Saturday, 14 March 2015

Why is the potential energy minimal when the repulsion and attraction force between molecules is 0?



According to this potentialcurve and the other curve(?) the potential energy is minimal when the intermolecular forces equal zero.


enter image description here


Why? How should I interprete negative potential energy in this context?



Answer



Remember that potential energy is defined up to a constant. Here it's defined so that when $r \to \infty$ the potential energy is worth 0, for convenience. So that there's nothing special about a negative potential energy per se.


What really matters here is that there's a "well", i.e. a region for $r$ where the potential energy has a minimum and that minimum is lower than for $r \to \infty$. This allows a region of stability for the molecule.


What happens is that the molecule will have discrete allowed values for its vibrational energy (an oscillator, similar to the harmonic oscillator if the energy is small compared to the depth of the potential well), and the energy of the molecule can't be greater than the depth of the potential well in your graph, else the molecule dissociates.


More details at this link.


To answer the title question, remember that $\vec F = - \nabla \Phi$ where $\Phi$ is the potential. So that $\vec F=\vec 0$ implies $\nabla \Phi =0 \Rightarrow \frac{d \Phi}{d r}=0$, this corresponds to a minimum for $\Phi (r)$.


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