Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach the surface of a 15-fmfm-diameter 238U238U nucleus

Explanation:

GIVEN

diameter = 15 fm  =m

we use here energy conservation

there will be some initial kinetic  energy but after collision kinetic energy will zero

on solving these equations we get kinetic energy initial

J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

the answer is to check all of them! they all apply well at least i think so!

prolly a beacuse the momentuem the car has increases while going down hill