International Journal of Scientific & Engineering Research Volume 2, Issue 4, April-2011 1 ISSN 2229-5518 New Formula of Nuclear Force Md. Kamal Uddin Abstract -It is well established that the forces between nucleons are transmitted by meson. The quantitative explanation of nuclear forces in terms of meson theory was extremely tentative & in complete but this theory supplies a valuable point of view . it is fairly certain now that the nucleons within nuclear matter are in a state made rather different from their free condition by the proximity of other nucleons charge independence of nuclear forces demand the existence of neutral meson as amongst the same type of nucleolus (P-P) or (N-N). this force demand the same spin & orbital angular momentum. The exchange interaction in produced by only a neutral meson. The involving mesons without electric charge, that it gives exchanges forces between proton & Neutron & also therefore maintains charge in dependence character. It is evident for the nature of the products that neutral mesons decay by strong & weak interaction both. It means that neutral mesons constituents responsible for the electromagnetic interaction. Dramatically neutral mesons plays important role for electromagnetic & nuclear force both. Index Terms - Restmass energy,Mesons,photons,protons,neutrons,velocity of light,Differentiation —————————— —————————— 1. INTRODUCTION T is well established that the forces application of quantum mechanics to Ibetween nucleons are transmitted by nuclear phenomena gives us confidence in meson. The quantitative explanation of general use of quantum mechanics for the nuclear forces in terms of meson theory description of the force between heavy was extremely tentative & incomplete, but particles in nuclei. this theory supplies a valuable point of Where dm0c2 = either rest mass energy of Δ0 view. Yukawa first pointed out that mesons(For neutral theory),or rest mass nuclear force can be explained by assuming energy of Δ+, Δ-& Δ0mesons( for symmetrical theory) that particle of mass about 200 times the dm0 =either mass of Δ0 mesons or mass of electron mass(mesons) exist & can be Δ+ , Δ-& Δ0 mesons emitted & absorbed by nuclear m0 = mass of nucleons particles(neutrons & protons) with such an m0cdc = rest mass energy of nucleons assumption a force between nuclear dr= Range of nuclear force, which can be particles of right range & right calculated from differentiation of Nuclear shape(rapid decrease at large distances is radius.( The force between two nucleons is now obtaining. attractive for distance r(radius) greater Now we have the rest mass energy = m0 c2 than dr (range) & is repulsive Differentiating with respect to r (Inner radius otherwise).This strongly suggests & well at which nuclear force comes into play) proved that to some degree of d(m0c2) approximation the total isotopic spin T is a dr constant of the motion & is conserved in = c2dm0 + m0d(c2) = c2dm0 + all processes, at least with a high dr dr dr probability. mo d(c2) dc = c2dm0 + 2m0c.dc dc= The average velocity of neutron & dc dr dr dr proton. A large velocity is used in nuclear . This force is short range, attractive & along disintegration. the line joining the two particles (central force).(The wide success of this first c = Velocity of light IJSER © 2011 http://www.ijser.org International Journal of Scientific & Engineering Research Volume 2, Issue 4, April-2011 2 ISSN 2229-5518 2 = multiplicity of interacting particles interact only with neutral mesons( particles is given by (2T+1), the isotopic spin neutral theory) or(2) that they interact has no such meaning for leptons or a gamma equally strongly with neutral, positive & rays negative mesons(symmetrical theory). It is 1 = either multiplicity of obvious that the part of the force which Δ0 mesons or Δ+, Δ- & Δ0 mesons (evidence of does not depend on the spin of the involving of mesons (all type)) nuclear does not fulfill any useful Where T = Vector sum of function in the theory. The force between isotopic spin of proton & proton, neutron& proton & neutron are result from the neutron, neutron & proton The success of transfer of positive meson from the former these applications supplies additional to the latter or a negative meson in the support for the hypothesis of the charge opposite direction. So there vector sum of independence & charge symmetry of component of isotopic spin of these particle nuclear force. As the nuclear interactions do must be zero. not extend to very large distances beyond The charges on charged the nuclear radius & this character is mesons must be equal in their magnitude. useful to solve the problem. The Full charge independence for any system in which the number of neutrons equals the Charge independence of nuclear forces numbers of protons, this formula give the demand the existence of Δ0 meson as amongst evidence the charge symmetry, merely the same type of nucleons (p-p) or (N-N).This means