Neglecting recoil of the gold nucleus how much kinetic energ

Neglecting recoil of the gold nucleus, how much kinetic energy must an alpha particle (charge = 2 times 1.6 times 10^19 C) have to approach to within 1.00 times 10^14 m of a gold nucleus (charge = 79 times 1.6 x 10^19 C)? (kc = 8.99 times 10^9 N-m^2/C^2 and 1 MeV = 1.6 times 10^13 J) 11.7 MeV 14.6 MeV 18.2 MeV 22.7 MeV

Solution

Charge of alpha particle q = 2e = 2x1.6 x10 ^-19 C

Charge of gold nucleus q \' = 79 e = 79 x1.6 x10 ^-19 C

Closest approach distance r = 1 x10 ^-14 m

Kinetic energy of the alpha particle = potential energy of alpha and gold nucleus at closest approach

                                                E = Kqq \' / r

Where K = Coulomb\'s constant = 8.99 x10 ⁹ Nm ²/C ²

Substitute values you get ,

E = (8.99x10 ⁹)(2x1.6 x10 ^-19 )(79x1.6 x10 ^-19 ) /(1 x10 ^-14 )

   = 3.636x10 ^-12 J

= [(3.636 x10 ^-12 ) /(1.6 x10 ^-13 )]MeV

=22.7 MeV