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 \(\theta=60^{\circ}\).