (i.e. deuteron ) possesses the shape and direction as shown in Fig. (c). Further, in deuteron ( ), since oth the nucleons ( P and N 1 ) possess their linear motion v in the same direction and parallel to each other, too possesses its linear motion v in the same direction, Fig. (c).. Why and how system di-proton (PP) does not exist in nature, and why and how system deuteron (NP) exists. In the system di-proton ( P P ), Fig. (d), proton 1 P1 is not ound in a neutron (i.e. P1 is not like proton P 1 of neutron N 1 ) ut it happens to e free as proton P is. nce, in system P 1 P, there takes place interaction etween two free protons P and P 1. The interaction etween the magnetic fields of protons P and P 1 takes place in the same manner, Fig. (), and a inding force F is generated etween protons P and P 1, ut this force F does not happen to e as strong as it happens in the system N 1P. The inding force F consists two components F 1 and F, i.e. F = F 1 ± F (see reference- ), where component F1 is generated due to interaction etween the magnetic fields of protons P and P 1, and it happens to e attractive. And component F is the Coulom force generated due to interaction etween the charges on the interacting protons P and P 1. In the case of system P 1 P, F happens to e repulsive ecause the charges on protons P and P 1 are similar. Therefore, the force F happens to e = F1 F = F P 1 P. Force FP 1 P proaly happens to e not sufficiently strong consequently system P P 1 does not persist and very shortly the protons P and P 1 are separated from each other. And hence the system P P does not exist in nature. [In system P P 1 1, if a neutron N is added, the resultant system ecomes stale (see section-5.) and it s E is increased to > (E ) (see section-5.)]

19 19 In the case of system N 1P, Fig. (a and ), due to charge e on electron E 1 (of neutron N 1 ), the effect of charge +e of proton P 1 is reduced quite a lot (ut not to zero, the net charge of neutron N 1 is of course reduced to zero). Consequently, the repulsive component F is reduced very much, say to F '. Therefore, the force F etween N 1P, i.e. FN 1 P ecomes = F ' 1 F, and it happens to e > F P 1 P. Force F N 1 P proaly happens to e sufficiently strong, consequently the protons P and P 1 (of neutron N 1 ) of system N1P are not eing separated from each other and system N 1P persists (i.e. ecomes stale). And hence system NP (deuteron) exists in nature.. Why and how the system di-neutron (NN) does not exist in nature In the system di-neutron ( N 1N ), Fig. (e), since oth the protons P 1 and P are of neutrons N 1 and N respectively, none is happened to e free (as proton P happens to e free in system N 1P ), there occurs no means to increase the velocities of protons P 1 and P so that the electrons E 1 and E respectively may not e separated from the protons P 1 and P. Therefore, after the mean life time (t) of neutrons N1 and N, the electrons E 1 and E are separated from their respective protons P 1 and P. Then the protons P 1 and P are left ehind, i.e. the situation ecomes exactly like the system P 1 P. And hence the interaction takes place etween protons P 1 and P, and the inding force F, generated due to their interaction, happens to e = F1 F = F P 1 P. Since the force FP 1 P happens to e not sufficiently strong (see section-.), very shortly the protons P and P 1 of the system are separated from each other and hence system N1N Consequently the system NN does not exist in nature. [In system N 1N does not persist., if a proton P is added, the resultant system ecomes stale (see section-5.1) and it s E is increased to > (E ) (see section- 5.).]

20 0 If oth the neutrons N1 and N do not decay simultaneously ut decay one y one, e.g. N decays first and afterwards N 1, then as soon as N decays, proton P (of neutron N ) ecomes almost free like proton P of system N1P and it (proton P ) increases the effect of attractive Coulom force of proton P 1 on electron E 1 (of neutron N 1 ) and the velocity of proton P 1 (of neutron N 1 ) y dragging proton P 1 along with it. ue to these increases, electron E 1 (of neutron N 1 ) is not separated from proton P 1, and neutron N 1 ecomes stale. And system N1N is converted into system N 1P.. iscussion No explanation is found why and how nature has provided us only deuteron ( NP ), while the ound states of two-nucleon systems, di-proton ( PP ), di-neutron ( NN ) too are theoretically possile ut do not exist in nature. If we try to explain these on the asis of the current strong short range nuclear interaction (which is assumed generated according to Yukawa s meson field theory [8]) and Coulom interaction, we succeed to explain only why and how nature has provided us only NP, ut fail to explain why and how PP and NN do not exist in nature. For non-existence of system PP, for time eing it can e assumed that the Coulom repulsive interaction etween PP reduces the strength of the current strong short range nuclear interaction etween them and consequently the strong short range nuclear interaction fails to keep PP ound together (though practically it is not possile ecause the Coulom repulsive interaction cannot reduce the aout 100 times stronger short range nuclear interaction so much). But in the case of system NN, where is no possiility of the Coulom repulsive interaction to come into play etween NN, system NN must exist. While on the contrary, it too does not exist. 5. EXPLANATION OF WHY AN HOW UE TO AITION OF ONE PROTON (P) AN ONE NEUTRON (N) IN THE SYSTEMS NN AN PP RESPECTIVELY, THE RESULTANT SYSTEMS, i.e. THE NUCLEI OF H AN BECOME STABLE AN THEIR E

21 1 BECOME > (E ), WHY AN HOW E ) > E ), WHY AN HOW H IS ( H ( RAIOACTIVE AN ECAYS INTO THROUGH β ECAY 5.1 Why and how due to addition of one P in the system NN, the resultant system, i.e. nucleus of H ecomes stale When a proton P is added in the system N1N and all the three nucleons are arranged in the same plane, very close and adjacent to each other, as shown in Fig. 5(a), and they are having their linear motion v in the same direction and parallel to each other, the resultant system i.e. the nucleus of H ecomes stale. Because then, due to interaction etween the magnetic fields of protons P, P 1 (of neutron N 1 ) and P (of neutron N ), all the three nucleons are ound together in the form of a group (see reference-), sayt 1. And further, since the linear velocity of proton P is happened to e > the linear velocities of protons P 1 and P [ecause the linear velocities of protons P 1 and P were reduced due to their collisions with their respective electrons E 1 (of neutron N 1 ) and E (of neutron N ), see section-], the protons P 1 and P are eing dragged along with proton P. Consequently, the linear velocities protons P 1 and P are increased while that of proton P is decreased and ultimately they all acquire the same velocity v. And due to +e charge on proton P, the effects of attractive Coulom force y the protons P 1 and P on the electrons E 1 and E respectively are increased. So, due to increase in the velocities of protons P 1 and P, and in the effects of attractive Coulom force, the electrons E 1 and E are not eing separated from their respective protons P 1 and P. And thus the neutrons N1and N ecome stale. When all the three nucleons are ound together in the form of a groupt 1 and the neutrons N1and N ecome stale, the whole system ecomes stale.

