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 \[{{d}_{0}}\].
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