can you guys please explain how this was calculated?

Example 22.2 Proton radiation therapy Proton radiation therapy destroys cancer cells by irradiating them with a beam of highenergy protons. Most of the proton energy disrupts bonds and damages DNA in tumor cells, but a small fraction of the protons come very close to the nucleus of an atom. Suppose a proton is aimed directly at the 5.4fm-diameter nucleus of a 12C atom, where 1 fm = 1 femtometer = 10'15 In. What initial speed must the proton have to just reach the surface of the nucleus? The proton mass is mp = 1.67 X 1027 kg. PREPARE Carbon has atomic number 6, so a 12C nucleus has six protons and charge qcarbon = +68. The interaction between the proton with qpmton = +6 and the nucleus is an interaction between two charged particles. The atom's electrons are much less massive than the proton and the nucleus, and they are spread out over a distance much greater than the nuclear diameter, so we will assume that the proton's interactions with electrons can be ignored. The proton starts from "far away,\" so we will assume that U, = 0. Figure 22.99 is a before-andafter pictorial representation. To "just reach\" the surface of the nucleus means that the proton comes to rest (vf = 0) as it reaches rf = R = 2.7 x 10'15 m, the radius of the nucleus. Figure 22.9 A proton approaching a nucleus. Figure 22.9 A proton approaching a nucleus. Vi Before: r ~0 so Uj = 0 R After: + rf = R Vf = 0 SOLVE Conservation of energy Kf + Uf = Ki + Uj is 0+ K qproton 9carbon 6Ke2 - mp Vi + 0 R We can solve for the proton's initial speed: 12Ke2 Vi = = 2.5 x 10 m/s mp R ASSESS This is a very high speed, nearly 10% of the speed of light, but it is typical of proton >