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In Rutherford\'s famous scattering experiments that led to the planetary model o

ID: 1329607 • Letter: I

Question

In Rutherford's famous scattering experiments that led to the planetary model of the atom, alpha particles (having charges of +2e and masses of 6.64 ? 10?27 kg) were fired toward a gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, is fired at 2.40 ? 107 m/s directly toward the nucleus, as in the figure below. How close does the alpha particle get to the gold nucleus before turning around? Assume the gold nucleus remains stationary. Please show your work (picture below).

3. +/10 points SerCP 1016, P023.WI. My Notes Ask Your Teacher In Rutherford's famous scattering experiments that led to the planetary model of the atom, alpha particles (having charges of +2e and masses of 6.64 x 10-27 kg) were fired toward a gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, is fired at 2.40 x 107 m/s directly toward the nucleus, as in the figure below How close does the alpha particle get to the gold nucleus before turning around? Assume the gold nucleus remains stationary 79e 2e Need Help?ReadIt Watch It

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 = 8.9 ×109 N·m2/C2
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

r = 2 k q1 q2 / mv^2

r = 2×8.9 ×109 × 2e×79e/6.64 1027 kg × (2.40 107 m/s)2

= 73.53 e2 1022 m/s = 73.53 0.162 10-36 1022 = 1.887 10-14 m

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