Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 33E / Solution

For a spherically symmetric potential, take the polar axis along the momentum transfer $q$. Its angular integral isThe first Born approximation consequently gives $f(\theta )=-2m({\hslash }^{2}q{)}^{-1}{\int }_0^{\infty }rV(r)\sin (qr)dr$. The incident and scattered wave vectors both have magnitude $k$, so $q^2=2k^2(1-\cos \theta )$ and $\boxed{q=2k\sin (\theta /2)}$.

For the exponential potential, ${\int }_0^{\infty }r{e}^{-(1-iq)r}dr=(1-iq{)}^{-2}$. Taking the imaginary part gives $2q/(1+q^2{)}^2$. ThereforeThe differential scattering cross-section is proportional to $[1+q^2{]}^{-4}$. At large $k$, substantial intensity requires $q=O(1)$, hence $\theta =O(k^{-1})$. For a precise convention, the half-intensity cone has half-angle ${\theta }_{1/2}\sim \sqrt{2^{1/4}-1}/k$. Its full angular width is twice that. The exponential's range has been set to one; restoring range $a$ gives width of order $1/(ka)$. As energy increases, the scattering therefore concentrates into the forward cone.