In a Rutherford scattering experiment an particle charge 2e

In a Rutherford scattering experiment, an -particle (charge = +2e) heads directly toward a mercury nucleus (charge = +80 e). The -particle had a kinetic energy of 5.0 MeV when very far (r ) from the nucleus. Assuming the mercury nucleus to be fixed in space, determine the distance of closest approach. (Hint: Use conservation of energy with PE = keq1q2/r.)

________fm

Solution

Charge of alpha particle q = 2e

Charge of mercury nucleus q \' = 80 e

Kinetic energy of the alpha particle K.E = 5 MeV

                                                      = 5 x1.6 x10 ^-13 J

At closest approach ,the kinetic energy of alpha particle = Potential energy of the alpha and mercury

                     K.E = Kqq \'/ r

                           =K(2e)(80e) / r

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

From above equation r = K(2e)(80e)/K.E

                                   = K(160 e ² ) /K.E

                                   = (8.99x10 ⁹)(160)(1.6x10 ^-19) ² /(5 x1.6 x10 ^-13 )

                                   = 4.6 x10 ^-14

                               r = 46 x10 ^-15 m

                                 = 46 fm