Answer to Question #105441 in Physics for Sophie

Question

In a rutherford scattering experiment and alpha particle is positive charge 2e is directed towards a gold nucleus with charge + 79e- alpha particle has a kinetic energy of 5.0MeV when very far from the nucleus. assuming the Gold nucleus to be fixed and space determine the distance of the closest approach. note 8.99 * 10*9 Newton per hour metre square column

Answer 1

From the conservation of energy:

$$K=\frac{kq_1q_2}{r}$$

$$5(1.6\cdot 10^{-13})=\frac{(8.99\cdot 10^{9})(2)(79)(1.6\cdot 10^{-19})^2}{r}$$

$$r=4.5\cdot 10^{-14}m$$