Question

Determine the magnitude of the electric field at the surface of a lead-206 nucleus, which contains 82 protons and 124 neutrons. Assume the lead nucleus has a volume 206times that of one proton and consider a proton to be a sphere of radius 1.20 ? 10^-15 m.  ??? N/C

Answer 1

Charge q = 82 e = 82 * 1.6 * 10 ^ -19 C

Radius of the proton r = 1.2* 10 ^ -15 m

Volume of the proton v = ( 4/ 3) (pi) r ^ 3

Volume of the nucleus V = 206 v

            ( 4/3 ) (pi) R ^ 3 = 206 * ( 4/ 3) (pi) r ^ 3

                             R ^ 3 = 206 * r ^3                                     

From this radius of the nucleus R = 5.906 * r

                                                      = 7.087* 10 ^ -15 m

the magnitude of the electric field at the surface of alead-208 nucleus is E = kq / R ^ 2

where K = coulomb's constant = 8.99 * 10 ^ 9 Nm^2/ C ^ 2

plug the values, we get the electric field as E = 2.35* 10 ^ 21 N / C