The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the nucleus with linear momentum P is d . The distance of closest approach of alpha particle to nucleus, if the linear momentum of the alpha particle is 1.5 P

Step-by-step solution

At sufficient distance, electric potential energy is zero. So, U ₁=0 at (1) K ₁= 1 2 mv ^2= P ^2 2 ~m at (2) K ₂=0 U ₂= 1 4 ₀ q _ q _ n d _ c By energy conservation aligned & U ₁+ K ₁= U ₂+ K ₂ & 0+ p ^2 2 ~m = 1 4 ₀ q _ q _ n d _ c +0 aligned Distance of closest approach, d_c 1 p^2 aligned & d ₂ ~d ₁ = p ₁^2 p ₂^2 & = d p ^2 (1.5 p )^2 = 4 ~d 9 aligned