In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an ex
ID: 2111527 • 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 x 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? Assume
the gold nucleus remains stationary.
Explanation / Answer
Use conservation of energy. The speed is low enough that you can get away with classical (non-relativistic) mechanics. We can neglect the impact of the gold's electrons if we end up near the center of their orbitals.
Initial kinetic energy = 1/2 mv^2
=
Final electrostatic potential energy = k q1 q2 / r
Solve for the distance:
r = 2 k q1 q2 / mv^2
k is coulomb's constant--look it up
q1 and q2 are the charges of the alpha and the gold nucleus, which are given in terms of the fundamental charge, e--look that up
m is the alpha particle's mass, which is given
v is the speed at which it is fired, also given