for a central potential, the angular momentum is a conserved quantity. Thus, we can expand the wave function by the angular momentum wave function:

$\sum a_l Y_{l , m=0} R_l(k, r)$

the m=0 is because the spherical symmetry. the R is the radial part of the wave function. and a is a constant. k is the linear momentum and r is the radial distance.

$R_l(k,r) \rightarrow J_{Bessel} (l, kr )$

which is reasonable when r is infinite and the nuclear potential is very short distance. when r goes to infinity,

$J_{Bessel} (l,kr) \rightarrow \frac {1}{kr} sin( k r - \frac{1}{2} l \pi )$

for elastic scattering, the probability of the current density is conserved in each angular wave function, thus,

the effect of the nuclear potential can only change the phase inside the sin function:

$\frac{1}{kr} sin( k r - \frac {1}{2} l \pi +\delta_l )$

with further treatment, the total cross section is proportional to $sin^2(\delta_l)$.

thus, by knowing the scattering phase shift, we can know the properties of the nuclear potential.

for more detail : check this website