What does this distance of an atom mean?

[Lotto]

TL;DR

I have let's say a hydrogen molecule and the optimum distance is $r_0$. When I know that the range of binding interactions is $r$ from the optimum distance, what does it mean? Is it a deviation of a vibrating hydrogen atom?

I can look at it as if a vibrational motion of the atoms was a simle harmonic motion. So I can consider one of the two atoms to be at rest and the second one to vibrate. Its deviation can be written as $x(t)=r(t)-r_0$.

When I know that the hydrogen molecule stops exiting when the range of binding interactions is $r$ from the optimum distance, does it mean that this $r$ is a deviation $x$? Or is it $r(t)$, the distance of the atoms with respect to time?

 

[Hyperfine]

I find your question rather difficult to understand, but allow me to take a stab at it anyway.

It sounds as if you are defining r₀ to be the equilibrium internuclear separation--essentially the bond length. Then r(t) becomes the change in the bond length during vibrational motion.

If I understand the question correctly, it is what is the physical significance of the maximum value of r(t), which you refer to as r. That would be the classical turning points of the vibrational motion. That is the values of the internuclear separation where the nucleus reaches maximum or minimum internuclear separation, and the nucleus "turns around" to move in the opposite direction. Where the nucleus bumps into the potential energy surface.

Wikipedia has an animated file illustrating the process that I have attempted to describe that might be helpful to you. https://en.wikipedia.org/wiki/File:Anharmonic_oscillator.gif

 

[Lotto]

I find your question rather difficult to understand, but allow me to take a stab at it anyway.

It sounds as if you are defining r₀ to be the equilibrium internuclear separation--essentially the bond length. Then r(t) becomes the change in the bond length during vibrational motion.

If I understand the question correctly, it is what is the physical significance of the maximum value of r(t), which you refer to as r. That would be the classical turning points of the vibrational motion. That is the values of the internuclear separation where the nucleus reaches maximum or minimum internuclear separation, and the nucleus "turns around" to move in the opposite direction. Where the nucleus bumps into the potential energy surface.

Wikipedia has an animated file illustrating the process that I have attempted to describe that might be helpful to you. https://en.wikipedia.org/wiki/File:Anharmonic_oscillator.gif

I don't understand the picture. Why is dissociation energy between $E_0$ and the line above? Why it isn't between the r-axis and the asymptote above?

 

[Hyperfine]

I don't understand the picture. Why is dissociation energy between $E_0$ and the line above? Why it isn't between the r-axis and the asymptote above?

Because the lowest possible energy state of the molecule is E₀.

 

[Lotto]

Because the lowest possible energy state of the molecule is E₀.

But in the picture, the atom is also in the equilibrium state when its energy is under $E_0$. So why is it the lowest energy, when accroding to the picture, it can have lower energy?

 

[Hyperfine]

But in the picture, the atom is also in the equilibrium state when its energy is under $E_0$. So why is it the lowest energy, when accroding to the picture, it can have lower energy?

The lowest energy level allowed by quantum mechanics is that with v=0 where v is the vibrational quantum number. That is E₀.

The illustration shows the mathematical form of the potential energy surface, but what matters is the vibrational states that are shown.

Has your original question been answered to your satisfaction?

 

[Lotto]

Has your original question been answered to your satisfaction?

Well, not really. I consider $r$ to be the range of binding interactions from the optimum distance $r_0$. I don't know if this distance is a displacement from the equilibrium state or a distance between the two atoms.

 

[Hyperfine]

Well, not really. I consider $r$ to be the range of binding interactions from the optimum distance $r_0$. I don't know if this distance is a displacement from the equilibrium or a distance between the two atoms.

The phrase "range of binding interactions from the optimum distance" is not well defined in my opinion.

$r_0$ is the equilibrium internuclear separation--what is typically called the bond length. I think, from your description, that $r$ is the distance between the two nuclei that varies with time as the molecule vibrates.