22 ue to interaction etween the magnetic fields of protons P, P1 and P, a magnetic field is created around them, Fig. 5(c). The outer portion of the magnetic field otained around them (i.e. groupt 1 ) possesses the shape and direction as shown in Fig. 5(g). Further, in groupt 1, since all the three nucleons P, N1and N possess their linear motion v in the same direction and parallel to each other, groupt 1 too possesses its linear motion v in the same direction. 5. Why and how E of the resultant system (nucleus of H ) ecomes > (E ) In the nucleus of H, when the neutrons N1and N ecome stale, two deuterons 1 (having nucleons P and N 1 ) and (having nucleons P and N ) are created, Fig. 5(), where proton P is common in oth the deuterons 1 and. These two deuterons 1 and are ound together y the inding force etween N 1 and N at their one ends, and at their other ends ound together due to having common P etween them. ue to having common P etween them, the inding generated etween the deuterons 1 and happens to e stronger than the inding generated etween them due to interaction etween N 1 and N. [In order to understand it more clearly, let us take an example. We consider two families A and B. Family A has two women (sisters), each having one son and family B has one woman having two sons. In family A, since the two women are sisters and two oys are cousins, the two women are ound with each other and similarly two oys are ound with each other. But in family B, since the mother is common etween the two oys, inding among the three memers happens to e stronger in comparison to inding among the four memers in family A] as follows: Therefore, the inding energy per nucleon ( E ) for the nucleus of H, i.e. E ) is otained ( H ( ) H E = [ ( i. e. total inding energy) of the nucleus of H ]/ E t

23 = [ ( E) + ( E) + ( E generated due to the inding force FN N ) generated due to common P etween the two deuterons)]/ = ( E ) ]/ + [ ( E generated due to theinding force FN N ) + ( E generated ecause ( E ) = ( E ) 1 = ( E ). [ 1 due to common P etween the two deuterons)]/ The aove expression can e expressed as follows too: ( E ( E ) = ( E H ) ( E ) / + [ ( E generated due to the inding force FN N ) + 1 generated due to common Petween the two deuterons)]/ (5.1) ( E > (E ). (5.) ecause E generated due to the inding force F N 1 N > ( E ) [see section-5.], and E generated due to common P etween the two deuterons > E generated due to the inding force FN 1 N and hence [( Egenerated due to the inding force F N 1 N ) + ( E generated due to common P etween the two deuterons)]/ happens to e > ( E ) /. 5. Why and how due to addition of one N in the system PP, the resultant system, i.e. nucleus of ecomes stale When a neutron N is added in the system P 1 P and all the three nucleons are arranged in the same plane, very close and adjacent to each other, as shown in Fig. 5(d), and they are having their linear motion in the same direction and parallel to each other, the resultant system i.e. the nucleus of ecomes stale. Because then, due to interaction etween the magnetic fields of protons P 1, P and P (of neutron N ), all the three nucleons are ound together in the form of a group (see reference-), sayt. And further, since the linear velocities of protons P 1 and P are happened to e >

24 the linear velocity of proton P, proton P is eing dragged along with the protons P 1 and P. Consequently, the linear velocities of protons P 1 and P are increased while that of proton P is decreased and ultimately they all acquire the same velocity v. And due to +e charges on protons P 1 and P, the effect of attractive Coulom force y proton P on electron E is increased. So, due to increase in the velocity of proton P and in the effect of attractive Coulom force y the protons P 1 and P on electron E, electron E is not eing separated from proton P. And thus neutron N ecomes stale. When all the three nucleons are ound together in the form of a groupt and the neutron N ecomes stale, the whole system ecomes stale. ue to interaction etween the magnetic fields of protons P, P1 and P, a magnetic field is created around them, Fig. 5(f). The outer portion of the magnetic field otained around them (i.e. groupt ) possesses the shape and direction as shown in Fig. 5(g). Further, in groupt, since all the three nucleons N, P1 and P possess their linear motion v in the same direction and parallel to each other, groupt too possesses its linear motion v in the same direction. 5. Why and how E of the resultant system (nucleus of ) ecomes > (E ) In the nucleus of, Fig. 5(d), neutron N is made stale y two protons P 1 and P exactly in the same manner as two neutrons N1and N are made stale y a single proton P in the nucleus of When neutron N ecomes stale, two deuterons 1 (having nucleons N and P 1 ) and (having nucleons N and P ) are created, Fig. 5(e), where neutron N is common in oth the deuterons 1 and. These two deuterons 1 and are ound together y the inding force etween P 1 and P at their one ends, and at their other ends ound together due to having common N etween them. ue H.

25 5 to having common N etween them, the inding generated etween the deuterons 1 and happens to e stronger than the inding generated etween them due to interaction etween P 1 and P. Therefore, E for the nucleus of, i.e. E ) is otained as follows: ( ( ) E = [ of the nucleus of ]/ E t = = [ ( E) + ( E) + ( E generated due to the inding force FPP ) generated due to common N etween the two deuterons)]/ + 1 ( E ) / + [( E generated due to the inding force FP P ) ( E to common N etween the two deuterons)]/ ( E generated due = ( E ) ( E) / + [ ( E generated due to the inding force FP P ) + 1 ( E generated due to common N etween the two deuterons)]/ (5.) > (E ). (5.) ecause, thoughe generated due to the inding force F P 1 P < ( E ) [see section-.], ut, since Egenerated due to common N etween the two deuterons > ( E ), ( E generated due to the inding force F P 1 P + E generated due to common N etween the two deuterons)/ may happen to e or < ( E ) / ut can never e > ( E ) /. 5.5 Why and how E ) > E ) ( H ( If we compare the expressions (5.1) and (5.), we find that, in expression (5.1), E generated due to the inding force F N 1 N > ( E ) [see elow], while in expression (5.), E generated due to the inding force F P 1 P < ( E ) [see section-.]. Therefore, E ) {=.87 MeV [9]}> E ) {=.567 MeV [9]}. ( H (

26 6 How E generated due to the inding force F N 1 N > ( E ), is as follows: When the neutrons N1and N ecome stale, due to charge e on each electron E 1 and E, the effects of +e charge on the protons P 1 and P, Fig. 5(a), are reduced quite a lot, and due to that, the component F of the inding force F( = F1 F ) etween the neutrons N1 N too is reduced accordingly, say to F ". Presently, since the effects of +e charge on oth the protons P 1 and P are reduced, the resultant component F " ecomes < F ' (see section-.), and hence the inding force F etween N 1, N, i.e. F N 1 N ( = F " 1 F ), ecomes > F PN ( = F ' 1 1 F ) and F PN (ecause F PN = 1 F PN ). Therefore, E generated due to the inding force F N 1 N > ( E ). 5.6 espite E ) > E ), why and howh is radioactive and decays into through β decay ( H ( In the nucleus of H, Fig. 5(a), since a single proton P dragging two protons P 1 (of neutron N 1) and P (of neutron N ) increases their velocity, and the charge +e of a single proton P increases the effects of attractive Coulom force of two protons P 1 and P on their respective electrons E 1 and E, proton P somehow succeeds only to keep protons P 1 and P just in contacts with their respective electrons E 1 ande. Consequently, the neutrons N1and N in the nucleus of H are happened to e just stale (loosely stale). Since the neutrons N1and N in the nucleus of H are happened to e just stale, these can decay if y some external means their staility is loosened a little it more (how their staility is loosened and they decay, see section-9..1). Further, since it is very rarely possile that oth the neutrons N1and N decay simultaneously, they decay one y one. But as soon as one neutron, e.g. N 1 decays, proton P 1 (of neutron N 1 ) ecomes like proton P, and then they ( P 1 and P ) together increase the velocity of proton P (of neutron N ) and the effect of attractive Coulom force of it on the

27 7 electron E (of neutron N ). Consequently the staility of neutron N is increased and that does not decay and the nucleus of H is converted (decayed) into the nucleus of 5.7 An important Conclusion [10]. The decay of H into through β decay, despite ( ) H E > E ), gives an important ( conclusion that the strength of staility of a nucleus does not depend only upon its E ut also upon the strength of staility of its neutrons. The half-lives of isotope Li (= s [11]) and of the synthesized isotope H (= s [10]) confirm the aove conclusion. The three-protons ( P 1, P, P ) of isotope Li make its oneneutron ( N ), Fig. 5(h and i), very strongly stale ut three interacting forces ( F P 1 P, F P P, F P 1 P ) make the inding among its three protons and hence among its four nucleons very weak consequently its half-life is reduced to s and emitting a proton it decays into [9]. And in isotope H, its one-proton ( P ) fails to make its three-neutrons ( N 1, N, N ), Fig. 5(j and k), stale consequently even having very strong inding among its all the four nucleons due to the interacting forces ( F N 1 N, F N N, F N 1 N ), isotope H does not exist in nature. The half-life of the synthesized isotope H too happens to e very short = s and emitting a neutron, it decays into H [10]. 6. EXPLANATION OF HOW TWO-NEUTRONS AN TWO-PROTONS ARE ARRANGE IN A α PARTICLE SUCH THAT IT PERSISTS AN BEHAVES LIKE A PARTICLE, HOW BEAMS OF α PARTICLES ARE OBTAINE ESPITE HAVING REPULSIVE COULOMB FORCE BETWEEN THEM, AN WHY AN HOW IT S E IS INCREASE TO > 6 (E ) INSTEA OF INCREASING TO (E )

