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: \( \int_0^\infty \sin(qr) e^{-\alpha r} \, dr = \frac{q}{\alpha^2 + q^2} \))
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: \( \int_0^\infty \sin(qr) e^{-\alpha r} \, dr = \frac{q}{\alpha^2 + q^2} \))
Show Hint
- \( q^{-2} \)
- \( q^{-1} \)
- \( q \)
- \( q^2 \)
The Correct Option is A
Solution and Explanation
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).
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