question_answer

In the Rutherford scattering experiment the distance of closest approach for an \[\alpha \]-particle is \[{{d}_{0}}\].If \[\alpha \]-particle is replaced by a proton, how much kinetic energy in comparison to \[\alpha \]-particle will it require to have the same distance of closest approach \[{{d}_{0}}\]?                                                    

Answer:

                At the distance of closest approach, \[{{K}_{\alpha }}=\frac{kZe.2e}{{{d}_{0}}}\] and \[{{K}_{p}}=\frac{kZe.e}{{{d}_{0}}}\] \[\therefore \]  \[{{K}_{p}}=\frac{1}{2}{{K}_{\alpha }}\] Thus, a proton would need half the initial K.E. of that of an \[\alpha \]-particle for the distance \[{{d}_{0}}\].