A proton of mass m and charge e is projected from a very large distance towards an α -particle with velocity v   . Initially α -particle is at rest, but it is free to move. If gravity is neglected, then the minimum separation along the straight line of their motion will be

Step-by-step solution

As α -particle is free to move, initial kinetic energy of system will be initial kinetic energy of system will be k i = 1 2 μ v 2 where, μ = reduced mass of system = m · 4 m m + 4 m Now, by energy conservation, we have Initial kinetic energy = Potential energy at minimum separation r 1 2 m · 4 m m + 4 m v 2 = 1 4 π ε 0 · 2 e 2 r ⇒    r = 5 e 2 4 π ε 0   m v 2