Problem 6.17

The effective cross section of a gold nucleus corresponding to the scattering of monoenergetic alpha particles within the angular interval from $90^{\circ}$ to $180^{\circ}$ is equal to $\Delta \sigma=0.50 \mathrm{~kb} .$ Find: (a) the energy of alpha particles; (b) the differential cross section of scattering $d \sigma / d \Omega(\mathrm{kb} / \mathrm{sr})$ corresponding to the angle $\theta=60^{\circ}$.