Question:easy

In Rutherford's scattering experiment, the path of a scattered \( \alpha \)-particle is:

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An unbound particle under a repulsive inverse-square Coulomb force traces an open conic; recall which conic corresponds to positive total energy.
Updated On: Jul 10, 2026
  • circular
  • parabolic
  • elliptical
  • hyperbolic
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The Correct Option is D

Solution and Explanation

Step 1: Compare with planetary motion.
In gravitation, a planet bound to the Sun by an attractive inverse-square force moves in a closed ellipse. The \(\alpha\)-particle problem has the same inverse-square mathematics, but with two key differences that change the shape of the orbit.

Step 2: Note the two differences.
(a) The Coulomb force here is repulsive, not attractive, because both the \(\alpha\)-particle and the nucleus are positive. (b) The \(\alpha\)-particle arrives from infinity with positive kinetic energy, so its total mechanical energy is positive, meaning it is unbound.

Step 3: Deduce the conic.
For a central inverse-square force, positive total energy always corresponds to a hyperbolic trajectory. The particle swings once around the nucleus and departs to infinity, never returning.

Step 4: State the result.
Therefore the scattered \(\alpha\)-particle follows a hyperbolic path with the repelling nucleus situated at the focus of the hyperbola. This is exactly what Rutherford used to analyse large-angle scattering.

\[\boxed{\text{Hyperbola}}\]
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