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Suppose I have an atom A (of mass $m$) moving randomly with some velocity $v$ in free space. Now suppose there is one more similar atom B but moving with a different velocity (smaller than $v$) on the same line as that of $A$ and in same direction but initially it is ahead of A.


Temperature of A is its kinetic energy and that of B is B's kinetic energy. So if they come in contact during collision, energy should flow from A to B and finally both should have same temperature i.e. same kinetic energy i.e. same velocity.


The above statement is true only when the collision is inelastic but I was told that atomic collisions are always elastic.

So does this mean that both the atoms will not have same velocity after collision? Or am I wrong somewhere.

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Temperature of A is its kinetic energy

Unfortunately, this is a very common misconception. Thermal energy is energy that is in all unmeasured internal degrees of freedom.

For an ideal gas the only internal degrees of freedom for energy are the kinetic energy, but that is not a general statement, it is specific only to ideal gasses. Real gasses and other materials have more degrees of freedom available to hold energy. For instance atomic orbitals can be excited, and there can be potential energy stored in vibrational, rotational, and torsional modes. In solids there can also be long range vibrational modes called phonons. So thermal energy includes energy in all of these degrees of freedom.

In your example, the collision could very well be inelastic. Then one of the atoms would be excited and the missing KE would be in the atomic orbitals. Although the energy would not be KE, it is still perfectly valid thermal energy. It could later relax and release the energy and it could radiate away or excite another atom thermally.

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  • $\begingroup$ so are you saying that collisions between two atoms can be inelastic ? $\endgroup$
    – Ankit
    Dec 20, 2020 at 18:45
  • $\begingroup$ @Ankit yes, definitely. That is exactly what I meant by “the collision could very well be inelastic” $\endgroup$
    – Dale
    Dec 20, 2020 at 18:52
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Temperature of A is its kinetic energy and that of B is B's kinetic energy

This is not true of particles. In ideal gases one can connect the average kinetic energy with temperature, and in a sloppy way on can say that the kinetic energy of a molecule or atom corresponds to a temperature, but in no way it can be treated in thermodynamic terms. Atoms and molecules belong to the realm of quantum mechanics and so does their scattering.

When two atoms scatter,they have a calculable quantum mechanically probability to scatter elastically and a calculable quantum mechanically probability to scatter inelastically, depending on the boundary conditions of the system and the type of atoms.

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Putting aside the discussion about the definition of temperature with only 1 degree of freedom (which doesn't change the question), the answer to your question is yes.

The collision can be inelastic, i.e. not conserving the kinetic energy, if the excess energy is stored in the internal degrees of freedom of the atoms. For example the atom(s) could become electronically excited or form a chemical bond.

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In addition to Dale's comment, you should be aware that a simple model of chemical reactions is built around atoms and molecules inelastically colliding to form new species. From Reichl "A Modern Course in Statistical Physics:

"Chemical reactions occur in systems containing several species of molecules,... which can transform into one another through inelastic collisions."

All chemical reactions have what is called an "activation energy". It is an amount of energy that must be given to the system so that the chemical reaction can start and comes from the total kinetic energy of the colliding species. Chapter 13, B.5 of Reichl, gives a detailed account of this.

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