Discussion
A particle with incoming wave vector k, after being scattered by the potential \(V(r) = \frac{c}{r^{2}}\), c real, goes out with wave vector k'. The differential scattering cross-section, calculated in the first Born approximation, depends on q = |k − k'|, as
Answer: A
The cross-section goes as \(1/q^{2}\).
In the first Born approximation, for a central potential the scattering amplitude is
\[ \begin{aligned} f(q) &= -\frac{2m}{\hbar^{2}q}\int_{0}^{\infty} rV(r)\sin(qr)\,dr \\ &= -\frac{2mc}{\hbar^{2}q}\int_{0}^{\infty}\frac{\sin qr}{r}\,dr \\ &= -\frac{2mc}{\hbar^{2}q}\cdot\frac{\pi}{2} \end{aligned} \]
The remaining integral is \(\pi/2\) for any \(q \gt 0\), so \(f \propto 1/q\) and \(d\sigma/d\Omega = |f|^{2} \propto 1/q^{2}\).
\(1/q^{4}\) is the Rutherford dependence of the Coulomb potential \(1/r\), not of \(1/r^{2}\). \(1/q\) is the amplitude, not the cross-section, and \(1/q^{3/2}\) does not arise.
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