that the neutron-neutron & proton force demand the same spin & orbital proton interaction are equal but says angular momentum. Positive pions are not nothing about the relations of neutron able to surmount the nuclear coulomb proton interaction to others. Nuclear forces barrier & there fore undergo spontaneous are symmetrical in neutrons & protons. i.e. decay while negative pions are captured the force between two protons are the by nuclei. The exchange of a pion is thus same as those between two neutrons. This equivalent to charge exchange. we can think identity refers to the magnitude as well as of nucleons as exchanging their space & the spin dependence of the forces. spin co-ordinates In the neutral theory, therefore neutron & protons are completely Now we can see the following reaction equivalent & indistinguishable as far as the P+ P ----- P+P+Δ0 associated meson fields are concerned. Such P+N -----P+N+Δ+ particle decay into two gamma rays. These N +P----- N+ P + Δ- gamma rays are Δ0 – rest systems are emitted These reactions are soon as Y + P -- in opposite direction & therefore spin Δ0 must P + Δ0 be Zero as the spin of photon is unity. It is The capture of photons can effect evident from the nature of the products the production of mesons by an that neutral mesons decay by the electromagnetic interaction, decay electromagnetic interaction while charged electromagnetically since these processes pions decay by strong & weak interaction involve no change of strangeness. both. It means that neutral mesons P + P --- P + P + Δ0 constituents responsible for the P + N ----- P + N + Δ0 electromagnetic interaction. We know that It is found that only two assumptions neutron & proton can change into one are in agreement with theoretical & another by meson capture. Protons &neutron experimental facts, notably the equality of can transform into each other by capture of the forces between two like & two unlike positive & negative pion respectively, or get nuclear particles in the singlet state. These transform into the same particle through assumption are either(1) that nuclear neutral meson interaction. During these IJSER © 2011 http://www.ijser.org International Journal of Scientific & Engineering Research Volume 2, Issue 4, April-2011 3 ISSN 2229-5518 transformation either an emission or an stable than hydrogen atom.Becouse binding absorption of meson is essential. The energy of helium atom is larger than that attraction between any nucleons can arise of hydrogen atom, so in this process large from the transfer of a neutral meson from energy is released.( here the negative pion one nucleon to the other. If the meson has significant role to produce nuclear were assumed to be charged ( positive or interaction. While charge pion main tain the negative) the resulting force between character of proton.) Now, as the hydrogen nuclear particles turned out to be of the nucleus converted into helium nucleus, exchange type which had been successful These are happen when resonances of in the interpretation in nuclear physics. The nucleons is in excited state. The scattering mesons must obey Bose statistics because cross- section can be interpreted by they are emitted in the transformation of a assuming that in the strong interactions the neutron into a proton( or vise versa) both total isospin is converted as well as the obey Fermi statistics. third component. The total isotopic spin of the system is 3/2.The ratio of radius & Important point-- range is about 2:1. It shows that those 1. The deuteron does posses nuclei has maximum number of nucleons measurable properties which might serve as are most stable than less mass number a guide in the search for the correct nuclei. nuclear interaction. The mass number of deuteron A is very minimum. From the 2*.Also, Strongly, The findings we must regard the deuteron as velocity of light depends loosely bound. The deuteron consist of two upon range of the nuclear particle roughly equal mass M, so that the force. The velocity of light reduced mass of the system is 1/2M.The equal like photons, lack deuteron has spin T=1, the neutron & mass & force carrying protons spins might be a parallel particle of other forces like combination. The magnetic moment of strong force. Because range deuteron will, therefore be sum of is variable, then the velocity magnetic moment of proton & neutron. of light must be variable. According to conclusion, As the range As, velocity of light= Range becomes larger in deuteron nucleus & they of nuclear force(distance becomes more unstable. travel by meson)/ Life time *When light nuclei of hydrogen atom of resonances. In this comes within the range of nuclear force relation we can see that the they can fuse together to form helium velocity of light must be nucleus. In this process (fusion) the range is variable. It is clear that not effected. However the force is twice of the fundamental particles the hydrogen’s nuclear force & so on.