28 8 6.1 How two-neutrons and two-protons are arranged in a α particle such that it persists and ehaves like a particle, and how eams of α particles are otained despite having repulsive Coulom force etween them In the structure ofα particle, two neutrons N 1, N and two protons P 1, P are arranged as shown in Fig. 6(a), and all the four nucleons possess their linear motion v in the same direction and parallel to each other. In this structure, we see: i- The pair of one-proton and one-neutron P 1 N 1 ehaves like deuteron 1, N1 P ehaves like deuteron, P N ehaves like deuteron, and N P 1 ehaves like deuteron, Fig. 6(); ii- Every comination of two-protons and one-neutron, P1 N 1 P and P N P 1 ehaves like a groupt, Fig. 6(c); and iii- Every comination of two-neutrons and one-proton, N1 P N and N P 1 N 1 ehaves like a groupt 1, Fig. 6(d). As in the case of nucleus of (i.e. groupt ), its neutron N happens to e strongly stale, similarly in the cominations- P 1 N 1 P and P N P 1 of α particle, Fig. 6(c), neutrons N1and N are strongly stale. And, as in the case of nucleus of H (i.e. groupt 1 ), since its all the three nucleons are strongly ound with each other, similarly in every comination- N1 P N and N P 1 N 1 ofα particle, Fig. 6(d), all the three nucleons are strongly ound with each other. And thus, in a α particle, it s oth the neutrons are strongly stale and all the four nucleons are strongly ound with each other. Consequently, aα particle, despite eing a comination of four nucleons, ehaves like a single particle. Further, in α particle, since all its four nucleons possess their linear motion v in the same direction, α particle too possesses its linear motion v in the same direction. In α particle, due to interaction etween magnetic fields of all its four nucleons, a magnetic field is created around them, Fig. 6(e). The outer portion of the magnetic field otained around them (i.e. α particle or group G) possesses the shape and direction as shown in Fig. 6(f). The shape and

29 9 direction of magnetic field otained around a α particle, Fig. 6(f), are happened to e similar as otained around an electron, Fig. 1(). Consequently, α particles ehave just like electrons, protons etc., and their (α particles) eams can e otained despite having repulsive Coulom force etween them, similarly as electron eams are otained despite having repulsive Coulom force etween them (see reference-). 6. Why and how E of α particle is increased to > 6 (E ) instead of increasing to (E ) In the structure of α particle, Fig. 6(): i- Two deuterons 1 and are ound together at their one ends y the inding force etween P 1 and P, and at their other ends due to having common N 1 etween them; ii- Two deuterons and are ound together at their one ends y the inding force etween N 1 and N, and at their other ends due to having common P etween them; iii- Two deuterons and are ound together at their one ends y the inding force etween P 1 and P, and at their other ends due to having common N etween them; iv- Two deuterons and 1 are ound together at their one ends y the inding force etween N 1 and N, and at their other ends due to having common P 1 etween them. Therefore, Efor α particle, i.e. ( E ) α is otained as follows: ( E ) α = [ E t ( total inding energy) of α particle]/ = [ 1 ( E ) + ( E) + ( E) + ( E) + ( E generated due to the inding force F PP + E 1 1 generated due to common N etween the two deuterons ) considering int eractionetween + ( E generated due the inding force F 1 N1N + E 1 generated dueto common P etweenthetwo deuterons ) considering int eractionetween + ( E E generated dueto common N generated dueto theinding force FP 1P +

30 0 etween thetwo deuterons ) considerin g int eractionetween + ( E generated dueto theinding force F N N + E 1 1 generated dueto common P etweenthetwo deuterons 1 ) considering int eractionetween 1 ]/ = [8 ( E ) + ( E generated due to the inding force F ) + ( E generated due N1N 1 to the inding force F ) 8 ( P1 P + E generated due to common nucleon etween the two deuterons )]/ (6.1) ecause ( E ) = ( E ) = ( E ) = ( E ) = ( E ) and ( E generated due to common P 1 etween the deuterons 1 ) = ( E generated due to common N 1 etween the deuterons 1 ) = ( E generated due to common P etween the deuterons ) = ( Egenerated due to common N etween the deuterons ) = ( E generated due to common nucleon etween the two deuterons). Further, since in eqn. (6.1), E generated due to the inding force F N 1 N > ( E ) [see section-5.5], E generated due to the inding force F P 1 P < ( E ) [see section-.], and E generated due to common nucleon etween the two deuterons > E generated due to the inding force F N 1 N [see section-5.], eqn. (6.1) reduces to ( E ) α > 6 ( ) (6.) E 7. EXPLANATION OF HOW NUCLEONS ARE ARRANGE IN NUCLEI HAVING A = INTEGER MULTIPLE OF SUCH THAT THEY (NUCLEI) ARE MOST STRONGLY STABLE, HOW THEIR E INCREASES AS THEIR A INCREASES IN MULTIPLE OF, AN HOW E OF Be 8 IS REUCE TO < E OF WHILE A OF Be 8 = A OF In nuclei with mass numer (A) integer multiple of, such as, Be, C, O, Ne the nucleons are arranged forming groups (G ), each group having nucleons, two protons and two neutrons. In every group, two protons and two neutrons are arranged exactly in the same manner as 0 etc.,

31 1 arranged in a α particle, Fig. 6(a), and the directions of linear motion of all the four nucleons, lie in the same direction and parallel to each other. And hence, every group G in every nucleus possesses its linear motion in the direction of linear motions of its nucleons. 7.1 How nucleons are arranged in a nucleus having A = ( ), and determination of its E In the nucleus of, there occurs only one group G, and two neutrons and two protons are arranged in that group in the same manner as they are arranged in a α particle. And hence, the nucleus of has same E as a α particle has. 7. How nucleons are arranged in a nucleus having A = 8 (Be 8 ), and determination of its E 8 In the nucleus of Be, there occur two groups G 1 (each having nucleons) and they are arranged, as shown in Fig. 7(a). Group G 1 possesses its linear motion along +X direction and group G along X direction. ue to having linear motions y the groups G 1 andg towards each other, they try to come close to each other, ut due to having +e charges on them, the repulsive Coulom force etween them does not allow them to do so. When the forces on them due to their linear motions and due to Coulom repulsion ecome equal to each other, they stop coming close to each other after setting a certain distance, say g, etween them. ue to having linear motions y the groups G 1 andg along +X and X directions, the magnetic fields around them occur in planes perpendicular to X axis (i.e. in Y Z plane) and parallel to each other. And the directions of magnetic fields around them lie in directions opposite to each other, as shown y round arrows in their centers, Fig. 7(a). Since the magnetic fields around the groups G 1 and G occur in planes parallel to each other, they (magnetic fields) do not interact with each other and hence no inding force is generated

