Question

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

• A. circular
• B. parabolic
• C. elliptical
• D. hyperbolic

Hint:

An unbound particle under a repulsive inverse-square Coulomb force traces an open conic; recall which conic corresponds to positive total energy.

Answer 1

The Correct Option is D

Step 1: Identify the force acting on the \( \alpha \)-particle.
An \( \alpha \)-particle carries a positive charge \( +2e \), and the nucleus of the target atom is also positive \( +Ze \). As the \( \alpha \)-particle approaches the nucleus, it experiences a repulsive Coulomb force
\[ F = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r^2} \]
which varies as the inverse square of the distance \( r \).

Step 2: Recognise the type of orbit produced by an inverse-square force.
A body moving under an inverse-square central force follows a conic section (circle, ellipse, parabola or hyperbola). Which one it is depends on whether the force is attractive or repulsive and on the particle's energy.

Step 3: Apply it to this repulsive, unbound case.
The \( \alpha \)-particle comes from far away with kinetic energy, is repelled by the nucleus, and flies off to infinity again. Because it is never trapped (it is an unbound trajectory) and the force is repulsive, the open path it traces is a hyperbola, with the nucleus at the outer focus.

Step 4: Choose the correct option.
Circular and elliptical paths (options i and iii) are closed, bound orbits and cannot describe a particle that escapes to infinity; a parabolic path (option ii) corresponds to exactly zero total energy, which is not the case here. Hence the path is hyperbolic, option (iv).

\[\boxed{\text{Hyperbolic path}}\]