Question #43827

A beam of particles of de Broglie wavelength $\lambda$ is scattered by a heavy nucleus located at the origin. Instead of a hard sphere, the nucleus acts as a 'soft' scattering center characterized by the Yukawa potential $V(r) = \frac{V_0}{r} e^{-r/R_0}$, where $R_0$ is the nuclear force range. The total scattered wave amplitude $A(\theta)$ at a scattering angle $\theta$ is the coherent superposition of waves scattered from all volume elements $dV$. The contribution to the amplitude from a volume element at position $\vec{r}$ is proportional to $V(r) e^{i \Delta \phi} dV$, where $\Delta \phi = (\vec{k}_i - \vec{k}_f) \cdot \vec{r}$ is the phase difference relative to the origin. Here $\vec{k}_i$ and $\vec{k}_f$ are the incident and scattered wave vectors respectively, with $|\vec{k}_i| = |\vec{k}_f| = \frac{2\pi}{\lambda}$. If the scattered intensity at $\theta = 0^\circ$ is $I_0$, what is the scattered intensity $I(\theta)$ at an angle $\theta$? \n(Hint: $\int_0^\infty e^{-ax} \sin(bx) dx = \frac{b}{a^2 + b^2}$)