32 etween the groups G 1 and G. Therefore, the inding energy per nucleon ( 8 E ) for the nucleus of Be, i.e. ( E ), is otained as follows: Be ( E ) Be = [ t E (i.e. total inding energy) of group G 1 + E t of group G ] /8 ecause = E t of group G / E t of all the groups G 1,.. are equal and say = ( E ) Be = E ) E of group G. Therefore, ( (7.1) t where ( E ) is the inding energy per nucleon ( E ) for the nucleus of Why and how E of nuclei of Be 8 < E of nuclei of To overcome the repulsive Coulom force etween the groups G 1 andg, since they (groups G 1 andg ) loose their some energy, their inding energies [ E of group G 1 and E of group G ] t t are reduced, and hence ( E ) Be is also reduced. Consequently, E ) Be ( happens to e a little < ( E ). 7. How nucleons are arranged in a nucleus having A = 1 (C 1 ) and its E is increased to > E of 1 In the nucleus of C, there occur three groups G 1 and they are arranged as shown in Fig. 7(), where the groups G 1 are arranged exactly as they are arranged in the nucleus of 8 Be. The additional group G possesses its linear motion along Y direction and hence the magnetic field around it occurs in X Z plane (perpendicular to Y axis). The magnetic field around group G possesses direction as shown y round arrow in its centre, Fig. 7(). The portion of the magnetic field occurring around group G and lying towards our left hand side, Fig. 7(), interacts with the magnetic field occurring around group G 1. And the portion of the magnetic field occurring around group G and lying towards our right hand side interacts with the

33 magnetic field occurring around group G. Because of having directions y the magnetic fields around the groups G 1, as shown y round arrows in their centers, Fig. 7(), the interactions etween the magnetic fields of the groups G 1, and etween the magnetic fields of the groups G, are happened to e attractive (for verification of its truth, see section-7.9). Consequently, the left side of group G is ound with group G 1, and similarly the right side of group G is ound with group G. And thus, the groups G 1 andg are ound with each other through group G. When, in the nucleus of C also ound and hence ( E ) is otained as follows: C 1, all the three groups are ound together, all the 1 nucleons are ( E ) C = [( t E of group G 1 + E of group G + t Et of group G ) + B.E. (inding energy) generated due to interaction etween the magnetic fields around groups G 1& G + B.E. generated due to interaction etween the magnetic fields around groups G & G ] /1 = [ E t of group G + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G )]/1 = E t of group G / + [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G )]/1 = E ) ( + [B.E. generated due to interaction etween the magnetic fields around (groups G 1& G & G )]/1 (7.) > E ) (. (7.) 7. How nucleons are arranged in a nucleus having A = 16 (O 16 ) and its E is increased to > E of C 1

34 16 In the nucleus of O, there occur four groups G 1 and they are arranged as shown in Fig. 7(c). The groups G 1 are arranged exactly as they are arranged in the nucleus of and hence the directions of their linear velocities, planes and directions of their magnetic fields too 1 C, 1 are arranged exactly as these are arranged in the nucleus of C. The groups G 1 are ound together due to interactions etween their magnetic fields exactly as they are ound together in the nucleus of C 1 due to interactions etween their magnetic fields. The additional group G possesses its linear motion along +Y direction, and the magnetic field around it occurs in X Z plane and possesses direction as shown y round arrow in the center of group G, Fig. 7(c). The portion of the magnetic field around group G, occurring on our left hand side, interacts with the magnetic field occurring around group G 1. And similarly the portion of the magnetic field around group G, occurring on our right hand side, interacts with the magnetic fields occurring around group G. Consequently, the left and right sides of group G are ound respectively with group G 1 and group G. Thus, all the four groups G 1 and hence all the 16 nucleons are ound together (for verification of its truth, see section-7.9). Therefore, ( E ) is otained as follows: O ( E ) O = [( t E of group G 1 + Et of group G + Et of group G + E of group G ) + B.E. t generated due to interaction etween the magnetic fields around (groups G 1 & G + groups G & G 1 & G & G )] /16 = Et of group G / + [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G 1 & G & G )]/16

35 5 = E ) ( + [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G 1 & G & G )]/16.. (7.) = [ ( E ) + {B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G )}/1] [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G + groupsg & G )]/8 + [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G + groups G & G )] /16 = E ) C ( [B.E. generated due to interaction etween the magnetic fields around (groups G 1& G & G )]/8 + [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G )] /16... (7.5) Because from expression (7.), ( E ) + (B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G )/1 = ( E ). Therefore, C ( E ) O > E ) C (. (7.6) ecause in expression (7.5), B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G )/16 can never e B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G )/8 7.5 How nucleons are arranged in a nucleus having A = 0 (Ne 0 ) and its E is increased to > E of O 16 0 In the nucleus of Ne, there occur five groups G 1 5 and they are arranged as shown in Fig. 7(d), where the groups G 1 are arranged exactly as they are arranged in the 16 nucleus of O. The additional group G 5 possesses its linear motion along +Z direction and the

36 6 magnetic field around it occurs in X Y plane and possesses direction as shown y round arrow in the center of group G 5, Fig. 7(d). The portions of the magnetic field around groupg 5, occurring on our left hand side, on our right hand side, on our front side, and on opposite to our front side interact with the magnetic fields occurring around group G 1, group G, group G, and group G respectively. Consequently, the sides of the group G 5, lying towards our left hand side, our right hand side, our front side, and opposite to our front side are ound respectively with group G 1, group G, group G, and group G. Thus, all the five groups G 1 5 and hence all the 0 nucleons are ound together. Therefore, ( E ) is otained as follows: Ne ( E ) Ne = [( t E of group G 1 + Et of group G + Et of group G + E of group G + E of group G 5 ) + {B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G 1 & G & G )} + {B.E. generated due to interaction etween the magnetic fields around (groups G 1 + groups G )}] /0 = Et of group G / + [{B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G 1 & G & G )} + { B.E. generated due to interaction etween the magnetic fields around (groups G 1 )}]/0 t t = E ) ( + [{B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G 1 & G & G )} + { B.E.

37 7 generated due to interaction etween the magnetic fields around (groups G 1 + groups G )}]/0 (7.7) = [ ( E ) + {B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G + groupsg 1 & G & G )}/16] [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G + groups G & G 1 & G & G )]/80 + [B.E. generated due to interaction etween the magnetic fields around (groups G 1 + groups G )]/0 = E ) O ( [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G + groupsg 1 & G & G )]/80 + [B.E. generated due to interaction etween the magnetic fields around (groups G 1 + groups G )]/0 (7.8) Because from expression (7.), [ ( E ) + {B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G 1 & G & G )}/16] = ( E ). Therefore, ( E ) Ne > E ) O (. (7.9) O ecause in expression (7.8), [B.E. generated due to interaction etween the magnetic fields around (groups G 1 )]/0 can never e [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G + groups G 1 & G & G )]/80

38 8 7.6 How nucleons are arranged in a nucleus having A = (Mg ) and its E is increased to > E of Ne 0 In the nucleus of Mg, there occur six groups G and they are arranged as shown in Fig. 7(e),where the groups G 1 5 are arranged exactly as they are arranged in 0 the nucleus of Ne.The additional group G 6 possesses its linear motion along -Z direction and the magnetic field around it occurs in X Y plane and possesses direction as shown y round arrow in the center of group G 6, Fig. 7(e). The portions of the magnetic field around groupg 6, occurring on our left hand side, on our right hand side, on our front side, and on opposite to our front side interact with the magnetic fields occurring around group G 1, group G, group G, and group G respectively. Consequently, the sides of the group G 6, lying towards our left hand side, our right hand side, our front side, and opposite to our front side are ound respectively with group G 1, group G, group G, and group G. In this way, all the six groups G and hence all the nucleons are ound together (for verification of its truth, see section-7.9). Therefore, ( E ) is otained as follows: Mg ( E ) Mg = [( t E of group G 1 + Et of group G + Et of group G + E of group G + E of t t group G 5 + Et of group G 6 ) + {B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G 1 & G + groups G & G )} + {B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G 5 )} + {B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G 6 & G 6 & G 6 & G 6 )}] /

