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- Thread starter eightsquare
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The R^6 term is an attractive part that arises from dispersion forces. The calculation, showing that R^7 is a better approximation, is outlined in the textbook Molecular Quantum Mechanics by PW Atkins.

The R^12 term is arbitrary. A better description of the repulsive part of the potential is Aexp(-Br)/r (Bohr potential). The main contribution is internuclear repulsion.

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The R^6 term is an attractive part that arises from dispersion forces. The calculation, showing that R^7 is a better approximation, is outlined in the textbook Molecular Quantum Mechanics by PW Atkins.

The R^12 term is arbitrary. A better description of the repulsive part of the potential is Aexp(-Br)/r (Bohr potential). The main contribution is internuclear repulsion.

gadong, I don't really have anything constructive to contribute here, with the exception that I think it would help you to know that he is trying to better understand contact forces. Here is the original thread...

https://www.physicsforums.com/showthread.php?t=700663

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In day to day life, the forces of contact become apparent when objects are compressed relative to their equilibrium separation, which represents a potential energy minimum. The closer they approach, the more the potential energy of the system (the objects in contact) increases, so a force is required in order to do the work of compression. That's the contact force that we experience.

Potential energy curves such as the LJ potential represent the system energy as a function of displacement from the equilibrium separation. The LJ potential is not suitable for describing solid materials, but is qualitatively correct. Interestingly, there is in fact a "universal" compression behaviour that applies to most solids (http://en.wikipedia.org/wiki/Rose–Vinet_equation_of_state).

However, the atomic displacements that we encounter in out day to day contact with metrials are usually small, and can mostly be understood by assuming a simpler harmonic potential. The properties of materials under small compressions are described using the so-called elastic constants.

In general, materials that have high cohesive energies (deep potential wells between atoms) tend to be difficult to compress (example: tungsten). However, the actual behaviour depends on the matter in which forces are applied to the material (e.g. compressive versus shear forces). In practical research, it is feasible to predict compressive forces for model systems (e.g. nanoscale tip indenting a surface), but frictional forces cannot be predicted with accuracy.

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What is the reason for the potential energy increasing?(I know it ultimately is the EM force, but why do electrons interact like that when they come very close?)

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eightsquare,

It is not correct to attribute repulsive interatomic forces to electronic effects. For example, the forces between two atoms are due to energy changes in the entire two-atom system. The dominant contribution is the repulsive interaction between the two nuclei. I will work through a simple example to demonstrate this - two hydrogen atoms approaching.

I make use of the following ionization potential data:

H #1: 13.6 eV (H+)

He #1: 24.6 eV (He+); #2: 54.4 eV (He2+).

Consider the situation when two H atoms are far apart. The nuclei do not interact significantly with each other, nor do the electrons, due to their large separation. The energy stored up in the electron-nuclear interactions is the same for each atom, or in total = -2*13.6 = -27.2 eV. This is the system energy at large internuclear separations.

Now bring the nuclei quite close together. Two things happen: (a) the Coulombic potential energy due to nuclear repulsion becomes significant; (b) the energy stored up in electron-nuclear interactions changes, since each electron can now interact with both nuclei; (c) there are also repulsive interactions between electrons.

For very small internuclear separations (*R*), the contribution from (a) goes up towards +infinity, while the contribution from (b) and (c) approaches a limiting value, which we can easily calculate: note that as *R* tends to zero, the system transforms into a helium atom (that is, a neutral atom with a +2 nuclear charge). So, the limiting value of (b) + (c) can be deduced from the sum of first and second ionization energies for He, or -79 eV.

From this, one can see that the energy associated with electronic effects does fall by up to 51.8 eV [i.e. from -27.2 to -79.0] as the nuclei approach. However, it is not difficult to see that this is overwhelmed by the repulsive internuclear interaction which rises much more rapidly.

At intermediate distances the overall interaction results in a negative energy change relative to large separations, so H atoms can form a stable molecule.

So, to sum up, the repulsive forces between atoms are mainly due to nuclear repulsion, which at small*R* is not counterbalanced by the more attractive electronic contributions.

It is not correct to attribute repulsive interatomic forces to electronic effects. For example, the forces between two atoms are due to energy changes in the entire two-atom system. The dominant contribution is the repulsive interaction between the two nuclei. I will work through a simple example to demonstrate this - two hydrogen atoms approaching.

I make use of the following ionization potential data:

H #1: 13.6 eV (H+)

He #1: 24.6 eV (He+); #2: 54.4 eV (He2+).

Consider the situation when two H atoms are far apart. The nuclei do not interact significantly with each other, nor do the electrons, due to their large separation. The energy stored up in the electron-nuclear interactions is the same for each atom, or in total = -2*13.6 = -27.2 eV. This is the system energy at large internuclear separations.

Now bring the nuclei quite close together. Two things happen: (a) the Coulombic potential energy due to nuclear repulsion becomes significant; (b) the energy stored up in electron-nuclear interactions changes, since each electron can now interact with both nuclei; (c) there are also repulsive interactions between electrons.

For very small internuclear separations (

From this, one can see that the energy associated with electronic effects does fall by up to 51.8 eV [i.e. from -27.2 to -79.0] as the nuclei approach. However, it is not difficult to see that this is overwhelmed by the repulsive internuclear interaction which rises much more rapidly.

At intermediate distances the overall interaction results in a negative energy change relative to large separations, so H atoms can form a stable molecule.

So, to sum up, the repulsive forces between atoms are mainly due to nuclear repulsion, which at small

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eightsquare,

Now bring the nuclei quite close together. Two things happen: (a) the Coulombic potential energy due to nuclear repulsion becomes significant; (b) the energy stored up in electron-nuclear interactions changes, since each electron can now interact with both nuclei; (c) there are also repulsive interactions between electrons.

For very small internuclear separations (R), the contribution from (a) goes up towards +infinity, while the contribution from (b) and (c) approaches a limiting value, which we can easily calculate: note that asRtends to zero, the system transforms into a helium atom (that is, a neutral atom with a +2 nuclear charge). So, the limiting value of (b) + (c) can be deduced from the sum of first and second ionization energies for He, or -79 eV.

So, to sum up, the repulsive forces between atoms are mainly due to nuclear repulsion, which at smallRis not counterbalanced by the more attractive electronic contributions.

Great explanation! I always wondered how how this can happen as I took the electron orbits to be kind of like walls which cannot be passed so I took that the nuclei cannot come closer than r1+r2.

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