Question

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: \( \int_0^\infty \sin(qr) e^{-\alpha r} \, dr = \frac{q}{\alpha^2 + q^2}  \))

• A. \( q^{-2} \)
• B. \( q^{-1} \)
• C. \( q \)
• D. \( q^2 \)

Hint:

In the first Born approximation, the scattering amplitude is proportional to the Fourier transform of the potential, and at low \( \alpha \), it follows a \( q^{-2} \) dependence.

Answer 1

The Correct Option is A

Step 1: The first Born approximation for elastic scattering amplitude involves calculating the integral of the potential with the sine function in the integrand. The key integral provided is: \[ \int_0^\infty \sin(qr) e^{-\alpha r} \, dr = \frac{q}{\alpha^2 + q^2}. \]

Step 2: The elastic scattering amplitude is proportional to the Fourier transform of the potential. This integral gives the dependence of the amplitude on the wave-vector \( q \) and the parameter \( \alpha \).

Step 3: As \( \alpha \to 0 \), the dominant term in the expression becomes proportional to \( q^{-2} \), which matches option (A).
Thus, the correct answer is (A).