39 9 = Et of group G / + [{B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G 1 & G & G )} + {B.E. generated due to interaction etween the magnetic fields around (groups G 1 )} + {B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G 6 + groups G & G 6 & G 6 & G 6 )}]/ = [ ( E ) + {B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G 1 & G & G ) + B.E. generated due to interaction etween the magnetic fields around (groups G 1 + groups G )}/0] [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G + groups G 1 & G & G ) + B.E. generated due to interaction etween the magnetic fields of (groups G 1 )]/10 + [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G 6 & G 6 & G 6 & G 6 )]/ = E ) Ne ( - [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G 1 & G & G ) + B.E. generated due to interaction etween the magnetic fields around (groups G 1 + groups G )]/10 + [B.E. generated due to

40 0 interaction etween the magnetic fields around (groups G 1 & G 6 & G 6 + groups G & G 6 & G 6 )]/.. (7.10) ecause from expression (7.7), ( E ) + {B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G 1 & G + groupsg & G ) + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G 5 )}/0 = ( E ). Therefore, Ne ( E ) Mg > Ne ( E ). (7.11) ecause in expression (7.10), [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G 6 & G 6 & G 6 & G 6 )]/ can never e [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G + groups G 1 & G & G ) + B.E. generated due to interaction etween the magnetic fields around (groups G 1 )]/ How nucleons are arranged in nuclei having A = 8,, 6, 0 etc. and their E increases In the nucleus having 8 nucleons, the nucleons are grouped in 7 groups G 1 5, G 6 7. Six groups G are arranged as shown in Fig. 7(e) and seventh group G 7 is arranged ehind any of the six groups G Let us assume group G 7 is arranged ehind group G 1. Then the direction of linear motion of group G 7 lies in the same direction in which the direction of linear motion of group G 1 lies, and the plane of magnetic field around group G 7 is parallel to the plane of magnetic field around group G 1 and the of direction of magnetic field around group G 7 lies in the same directions in which the direction of magnetic field around group G 1 lies. If group G 7 is arranged ehind group G, the direction of linear motion of group G 7 lies in the same

41 1 direction in which the direction of linear motion of group G lies, and the plane of magnetic field around group G 7 is parallel to the plane of magnetic field around group G and the direction of magnetic field around groupg 7 lies in the same directions in which the direction of magnetic field around group G lies. ue to having repulsive Coulom force etween the groups G 1 and G 7 (when group G 7 is arranged ehind group G 1 ), group G 7 is not set ehind group G 1 touching that, ut is set at some distance apart from that. Since group G 7 is set ehind group G 1, outer portions (or can say, weaker portions) of the magnetic fields occurring around the groups G 5 6, which are left free from the interaction with the magnetic field occurring around group G 1, are availale to interact with the magnetic field occurring around group G 7. Consequently, group G 7 is ound with the groups G 5 6, ut with lesser inding force in comparison to that with which group G 1 is ound. And hence, E of the nucleus (having A = 8) is increased ut with reduced magnitude. Similarly as A of the nucleus increases y integer multiple of, the numer of G type groups goes on increasing one y one and they go on setting ehind the groups G 5 6 and so on, and accordingly E of the nucleus goes on increasing. It goes on till A ecomes = 8. When A increases eyond 8 (y integer multiple of ), the groups start setting ehind the groups G and so on till A ecomes = 7. But now (i.e. when A increases eyond 8), the increase in E is reduced further and near A = 6, it (increase in E ) ecomes minimum. So, near A = 6, E, after attaining its maximum value, starts decreasing as A increases (why and how it starts decreasing, see section-9). 7.8 Why and how nuclei having A = integer multiple of, are most strongly stale As have een descried aove in sections-7.1 to 7.7, in all the nuclei of, Be, C, O, 0 Ne etc., the nucleons are arranged forming groups G, which (groups G ) are very strongly stale.

42 And secondly, when G groups (two, three, four and so on) interact, due to interaction etween their magnetic fields, a strong inding is generated etween them, while if other groups, T 1, T interact etween themselves or withg (see section-8), due to interaction etween their magnetic fields, loose indings are generated etween them. Consequently, nuclei having A = integer multiple of, are most strongly stale. 7.9 Experimental verification of the truth of inding of groups G 1 etc. with each other due to interaction etween their magnetic fields Let us take three groups H 1, H and H, each group having four electric current carrying very small pieces of rods. Let us arrange all these three groups as the groups G 1 are arranged in Fig. 7(). The directions of flow of current through the rods of groups H 1, H, H are respectively along X, +X, +Y directions. Then the electrons shall flow through the rods of groups H 1, H, H respectively along +X, -X, -Y directions, i.e. the electrons shall have their linear motion through the rods of groups H 1, H, H along the same directions, along which the nucleons of groups G 1, G have directions of their linear motions. Now if a current is allowed to flow through the rods of all the three groups H 1, H, H, all the three groups are ound together. Similarly the three groups G 1 are also ound together in the nucleus of 1 C (see section-7.). If we take one more similar group H, having the direction of flow of current through its rods along Y, and arrange it along with the groups H 1, H, H as group G is arranged along with the groups G 1, Fig. 7(c), all the four groups H 1, H, H, H are ound together as soon as a current is allowed to flow through the rods of all the four groups. Similarly the four 16 groups G 1 are also ound together in the nucleus of O (see section-7.).

43 If we take one more similar group H 5, having the direction of flow of current through its rods along -Z, and arrange it along with the groups H 1, H, H, H as group G 5 is arranged along with the groups G 1, Fig. 7(d), all the five groups H 1, H, H, H, H5 are ound together as soon as a current is allowed to flow through the rods of all the five groups. Similarly the five groups G 1 5 are also ound together in the nucleus of Ne 0 (see section-7.5). If we take one more similar group H 6, having the direction of flow of current through its rods along + Z, and arrange it along with the groups H 1, H, H, H, H5 as group G 6 is arranged along with the groups G 1 5, Fig. 7(e), all the six groups H 1, H, H, H, H 5, H6 are ound together as soon as a current is allowed to flow through the rods of all the six groups. Similarly the six groups G are also ound together in the nucleus of Mg (see section-7.6). 8. HOW NUCLEONS ARE ARRANGE IN NUCLEI HAVING A INTEGER MULTIPLE OF (e.g. Li 6, Li 7, B 11 and N 1 ) SUCH THAT THESE ARE NOT STRONGLY STABLE, HOW E OF Li 6 AN Li 7 ARE REUCE TO < E OF, E OF B 11 TO < E OF Be 8, AN E OF N 1 TO < E OF C 1 WHILE E OF NUCLEI INCREASES AS THEIR A INCREASES 8.1 How nucleons are arranged in the nuclei of Li 6, Li 7 such that their (Li 6, Li 7 ) E increase as their A increases ut are reduced to < E of 6 In the nucleus of Li, 6 nucleons are arranged in two groups, two-neutrons and two-protons in group G and one-neutron and one-proton in group, and the groups G and are arranged as shown in Fig. 8(a). GroupG possesses its linear motion along +X direction, and the magnetic field around it occurs in YZ plane and in direction as shown y round arrow in its middle, Fig. 8(a). And group possesses its linear motion along -X direction, and the magnetic field around it occurs in YZ plane and in direction as shown y round arrow in its centre, Fig. 8(a).

