= Solution
For a spherically symmetric potential, take the polar axis along the momentum transfer $q$. Its angular integral is
$$
\int e^{-iqr\cos\vartheta}\,d\Omega=2\pi\int_{-1}^1e^{-iqr v}\,dv=4\pi\frac{\sin(qr)}{qr}.
$$
The <first Born approximation> consequently gives $f(\theta)=-2m(\hbar^2q)^{-1}\int_0^\infty rV(r)\sin(qr)\,dr$. 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, $\int_0^\infty re^{-(1-iq)r}\,dr=(1-iq)^{-2}$. Taking the imaginary part gives $2q/(1+q^2)^2$. Therefore
$$
\boxed{f(\theta)=-\frac{4mV_0}{\hbar^2[1+4k^2\sin^2(\theta/2)]^2}.}
$$
The <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.
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