# Relation between Nuclear potential and Range of nuclear force

The nuclear potential is the potential energy associated with the nuclear force, which is the strong force that binds protons and neutrons together in the nucleus of an atom. The nuclear force is a short-range force that acts over distances of only a few femtometers (10^-15 meters). It is one of the four fundamental forces of nature, along with gravity, electromagnetism, and the weak force.

The range of the nuclear force is determined by the exchange of particles called mesons between nucleons (protons and neutrons). The mesons transmit the strong force between nucleons, and their exchange gives rise to the attractive force that binds them together. The range of the nuclear force is limited by the fact that mesons are massive particles, and their exchange is therefore limited by the uncertainty principle of quantum mechanics.

The deuterium nucleus is the simplest bound state problem having only two nucleons i.e one proton and one neutron with only one n-p interaction so if the potential of the deuterium nucleus can be determined then the nature of nuclear force can be obtained.

The nuclear potential for deuterium is assumed is assumed to be square well of finite depth as given as

V(r) = { - Vo for 0<r<b

{ 0 for r > b

The Schrodinger wave equation is solved for the ground state of deuterium nucleus having reduced mass

µ = M/2

i.e the equation becomes,

where l = orbital quantum number

For ground state, l = 0 (s tate)

Multiplying R/r²

For the wave function to be finite

R (r=0) = R (r -> ∞ ) =0

=> C=0 and C'=0

Equation 7 reduces to

Using boundary condition

Dividing we get kcot kb = -k'

or, cot kb = -k'/k

Graphical solution

If a graph of y = cot kb and y = - k'/ k then intersecting points gives solution. Since intersecting points are slightly greater then 𝜋/2, 3𝜋/2, 5𝜋/2 only possible solution is 𝜋/2 then

cot kb = 0 = cot 𝜋/2

kb = 𝜋/2

where Vo = potential

b = range of nuclear force

This note is a part of the Physics Repository.