44 The groups G and, ecause of having their linear motions along +X and X directions respectively, move towards each other, ut due to Coulom repulsive force etween them (ecause of having charges +e and +e y the groups G and respectively), they stop moving forward towards each other after a certain distance, say d, etween them. Since the magnetic fields around the groups G and occur in planes parallel to each other, they (magnetic fields) do not interact with each other and hence no inding force is generated 6 etween the groups G and. Therefore, the inding energy per nucleon ( E ) for the nucleus of Li, i.e. E ) 6, is otained as follows: ( Li ( ) Li E 6 = [ t E (i.e. total inding energy) of group G + E t of group ] /6 = [ ( E ) + ( E ) ]/6 where ( E ) is E of deuteron [ ( E ) + ( E ) /6]/6 ecause ( E ) [ ( E ) ]/6 0.7 ( E ). (8.1) But against the repulsive Coulom force etween the groups G and, since the groups G and loose their some energy, E of groups G and are reduced, and hence E ) 6 is also reduced. t ( Li 7 In the nucleus of Li, 7 nucleons are arranged in two groups, two-neutrons and two-protons in group G and two-neutrons and one-proton in groupt 1, and the groups G andt 1 are arranged, as shown in Fig. 8(). GroupG possesses its linear motion along +X direction, and the magnetic field around it occurs in Y Z plane and in direction shown y round arrow in its centre, Fig. 8(). And groupt 1 possesses its linear motion along -X direction, and the magnetic field around it occurs in Y Z plane and in direction shown y round arrow in its centre, Fig. 8(). The groups G andt 1, ecause of having linear motion along +X and X directions respectively, move towards each other, ut due to Coulom repulsive force etween them (ecause

45 5 of having charges +e and +e y the groupsg andt 1 respectively), they stop moving forward towards each other after a certain distance, say d, etween them. Since the magnetic fields around the groups G andt 1 occur in planes parallel to each other, they (magnetic fields) do not interact with each other and hence no inding force is generated 7 etween the groups G andt 1. Therefore, the inding energy per nucleon ( E ) for the nucleus of Li, i.e. E ) 7, is otained as follows: ( Li ( ) Li E 7 = [ t E of group G + Et of groupt 1 ] /7 = [ ( E ) + ( E ) T ] /7 where ( E ) 1 T is E 1 of groupt 1 = [ ( E ) +.55 ( E ) ] /7 ecause ( E ) T =.55 ( E 1 ) [7] [ ( E ) +.55 ( E ) /6] /7 ecause ( E ) [ ( E ) ]/ ( E ) (8.) But against the repulsive Coulom force etween the groups G andt 1, since the groups G andt 1 loose their some energy, Et of groups G and 1 T are reduced, and hence E ) 7 is also reduced. ( Li From the expressions (8.1) and (8.) we see, E ) 7 > E ) 6, i.e. as A increases (i.e. from ( Li ( Li 6 to 7), E of nuclei increases, ut E ) 7 and E ) 6 oth are < ( E ). ( Li ( Li 8. How nucleons are arranged in the nucleus of B 11 such that its E is reduced to < E of Be 8 though A of B 11 > A of Be 8 11 In the nucleus of B, 11 nucleons are arranged in three groups, two-neutrons and twoprotons in group G 1, two-neutrons and two-protons in group G, and two-neutrons and on-proton in groupt 1, and the groups G 1 andt 1 are arranged as shown in Fig. 8(c). The groups G 1 and G possess their linear motions along +X and X directions respectively, and the magnetic fields around

46 6 them occur in Y Z plane and in directions opposite to each other, shown y round arrows in their centers in Fig. 8(c). GroupT 1 possesses its linear motion along -Y direction, and the magnetic field around it occurs in X Z plane and in direction shown y round arrow in its centre, Fig. 8(c). Since the magnetic field around groupt 1 happens to e of triangular shape [see Fig. 5(g)], a very little portion of the magnetic field around groupt 1, lying towards our left hand side, interacts with the magnetic field around group G 1. And similarly, a very little portion of the magnetic field around groupt 1, lying towards our right hand side, interacts with the magnetic field around group G. Consequently, the left and right sides of groupt 1 are loosely ound with the groups G 1 andg respectively, and hence the groups G 1 andg are loosely ound with each other through the groupt 1 (for confirmation of its truth, see section-8.). Therefore, 11 E for the nucleus of B, i.e. E ) 11, is otained as follows: ( B ( B E ) 11 = [( t E of group G 1 + E of group G + E of groupt 1 ) + B.E. generated due to t t interaction etween the magnetic fields of (groups G 1 & T 1 & T 1)]/11 = [( E t of groupsg + E t of groupt 1 ) + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & T 1 & T 1 )] /11 = [ { ( E ) } + ( E ) T + B.E. generated due to interaction etween the 1 magnetic fields around (groups G 1 & T 1 & T 1 )]/11 [8 ( E ) +.55 ( E ) /6+ B.E. generated due to interaction etween the magnetic fields around (groups G 1 & T 1 & T 1 )]/ ( E ) / + [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & T 1 & T 1 )]/11

47 7 ( E ) - {.5 E ) ( }/ + [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & T 1 & T 1 )]/11 (8.) < E ) Be (. (8.) Because in expression (8.), the value of B.E. generated due to interaction etween the magnetic fields around (groups G 1 & T 1 & T 1 )/11 happens to e very small, and hence can never e {.5 ( E ) }/. 8. How nucleons are arranged in the nucleus of N 1 such that its E reduced to < E of C 1 though A of N 1 > A of C 1 In the nucleus of N 1, 1 nucleons are arranged in four groups, two-neutrons and twoprotons in each group G 1, and one-neutron and one-proton in group, and all the four groups G 1, are arranged as shown in Fig. 8(d). The groups G 1 and G possess their linear motions along +X, X and -Y directions respectively, and the magnetic fields around the groups G 1 occur in Y Z plane and around group G occurs in X Z plane. And the directions of magnetic fields around them occur as have een shown y round arrows in their centers in Fig. 8(d). Group possesses its linear motion along +Y direction, and the magnetic field around it occurs in X Z plane. The direction of magnetic field around it occurs as has een shown y round arrow in its centers in Fig. 8(d). Since the magnetic field around group happens to e of rectangular shape, Fig. (c), a very little portion of the magnetic field around group, lying towards our left hand side, interacts with the magnetic field around group G 1. And similarly, a very little portion of the magnetic field around group, lying towards our right hand side, interacts with the magnetic field around group G. nce, the groups G 1 andg are ound with each other through the group, ut their inding happens to e loose (for its confirmation, see section-8.).

48 8 1 The inding energy per nucleon for the nucleus of N, i.e. E ) 1, is otained as follows: ( N ( N E ) 1 = [( t E of group G 1 + E of group G + E of group G + t t Et of group ) + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G + groups G & G ) + B.E. generated due to interaction etween the magnetic fields around (groups G 1& & )] /1 = [ E t of groups G + E t of group + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G ) + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & + groups G & )]/1 = [ { ( E ) } + ( E ) + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G ) + B.E. generated due to interaction etween the magnetic fields of (groups G 1 & & )]/1 [1 ( E ) + ( E ) /6 + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G ) + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & & )]/1 7 ( E ) / + [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G ) + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & & )]/1

49 9 [ ( E ) + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G )/1] [5 ( E ) / + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G )/8] + [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & & ]/1 ( E ) C [5 E ) ( / + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G )/8] + [B.E. generated due to interaction etween the magnetic fields around (groups G 1 & & ]/1 (8.5) Because from expression (7.), ( E ) + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G )/1 = ( E ). Therefore, C ( N E ) 1 < E ) C (... (8.6) ecause in expression (8.5), the value of B.E. generated due to interaction etween the magnetic fields around (groups G 1 & + of groups G & )/1 happens to e very small and hence can never e [5 ( E ) / + B.E. generated due to interaction etween the magnetic fields around (groups G 1 & G & G )/8]. 8. Experimental confirmation of the truth of loose inding when two or more G groups are ound together through a or T 1 or T group We take two groups H 1, H and one group J (having three current carrying very small pieces of rods) and arrange group J along with the groups H 1, H as group G is arranged along with the groups G 1, Fig. 7(). If the flow of current through the rods of groups H 1, H and J are

