Question

In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles (nuclei of helium atoms) from thin sheets of gold. An alpha particle, having charge 2e and mass 6.64 10-27 kg, is a product of certain radioactive decays. The results of the experiment led Rutherford to the idea that most of an atom's mass is in a very small nucleus, with electrons in orbit around it. Assume an alpha particle, initially very far from a stationary gold nucleus, is fired with a velocity of 1.50 107 m/s directly toward the nucleus (charge 79e). What is the smallest distance between the alpha particle and the nucleus before the alpha particle reverses direction?

Answer 1

Answer:

Explanation:

We shall apply conservation of mechanical energy

kinetic energy of alpha particle is converted into electric potential energy.

1/2 mv² = k q₁q₂/d , d is closest distance

d = 2kq₁q₂ / mv²

= 2 x 9 x 10⁹ x 79e x 2e / 4mv²

= 1422 x2x (1.6 x 10⁻¹⁹)² x 10⁹ /4x 1.67 x 10⁻²⁷ x (1.5 x 10⁷)²

= 3640.32 x 10⁻²⁹ /2x 3.7575 x 10⁻¹³

= 484.4 x 10⁻¹⁶

=48.4  x 10⁻¹⁵ m