Given- ∗ Energy of α -particle =5MeV By conservation of energy, we can say that for the distance of closest approach all the kinetic energy is converted into potential energy. ⇒KE initial =PE final
⇒12mv2=kq1q2r,q1= charge of uranium nucleus q2= charge of α− particle
⇒5×106×1.6×10−19 J=9×109×2×1.6×10−19×92×1.6×10−191⇒r0=5.3×10−14 mr0=5.3×10−12 cm Hence, radius of closest approach is of the order of 10−12 cm . Hence, option (c) is correct.