(a) Use energy methods to calculate the distance of closest...

(a) Use energy methods to calculate the distance of closest approach for a head-on collision between an alpha particle with an initial energy of 0.5 MeV and a gold nucleus $\left(^{197} \mathrm{Au}\right)$ at rest. (Assume that the gold nucleus remains at rest during the collision.) (b) What minimum initial speed must the alpha particle have in order to approach to a distance of $300 \mathrm{fm} ?$

Key Concepts

Energy Conservation

In collision problems, the total mechanical energy (the sum of kinetic and potential energies) remains constant if there are no non-conservative forces acting on the system. This principle allows us to relate the initial kinetic energy of a particle to the potential energy at the point of closest approach, where all the kinetic energy has been converted into potential energy.

Kinetic to Potential Energy Conversion

For a head-on collision under repulsive forces, the initial kinetic energy of the incoming particle is completely transformed into potential energy at the point of closest approach. Setting the kinetic energy equal to the electrostatic potential energy provides a method to calculate the minimum separation distance between colliding nuclei.

Coulomb Potential Energy

Coulomb's law describes the electrostatic interaction between two charged particles. The potential energy of the interaction is inversely proportional to the distance between them, and by equating this potential energy to the initial kinetic energy, one can determine the distance at which the repulsion prevents further approach.

Distance of Closest Approach

The distance of closest approach is the minimum separation between two charged particles during a collision, reached when the initial kinetic energy of the projectile is exactly balanced by the repulsive Coulomb potential energy. This concept is crucial in understanding the dynamics of collisions in scattering experiments and in nuclear physics.

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