Using the Born approximation calculate the cross section for the shielded Coulomb potential V(r)=-ze^2*exp(-r/a)*1/r

In the Born approximation we have

kB(,) = kB(k,k') = [2/(424)]d3r' exp(-iqr')U(r')2,

where q = k' - k, k = v0/h, k' = v0/ (k'/k'), and  is the reduced mass.

Here  = me.

k is the incident wave vector, and (k'/k') is a unit vector pointing in the direction (,).

With v(r) = v(r) = (-Ze2/r)exp(-r/a) we have

d3r' exp(-iqr')V(r') = -Ze2r'dr'sin'd'd'exp(-r'/a)exp(-iqr'cos')

= -Ze220r'dr'exp(-r'/a)-11exp(-iqr'cos')dcos'

= -(Ze22/q)0dr'exp(-r'/a)-qr'qr'exp(-ix)dx

= -(Ze22/q)0dr'exp(-r'/a)[-i(eiqr' - e-iqr')]

= -(Ze24/q)0dr'exp(-r'/a)sin(qr') = -(Ze24/q2)q2a2/(1 + q2a2).

Therefore kB(,) = [42Z2e4/4][1/((1/a2) + q2)]2.

With  q = 2ksin(/2) we have kB(,) = [42Z2e4/4][1/((1/a2) + 4k2sin2(/2))]2.

(b)  Let a --> , then  v(r) = (-Ze2/r)exp(-r/a) -->  v0(r) = -Ze2/r.

Then  kB(,) = [42Z2e4/4]/(16k4sin4(/2)) = [Z2e4]/(16E2sin4(/2)).

This is the Rutherford cross section.

(c)  Ratio: kB(,)part a/kB(,)part b = (q2a2/(1 + q2a2))2 = (x2/(1 + x2))2 with x = qa.

As x --> 0, the ratio approaches 0.

As a --> 0, x --> 0, the nuclear charge is completely screened, we have a neutral object.

As x --> , the ratio approaches 1.

As a --> , x --> , the screening vanishes, we have a bare nucleus.