Physics Problem: Coulomb force, acceleration, and Potential energy on an alpha particle...?

Question

An Alpha particle (Z=2, mass 6.64 X 10^-27 kg) approaches to within 1.00 X 10^-14 m of a carbon nucleus (z=6). What are (a) the maximum Coulomb force on the alpha particle, (b) the acceleration of the alpha particle at this point, and (c) the potential energy of the alpha particle at this point?

I have the answers

(a) 27.6 N away from the carbon nucleous

(b) 4.16 X 10^27 m/s away from the nucleous

(c) 1.73 MeV

The problem is that I do not know how to solve these or what equations to use. My book doesnt use alot of explainations so if you can provde me some that will be very helpful. Thanks!

Answer 1

A carbon nucleus has a mass of 12 u, which is 1.99 x 10^-26 kg. The nucleus has 6 protons, which means it has a charge of 9.60 x 10^-19 C.

Alpha particles have a charge of 3.20 x 10^-19 C and a mass of 6.64 x 10^-27 kg.

(a) F = kq1q2 / r²

= ((8.99 x 10^9 Nm²/C²) x (9.60 x 10^-19 C) x (3.20 x 10^-19 C)) / (1.00 x 10^-14 m)²

= 27.6 N repulsive

(b) a = F / m

= (27.6 N) / (6.64 x 10^-27 kg)

= 4.16 x 10^27 m/s² away

(c) U = kq1q2 / r

= ((8.99 x 10^9 Nm²/C²) x (9.60 x 10^-19 C) x (3.20 x 10^-19 C)) / (1.00 x 10^-14 m)

= 2.76 x 10^-13 J

= (2.76 x 10^-13 J) / (1.60 x 10^-19 J) = 1.73 x 10^6 eV or 1.73 MeV