50 50 respectively along X, +X +Y directions, the two groups H 1, H are ound together through group J as soon as a current is allowed to flow through the rods of all the three groups. But their inding happens to e loose as compared to if in place of group J, a group H is placed. Similarly the groups G 1 and G too are loosely ound with each other through the groupt 1 (section-8.). We take three groups H 1, H, H and one group K (having two current carrying very small pieces of rods) and arrange group K along with the groups H 1, H, H as group G is arranged along with the groups G 1, Fig. 7(c). If the flow of current through the rods of groups H 1, H, H and K are respectively along X, +X, +Y, -Y directions, all the four groups H 1, H, H and K are ound together as soon as a current is allowed to flow through the rods of all the four groups. But their inding happens to e loose as compared to if in place of group K, a group H is placed. Similarly the groups G 1 and too are loosely ound with each other (section-8.). 9. EXPLANATION OF WHY AN HOW E OF NUCLEI, AFTER BECOMING MAXIMUM NEAR A = 6RAUALLY STARTS ECREASING AS A INCREASES, AN FOR A > 00, NUCLEI BECOME RAIOACTIVE AN THEα, β, γ AN ν ARE EMITTE FROM THEM, AN THE γ AN ν OBTAIN PARTICLE LIKE EXISTANCE AN SO HIGH ENERGY 9.1 Why and how E of nuclei, after ecoming maximum near A = 6 [9], gradually starts decreasing as A increases. In the nucleus, nucleons do not reside independently (except hydrogen, where one proton resides independently in the nucleus) ut reside in the form of different types of groups G (having two-neutrons and two-protons, see section-7), T 1 (having one-proton and two-neutrons, see section- 5), T (having two-protons and one-neutron, see section-5), and (having one-proton and one-

51 51 neutron, see section-). The nucleons are arranged in the form of aove types of groups in nuclei of different mass numer (A), as have een descried, e.g., in sections-, 5, 6, 7, 8. In a nucleus, when the nucleons are arranged in the form of aove types of groups, among them, there act three types of forces: 1- The force which they possess due to having their linear velocity; - The force due to interaction etween their magnetic fields which (magnetic fields) occur around them; - The Coulom repulsive force due to having positive charge on them (+e on groups andt 1, +e on groupst andg ). The first two forces try to ring the groups close to each other while the third Coulom repulsive force tries to oppose the first two forces. The first force remains always constant and does not depend on A, ut the second and third forces go on increasing as A increases, ecause as A increases, the closeness among the groups increases. Therefore, E of the nucleus depends upon the resultant of second and third forces. Up to A =, the rate of increase in the magnitude of second force happens to e quite fast (see sections-7.1 to 7.6 and also section-8). The rate of increase in the magnitude of third force happens to e comparatively negligile. Consequently E incases quite fast. Beyond A = as A increases, the rate of increase in the magnitude of second force is reduced (see section-7.7). The rate of increase in the magnitude of third force also proaly now starts ecoming gradually significant. Consequently, increase in E ecomes slow. And eyond A = 8 as A increases, the rate of increase in the magnitude of second force is reduced further, and the rate of increase in the magnitude of third force proaly now ecomes significant. Consequently, near A = 6, after attaining its maximum value E starts gradually decreasing. 9. Why and how the nuclei ecome radioactive when A > 00 and theα, β, γ andν are emitted from them 9..1 Why and how the nuclei ecome radioactive when A > 00 and the α and β are emitted

52 5 Beyond A is nearly = 6 as A increases, the closeness among the groups is proaly increased as much that the third force gradually starts ecoming more and more significant. Consequently E gradually starts decreasing and goes on decreasing continuously. When A > 00, E is proaly reduced as much that the indings of the outer most groups with the joint group (sayg ) of the rest of all the other inner groups, start getting gradually loose. R Beyond Z is nearly = 5 as A increases, in nuclei, the percentage of N starts increasing in comparison to Z, and in heavy nuclei, it ecomes nearly 50% higher than Z. It means, eyond Z is nearly = 5 as A increases, in nuclei, the percentage of groupst 1 starts increasing in comparison to the other groups (ecause in a groupt 1, there occur two neutrons and one proton, see section-5.1 and 5.). Therefore, in heavy nuclei, among the outer groups, the percentage of groupst 1 happens to e higher in comparison to other groups. In a heavy nucleus, suppose a group ( T 1 ) m is set ehind a group G n. In heavy nuclei, since all the groups ecome very close to each other, due to having +e more charge on group G, group G ecomes ale to influence the electrons of the neutrons of group ( T 1 ) m. Since the neutrons in groupst 1 are happened to e loosely stale (see section-5.6), the influence of +e more charge of groupg n, loosens the staility of neutrons of group ( T 1 ) m even more. Now if the inding of group ( T 1 ) m with group G R is roken [which ecomes possile in heavy nuclei, ecause in heavy nuclei, their E is proaly reduced as much that the indings of the outer most groups with group G start getting gradually loose], the electrons of the neutrons of group ( T 1 ) m may e separated from their respective protons. When they are separated, oth the electrons are proaly not separated simultaneously ut one electron is separated at a time. When one electron, say E 1 (of neutron N 1 ) is n n R

53 5 separated from proton P 1, Fig. 5(a), proton P 1 ecomes like proton P, and hence they together increase the velocity of proton P (of neutron N ). Consequently, the staility of neutron N is increased. Then group ( T 1 ) m is converted into groupt (see section-5.6). After separation of electron E 1 from neutron N 1 of ( T 1 ) m, the conversion process of group ( T 1 ) m into groupt is not completed immediately. It takes time and may e completed at any instant during the course of time t + t (see section-.6) of the travel of electron E 1. If even during the time t + t, the conversion process of group ( T 1 ) m into groupt is not eing completed, electron E 1 collides again with proton P 1 (of neutron N 1 ). Then at any instant during time t of neutron N 1, electron E (of neutron N ) may e separated from proton P (of neutron N ), and with electron E same process may e repeated. In this way, this process goes on. uring the course of time t + t of electron E1 or of E if that is found in position, as shown in Fig. (), with a group G, that electron is ejected from the nucleus as the consequence of repulsive force on that y group G, or electron and group G oth are ejected from the nucleus. The ejection of electron, or of electron and groupg oth, depends upon whether group G is free or ound with group G. If groupg is free, then electron and group G oth R may e ejected simultaneously, i.e. α and β oth the decays may take place. And if group G is ound, then only the electron is ejected, i.e. only β decay takes place. The energy of the ejected electron depends upon, at which instant of the course of time t + t the electron is found in position as shown in Fig. () with a group G and is ejected, ecause the ejected electron possesses energy what it had at the instant when it was ejected (for detail, see section-.6). uring the course of time t of electron E1 or of E if that is found in position, as shown in Fig. (), with a group G, group G is + only ejected, i.e.α decay takes place. (How two types of β decay, i.e. β and β decays take place, it

54 5 shall e sumitted for pulication later on ut shortly ecause it needs a lot of discussion. That is presently not possile ecause that is eyond the scope of the present paper.) The groupst 1, T, and are not ejected (emitted) from the nuclei ecause: 1- In nuclei having A > 00, since the percentage of N ecomes nearly 50% higher than Z, the possiility of occurrence of groupst and among the outer groups is proaly reduced to zero, and hence their ejection (emission) from the nuclei is also reduced to zero; - Since the nucleons in the groupst 1, T, and are not so strongly ound as are ound in groupg, the groupst 1, T, and do not ehave like particles, and also, the magnetic fields around the groupst 1, T are happened to e of triangular shape, Fig. 5(g), and around group that happens to e of rectangular shape, Fig. (c), hence when any one of these groups is found in situation with an electron of neutron of groupt 1, as group G is found, Fig. (), due to interaction etween their (electron and groupt 1 /groupt /group ) magnetic fields, the repulsive force generated etween them proaly happens to e weak and not sufficient to eject groupt 1 /groupt /group from the nuclei. The groups G, T 1, T, cannot e ejected (emitted) from the nuclei due to interaction etween their magnetic fields, ecause for their ejection, it is essential that when their magnetic fields interact, the magnetic fields must e in same plane and opposite in directions, as shown in Fig. () where the magnetic fields around the electron and group G are lying in the same plane ut opposite in directions. The magnetic fields around two groups (e.g. groupsg andt 1, groups G andt, groups T1 andt etc.) can come in situation as shown in Fig. () only if they (two groups) are moving parallel to each other ut opposite in directions, and it can e possile only if they are free to move in nuclei. While in nuclei, all the groups are ound in a group and are not free to move freely.

