Consider the potential $U(r)$ defined as $$U(r) = -U_0 \frac{e^{- \alpha r}}{r}$$ where $ \alpha$ and $U_0$ are real constants of appropriate dimensions. According to the first Born approximation, the elastic scattering amplitude calculated with $U(r)$ for a (wave-vector) momentum transfer $q$ and $ \alpha \to 0$, is proportional to (Useful integral: $ \int_0^{ \infty} \sin(qr)e^{- \alpha r} dr = \frac{q}{ \alpha^2+q^2}$)
Calculating Scattering Amplitude with First Born Approximation
The question asks for the behavior of the elastic scattering amplitude in the first Born approximation, specifically in the limit where the parameter $ \alpha $ approaches zero. The scattering amplitude $ f(\mathbf{q}) $ is proportional to the Fourier transform of the interaction potential $ U(\mathbf{r}) $, where $ \mathbf{q} $ is the momentum transfer vector.
Interaction Potential
The potential is given by:
$U(r) = -U_0 \frac{e^{- \alpha r}}{r}$
Here, $ U_0 $ and $ \alpha $ are real constants.
First Born Approximation
In the first Born approximation, the scattering amplitude $ f(\mathbf{q}) $ is proportional to the Fourier transform of the potential $ U(\mathbf{r}) $:
$f(\mathbf{q}) \propto \int d^3r \, e^{-i\mathbf{q}\cdot\mathbf{r}} U(\mathbf{r})$
Substituting the given potential:
$f(\mathbf{q}) \propto \int d^3r \, e^{-i\mathbf{q}\cdot\mathbf{r}} \left( -U_0 \frac{e^{- \alpha r}}{r} \right)$
The integral term is the Fourier transform of the screened Coulomb potential (or Yukawa potential kernel) $ \frac{e^{- \alpha r}}{r} $. The Fourier transform of $ \frac{e^{- \alpha r}}{r} $ is known to be $ \frac{4\pi}{\alpha^2 + q^2} $, where $ q = |\mathbf{q}| $.
Therefore, the scattering amplitude is proportional to:
$f(q) \propto -U_0 \left( \frac{4\pi}{\alpha^2 + q^2} \right)$
Ignoring the constants $ -U_0 $ and $ 4\pi $, we have:
$f(q) \propto \frac{1}{\alpha^2 + q^2}$
Limit $ \alpha \to 0 $
We need to find the dependence on $ q $ as $ \alpha \to 0 $. Substituting $ \alpha = 0 $ into the expression for $ f(q) $:
$f(q) \propto \frac{1}{0^2 + q^2}$
$f(q) \propto \frac{1}{q^2}$
Thus, the scattering amplitude is proportional to $ q^{-2} $.
Conclusion
According to the first Born approximation, for the potential $ U(r) = -U_0 \frac{e^{- \alpha r}}{r} $, the elastic scattering amplitude for a momentum transfer $ q $, in the limit $ \alpha \to 0 $, is proportional to $ q^{-2} $.
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