Theory of nuclear force

The attractive force stronger than the electric force acting between the nucleons and which binds them together in the nucleus is known as nuclear force. This force is very complex in comparison to gravitational and electromagnetic forces. In this particular article Theory of nuclear force, we are going to discuss about properties of nuclear force and meson theory. Nuclear force have following properties-

Properties of nuclear force

1) Attractive force:

The nuclear force is an attractive force and strongest among all other fundamental force if its strength is assumed to be 1, then the strength of EM force would be about the order of 10⁻²  and the strength of gravitational force would be the order of 10⁻³⁹.

 2) Short Range:

The nuclear force is a short range of the order of 1.5 to 2.0 Fermi it is observed that in some nuclear states repulsive love on nuclear force also exist within as very short range (≈ 0.5 fermi ) around which there is strong attractive force range up to 1.5 to 2.0 FM.

3)  Spin dependence:

The nuclear force depends upon the spin of the nucleons interacting with each other in the nucleus, it is observed that the spin of P and N is in the electron are parallel ( s=1).
If we assume that the spin is antiparallel ( s=0) then in this arrangement electron can’t bind them together in the stable state, thus M.F depends on the spin of interacting nucleons.

4) Charge Independence:

The nuclear force acting between two nucleons seems to be independent of the charge of nucleons. It is the same for all the three types of nucleons pairs, namely n-n, p-p, and n-p. This means nuclear forces are of a non- electric nature.

5) Charge Symmetry:

Let us consider two mirror nuclei ₁H³ and ₂He³ for proving charge symmetry of nuclear force. Nucleus ₁H³ contains 1P and 2N i.e one nuclear force n-n and two nuclear force n-p acting in it. Whereas nucleus 2He³ contains 2P and 1N i.e., one nuclear force p-p and two nuclear forces acting on it. B.E (Bond energy)of ₁H³ is 8.482 MeV and B.E. of ₂He³ is 7.711MeV.

The difference in B.E. is  0.711MeV. If we assume the nuclear force between similar nucleons i.e. p-p and n-n have the same strength then the difference in bond energy is only due to the repulsive force. If we assume that ₂He³ is charged sphere of radius R, then the Coulomb energy is given by E=6Ke²/5R where E=0.711 MeV, K= 9×10 Nm²/c², e= 1.6×10 c then R= 2.24 × 10m or 2.24 fm which is of the same order of the size of the nucleus. Thus, the assumption is correct i.e. the nuclear force between similar nucleons i.e. P-P and n-n have the same strength and the difference Bond energy is only due to the repulsive force.

6) Saturation property:

Nuclear force has a saturation property i.e, the Nuclear force acts only between two nearest nucleons and it is negligible for other nucleons.

7) Non-central Component of Nuclear force:

The Magnetic moment of deuteron measured experimentally Md=0.857411 whereas total magnetic moment of N and P is- Mn+Mp= -1.91315 + 2.79271= o.87956. This difference of Magnetic moment cant is explained by assuming nuclear force as the central force. To explain this small difference of magnetic moment the small fraction of nuclear force has to be associated.

Meson theory of Nuclear forces

Japanese scientist Yukawa proposed a Meson theory of Nuclear force. According to this theory, each nucleon continuously emits and simultaneously absorb pion particle. Pion may be of three types-

  1. π⁺ meson
  2. π⁻ meson
  3. π⁰ meson

In this process, momentum is transferred as a result of which nuclear force is generated.

The change of pion between the nucleon can be explained as follows-

  1.  When a proton emits a π⁺ meson, then it is converted into the neutron. If there is another neutron near to the proton. It absorbs the emitted meson and converts into the proton. p+n→(n+π⁺)+n→n+(π⁺+n)→n+p
  2. When a neutron emits a π⁻ meson, then it is converted into the proton. If there is another proton near to the neutron, it absorbs the emitted meson and converts into the neutron. n+p→(p+π⁻)+p→p+(π⁻+p)→p+n
  3. exchange of π⁰⁻ meson can also generate nuclear force. p+p→(p+π⁰)+p→p+(π⁰+p)→p+p

n+n→(π+π⁰)+n→n+(π⁰+n)→n+n

 

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