In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an ex
ID: 1398714 • Letter: I
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 x 10^-27 kg, is a product of certain radioactive decays. The results of the experiment led Rutherford to the idea that most of the atom?s mass is in a very small nucleus, with electrons in orbit around it. (This is the planetary classic model.) Assume an alpha particle, initially very far from a stationary gold nucleus, is fired with a velocity of 2.00 x10^7 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? Assume the gold nucleus remains stationary.Explanation / Answer
Given
Charge of alpha particle q = +2e
Charge of Gold nucleus Q = +79e
Initial velocity of the alpha particle v = 2.00 x 107 m/s
Mass of the alpha particle m = 6.64 x 10-27 Kg
Solution
The kinetic energy of the alpha particle
KE = ½ mv2
KE = ½ (6.64 x 10-27) x (2.00 x 107)2
KE = 13.28 x 10-13 J
This energy is spend in doing work against the electrostatic force of repulsion, let’s say at a distance r from the nucleus, all of the kinetic energy is spent.
As per the conservation of energy, the kinetic energy lost by alpha particle in doing work is saved in it in the form of potential energy
The potential energy
PE = kqQ/r
13.28 x 10-13 = 9 x 109 x 2e x 79e /r
r = 9 x 109 x 2e x 79e /13.28 x 10-13
r = 107.08 x e2 x 1022
r = 107.08 x (1.6 x 10-19)2 x 10-22
r = 2.74.x 10-14 m
at this point the alpha particle will change direction since it does not have any kinetic energy left to resist the repulsive force of gold nucleus