In Rutherford’s famous scattering experi- ments (which led to the planetary mode

Question

In Rutherford’s famous scattering experi-
ments (which led to the planetary model of
the atom), alpha particles (having charges of
+2 e and masses of 6.64×10-27 kg) were fired
toward a gold nucleus with charge +79 e. An
alpha particle, initially very far from the gold
nucleus, is fired at 2.77 × 107 m/s directly
toward the gold nucleus.

2e v=0 (no charges shown) 79e
(++)>>>>>>><<<<<<<()----------------d------------- (79+)
(what's shown at the top () is a circle)

The Coulomb constant is 8.98755 ×
109 N · m2/C2.
How close does the alpha particle get to the
gold nucleus before turning around? Assume
the gold nucleus remains stationary.
Answer in units of m.

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 Plugnchug

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