"In atomic or nuclear scattering, the collisions are typically elastic because the repulsive Coulomb force keeps the particles out of contact with each other"

How does this allow the interaction to be elastic, because wouldn't the electrostatic force alter the speed as particles are accelerated away from each other?

  • $\begingroup$ again, I am concerned about the source. What is it? It seems very misleading. There really is no such thing as "contact" in particle scattering, as explained in Sammy Gerbil's answer. $\endgroup$
    – JEB
    Mar 20, 2020 at 13:44

1 Answer 1


The speed (and therefore also kinetic energy) of both of the particles changes during the interaction. What the term elastic means is that the total kinetic energy is conserved. It is the same after the interaction as it was before.

As with a spring, the kinetic energy of the particles is converted to electrostatic potential energy, stored in the electric field between them, as the particles get closer together. Their speeds reduce, the electrostatic PE (similar to elastic energy stored in a spring) increases. At the point of closest approach the speeds are zero in a head-on collision; then the electrostatic PE is maximum and equal to the total initial KE of both particles.

The process reverses as the particles push each other apart. The electrostatic PE reduces, the KE of each increases. Finally there is no electrostatic PE; it has all been converted back to KE which is the same as it was initially.

The interaction is called elastic because total KE is the same before the interaction as it is after.

The reason why atomic and nuclear interactions are usually elastic is because these are quantum objects. They have internal energy levels which are much further apart than their total kinetic energy - provided that their speeds are low enough. Then there is not enough KE to make any change to the internal energy of the atom or nucleus. So the only way that KE can be stored is as electrostatic energy, and this mechanism is perfectly elastic as explained above.

For atoms the internal energy levels are separated by about 1 electron volt ($1 \text{eV}$). At room temperature the KE of an atom is typically $\frac{1}{40}\text{eV}$. So collisions at room temperature are rarely enough to cause changes in internal energy. Such collisions are elastic.

At much higher temperatures around $8000K$ the KE of atoms is high enough to change their internal energy during a collision. At these temperatures collisions are not elastic because after the atoms have separated some of the initial KE remains as internal energy in one or both atoms.

For nuclei the energy levels are separated by around $1\text{MeV}$. The KE required to change the internal energy is very much higher than for atoms, corresponding to a temperature of about $6\times 10^9 K$. Such temperatures are found in nature only at the centre of stars. So nuclear interactions are usually elastic : they do not change internal energy.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.