# How many different types of energy can a particle have? [closed]

I was just wondering if somebody could tell me if my reasoning here is correct:

An atom or molecule as a whole can have three types of energy: translational, rotational, and vibrational.

If the atom or molecule is a free particle, then its translational motion obeys classical laws. If it is not free, then it must be analyzed using the "particle in a box" model in one, two, or three dimensions, depending upon the situation.

Diatomic molecules have rotational energy, which can be analyzed using the rigid rotor model.

Atoms or molecules can both have vibrational energy, which can be analyzed using the simple harmonic oscillator idealization.

Finally, in atoms with only one electron (mainly hydrogen), the electron is analyzed using the "particle in a box" model, where the potential energy well is the nucleus itself.

Thanks!

## closed as unclear what you're asking by lemon, Jon Custer, stafusa, Kyle Kanos, sammy gerbilOct 24 '17 at 14:16

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• This question is very unclear. You ask about 'types of energy' but then discuss degrees of freedom. You also seem to draw some vague and arbitrary divisions between how these systems may be 'analyzed' without any obvious relevance to an actual question. – lemon Oct 20 '17 at 14:06
• Alright then. Is it correct to say that a diatomic molecule has six degrees of freedom (three translational, two rotational, and three vibrational) and that a single atom has only three (all translational, with no rotational or vibrational degrees of freedom)? – user172785 Oct 20 '17 at 15:12

An atom or molecule as a whole can have three types of energy: translational, rotational, and vibrational.

No, these are only three types of energy, but actually as we will see the the ways in which energy can be "stored" inside an atom or a molecule are much more...

## General definition of energy

Usually, the total energy of a system is subdivided into kinetic energies of all particles and potential energies of the interactions between them, where the latter terms are most often limited to the strongest Coulomb interactions only. In atomic units we get then:

$$E = -\sum_{\alpha=1}^{\nu}\frac{1}{2m_{\alpha}}\nabla_{\alpha}^2 +\sum_{i=1}^{n}\frac{1}{2}\nabla_{i}^2 -\sum_{\alpha=1}^{\nu}\sum_{i=1}^{n}\frac{Z_{\alpha}}{r_{\alpha i}} +\sum_{\alpha=1}^{\nu}\sum_{\beta>0}\frac{Z_{\alpha}Z_{\beta}}{r_{\alpha\beta}} +\sum_{i=1}^{n}\sum_{j=i+1}^{n}\frac{1}{r_{ij}},$$

where $m_{\alpha}$ is the rest mass of nucleus $\alpha$ and $Z_{\alpha}$ is its atomic number, $r_{\alpha i}=|r_{\alpha}-r{i}|$ is the distance between nucleus $\alpha$ and electron $i$, $r_{\alpha \beta}=|r_{\alpha}-r{\beta}|$ is the distance between two nuclei $\alpha$ and $\beta$, and $r_{ij}=|r_{i}-r_{j}|$ is the distance between two electrons $i$ and $j$.

Now, the zero of energy corresponds to the case when all the terms are zero. Kinetic energy is zero when velocity is zero, while the Coulomb interaction energy is zero when particles are infinitely separated from each other. Thus, the zero of energy corresponds to the case when all the particles are infinitely far away from each other and not moving. In accordance with that choice of the zero of energy, the energy of any stable molecular system in the modeled environment (usually, an isolated molecule in the gas phase) has to be negative.

## Molecular mechanics definition of energy

Diatomic molecules have rotational energy, which can be analyzed using the rigid rotor model.

True, but only with a very rough degree of approximation. Obviously tho atoms taken as a system of two atoms can be seen as a rigid rotor model, but that doesn't take into consideration electrons and chemical potentials.

Molecular mechanics (MM) operates with a different kind of energy. To start from, usually, it is just the potential energy. Secondly, in MM we think about a molecule in terms of atoms and bonds (rather than nuclei and electrons) and the potential energy is divided into contributions from bond stretching, angle bending, torsional rotations around single bonds, non-bonded interactions, etc.

$$E=E_{stretch}+E_{bend}+E_{torsion}+E_{nb}+\dots,$$

where the first term, for instance, is typically proportional to the sum of the squares of extensions from the equilibrium length for all the bonds:

$$E_{stretch}=\sum_{bonds}k(r−re)^2.$$ Now, we could take the energy corresponding to the equilibrium length re as the zero of energy and proceed similarly for the other bonded terms, which would mean that (ignoring usually small non-bonded interactions) the MM potential energy is zero when the molecular system is completely "unstrained" (i.e. when all geometrical parameters have their "natural" equilibrium values). However, different molecules are bonded differently, and thus, the zero of energy would, in general, be defined differently. Each potential energy will thus be computed relative to a different zero of energy which makes the comparison of energies for any two different molecules meaningless.

Is it correct to say that a diatomic molecule has six degrees of freedom (three translational, two rotational, and three vibrational) and that a single atom has only three (all translational, with no rotational or vibrational degrees of freedom)?

A diatomic molecule has 5 (respectively 6) degrees of freedom (3 spatial and 2 rotational + 1 vibrational)

A monoatomic molecule has 3 spatial degrees of freedom, but see above for a detailed discussion.