(Solved): An alpha particle with kinetic energy 10.5MeV makes a collision with lead nucleus, but it is not "a ...

An alpha particle with kinetic energy $$10.5\mathrm{MeV}$$ makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude $$L=p_{0}b$$, where $$p_0$$ is the magnitude of the initial momentum of the alpha particle and $$b=1.20\times 10^{?12}\mathrm{m}$$. (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82 . The alpha particle is a helium nucleus, with atomic number 2.) What is the distance of closest approach? $$r$$ Part B Repeat for $$b=1.40\times 10^{?13}\mathrm{m}$$. Repeat for $$b=1.50\times 10^{?14}\mathrm{m}$$.