55 55 When A of nuclei ecomes > 00, the indings of the outer groups ecome loose only; they do not ecome free to move freely. 9.. Why and how the γ and neutrino ν are emitted and they otain their particle like existence In Compton s scattering experiment, a γ photon collides with an electron. It can e possile only if the γ photon too physically exists similarly as the electron exists. The charge of electron provides physical existence to it, and hence, to provide physical existence to γ particle, there too must e some thing. That thing is a undle of radiant energy. This undle of radiant energy provides a particle like physical existence to γ photon. The infrared, visile, or ultraviolet photons emitted from the oriting (or can say, atomic) electrons are the undles of radiant energy. They are emitted from the oriting electrons as the consequence of their expansion and sudden compression (for detail, see reference-5). In nuclei, the electrons do not travel in orits, ut in neutron structure when they collide with their respective protons (e.g. electron E 1 of neutron N 1 with its proton P 1 ), they suffer sudden jerk, that produces almost the same effect the compression produces and with much strength. Consequently, a undle of radiant energy is emitted out from each of the colliding electrons. The undles of radiant energy thus otained are theγ photons we oserve emitting during the nuclear decays. (The compression of electrons cannot e ruled out or denied ecause if a proton can shrink [1, 1], the electron too can shrink.) When the electrons collide with their respective protons, since the protons too suffer jerk, a undle of energy is emitted from each of the colliding protons. Those undles are the neutrinos. Currently it is elieved that, when a nucleus emits α or (and) β particle(s), the daughter nucleus is usually left in an excited state. It can then move to a lower energy state y emitting aγ photon, in much the same way that an atomic electron can jump to a lower energy state y emitting

56 56 infrared, visile, or ultraviolet photons. But y such mechanism, neither the infrared, visile, ultraviolet photons are emitted (see reference-5) nor can theγ photons e emitted. Because, y such mechanism, the difference of energy etween the two states can never e emitted in the form of a undle having a particle like physical existence and such a high penetrating power. Further, the emission of aγ photon from a nucleus typically requires only 1 10 seconds [1], and is thus nearly instantaneous. The emission of a γ photon as the consequence of collision of an electron with a proton is instantaneous. 9.. Why and how γ and ν otain so high energy and momentum After collision with proton P 1 when the velocity of electron E 1 is reduced to zero, a undle of radiant energy is emitted from the electron E 1 in the form of aγ particle and the difference, M.E. of electron E 1 efore its collision - M.E. of electron E 1 after its collision radiant energy contained in theγ photon, is imparted to the γ particle as its M.E. in order to conserve the M.E. of electron E 1. The difference, M.M. of electron E 1 efore its collision - M.M. of electron E 1 after its collision, is imparted to the γ photon as its M.M. in order to conserve the M.M. of electron E 1. And the difference, L s (resultant spin angular momentum = L sc ± L sm ) of electron E 1 efore its collision - Ls of electron E 1 after its collision, is imparted to the γ photon as it s L s in order to conserve electron E 1 (for detail, see reference-5). Ls of Similarly, after collision with electron E 1 when the velocity of proton P 1 is reduced, a undle of energy is emitted from proton P 1 in the form of a neutrino (ν ) and the difference, M.E. of proton P 1 efore its collision - M.E. of proton P 1 after its collision energy contained in ν, is imparted to ν as its M.E. in order to conserve the M.E. of proton P 1. The difference, M.M. of

57 57 proton P 1 efore its collision - M.M. of proton P 1 after its collision, is imparted to ν as its M.M. in order to conserve the M.M. of proton P 1. And the difference, L s (resultant spin angular momentum = L sc ± L sm ) of proton P 1 efore its collision - it s L s in order to conserve Ls of proton P 1. Ls of proton P 1 after its collision, is imparted to ν as After the collision, e.g. of electron E 1 with proton P 1, since the velocity of electron E 1 is reduced to zero, the difference in its M.E. and in M.M. etween its two states (efore collision and after collision) are happened to e quite large. Consequently, the M.E. and M.M. of the emitted γ photon from electron E 1 too are happened to e quite large. ue to very large M.M. of γ photons, they possess very high penetrating power. (How M.M. of photons provides penetrating power to them, see reference-5). The difference in M.E. and in M.M. etween the two states, e.g. of proton P 1 too are happened to e quite large, and consequently the M.E. and M.M. of the emitted ν from proton P 1 are happened to e quite large. 9.. Question mark over the existence of antineutrino Anti means opposite or reverse, and if the neutrino is a undle of energy, the antineutrino must e a undle of anti energy, or can say of negative energy, as, e.g., an anti electron (i.e. positron) is assumed to e a undle of +e. But the anti energy does not exist in nature and hence the antineutrino cannot e a undle of anti energy. Currently it is assumed that due to having anti spin states, the antineutrino otains the anti state of neutrino. But this assumption gives rise to several questions, e.g., how and from where do the neutrino and antineutrino otain their spin motions, spin states, and why and how are their spin states anti to each other? If it is argued that they otain their spin motions, anti spin states from the

58 58 neutron, the questions arise, how and from where does the neutron otain its spin motion, how does that persist, and how and from where does the neutron otain energy for their persistent spin motion? Following the current concepts, since no answer to these questions can e given, the spin motion of neutron and hence of neutrino and antineutrino, and their anti spin states cannot e assumed. So, the current assumption is ruled out. Then how is antineutrino anti to neutrino? And further, how is it produced? These questions raise question mark over the existence of antineutrino. ACKNOWLEGEMENT The author is grateful to his respected teacher, Prof. Ashok Kumar Gupta, Professor of Physics (Retd.), Allahaad University, Allahaad (U.P.), INIA, for his continuous encouragement, time to time discussion, sincere advice and help. The author is also thankful to his nephew Mr. Nagendra Kumar for preparing all the diagrams.

59 59 REFERENCES [1] [] M. Gell-Mann, A schematic model of aryons and mesons, Phys. Let. 8 () (196) p [] [] Kunwar Jagdish Narain, Theory which explains several greatly important mysterious phenomena and it determines a new force having characteristics of nuclear force, vixra: (condensed matter), [5] Kunwar Jagdish Narain, A new interpretation to quantum theory, Phys. Essays (009) p. 7. Also availale at [6] [7] [8] H. Yukawa, Proc. Phys.-Math. Soc. Japan 17 (195) p. 8. [9] [10] [11] [1] [1] Geoff Brumfiel (July 7, 010). The proton shrinks in a size. Nature. oi:10.108/news [1]

60 60 (a) () Fig. 1: (a) Spherical all, dark solid line circle and concentric roken line circles respectively represent the charge, magnetism and magnetic field of electron. () Cross sectional view of electron where, in order to introduce arrow marks with the all of charge to show the direction of its spin motion, the all of charge has een shown y a dark thick solid line circle.

61 61 (a) () Fig. : (a)transverse cross sectional view of motion of an electron E and a group G at the instant when they are in the same plane and at distance d apart while moving parallel to each other with same velocity v ut opposite in directions. () Transverse cross sectional view of interaction etween their magnetic fields when the distance d etween them is < ( r 1 + r ).

62 6 Fig. : The Feynman diagram for β decay of a neutron into a proton, electron, and antineutrino via an intermediate W oson.

63 6 (a) () (c) (d) (e) Fig. : In the structure of a deuteron: (a) arrangement of one-neutron and one-proton each having linear velocity v; () interaction etween the magnetic fields of nucleons; (c) shape of the outer portion of the magnetic field otained around the deuteron. (d) In the structure of di-proton, arrangement of two-protons. (e) In the structure of di-neutron, arrangement of two-neutrons.

64 6 (a) () (c) (d) (e) (f) (g)

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