A proton is fired from far away toward the nucleus of a mercury atom. Mercury is element number 80, and the diameter of the nucleus is 14.0 fm. The proton is fired at a speed of 1.80×107 m/s. When it passes the nucleus, how close will the proton be to the surface of the nucleus? Assume the nucleus remains at rest.

Answer 1

KE = Fd (work required to stop the proton)
1/2 m v^2 = kq(80q)/r
elementary charge is 1.6e-19 but I don't know the mass of a proton offhand.

Answer 2

mass pf proton is 1.67x10^-27 kg