How force on atom in diatomic molecule varies with distance

• I_Try_Math
I_Try_Math
Homework Statement
The potential energy function for either one of the two atoms in a diatomic molecule is often approximated by where $$U(x)=\frac {a} {x^{12}} - \frac {b} {x^6}$$ x is the distance between the atoms. (a) At what distance of separation does the potential energy have a local minimum (not at ∞ (b) What is the force on an atom at this separation? (c) How does the force vary with the separation distance?
Relevant Equations
$$U(x)=\frac {a} {x^{12}} - \frac {b} {x^6}$$
##U(x)=\frac {a} {x^{12}} - \frac {b} {x^6}##

##=ax^{-12} - bx^{-6}##

##U'(x)=-12ax^{-13} + 6bx^{-7}##

##-U'(x) = F(x) = 12ax^{-13} - 6bx^{-7}##

The answer for part (c) is supposedly x^6. If I found F(x) correctly then wouldn't the force drop to near zero as the distance gets larger? I can't see how it varies as x^6.

Not at all sure what the question means. If it means the dominant term, that depends on x. For very small x it varies as ##x^{-13}##, for very large x as ##x^{-7}##.

I_Try_Math
haruspex said:
Not at all sure what the question means. If it means the dominant term, that depends on x. For very small x it varies as ##x^{-13}##, for very large x as ##x^{-7}##.
Just trying to make sure I wasn't missing something. I'm not really sure what it's asking for either.
I guess it's not really helpful/relevant but for the record the answers given for (a) and (b) are (2a/b)^(1/6) and 0 respectively which appear to be correct. Maybe it's some kind of typo.

I_Try_Math said:
Just trying to make sure I wasn't missing something. I'm not really sure what it's asking for either.
I guess it's not really helpful/relevant but for the record the answers given for (a) and (b) are (2a/b)^(1/6) and 0 respectively which appear to be correct. Maybe it's some kind of typo.
I confirm those answers for a and b.

I tried to track down the origin of the question. Seems to be here:
https://pressbooks.online.ucf.edu/osuniversityphysics/
https://pressbooks.online.ucf.edu/osuniversityphysics/chapter/8-chapter-review/,

Note the blunder: both terms are negative in the potential!
Lots of websites with paywalled solutions have copied it, some correcting the sign, but none altering question c. It would be interesting to see how they solve it, but I won't waste money on such shonks.

I_Try_Math
I_Try_Math said:
Homework Statement: The potential energy function for either one of the two atoms in a diatomic molecule is often approximated by where $$U(x)=\frac {a} {x^{12}} - \frac {b} {x^6}$$ x is the distance between the atoms. (a) At what distance of separation does the potential energy have a local minimum (not at ∞ (b) What is the force on an atom at this separation? (c) How does the force vary with the separation distance?
Relevant Equations: $$U(x)=\frac {a} {x^{12}} - \frac {b} {x^6}$$

##U(x)=\frac {a} {x^{12}} - \frac {b} {x^6}##

##=ax^{-12} - bx^{-6}##

##U'(x)=-12ax^{-13} + 6bx^{-7}##

##-U'(x) = F(x) = 12ax^{-13} - 6bx^{-7}##

The answer for part (c) is supposedly x^6. If I found F(x) correctly then wouldn't the force drop to near zero as the distance gets larger? I can't see how it varies as x^6.
Google Lennard-Jones potential.
Answer for a seems ok. For b, too.
For c, I think the idea is to develop a power series around ##r_{\text{min}}##. First nonzero term is quadratic.

 add plot of
$$U = {1\over x^{12}}-{1\over x^6}$$

##\ ##

Last edited:
BvU said:
For c, I think the idea is to develop a power series around rmin.
I tried that, but the first term would necessarily be an odd power of the displacement. I found it to be linear.

haruspex said:
I tried that, but the first term would necessarily be an odd power of the displacement. I found it to be linear.
Can you explain what you mean by this? The expansion about the equilibrium point ##x_m## would be something like$$U(x)\approx U(x_m)+U'(x_m)(x-x_m)+\frac{1}{2}U''(x_m)(x-x_m)^2+\dots$$The equilibrium point is determined by solving ##U'(x_m)=0## for ##x_m##. Therefore the linear term in the expansion is zero.

Are we talking about the same thing?

I was talking about the potential at ##r_{\text{min}}##.
Force is zero there (and changes sign). First order of force is linear. ##U''## should provide the coefficient.
Sorry for the confusion.

 the question asks for the force, I know. Clear. It doesn't say where -- sadly.

I do the potential at ##r_{\text{min}}##
because it's fun with ##x_0 = \sqrt[6]2 ## \begin{align*} \left ( 1/x_0 \over 1+ x/x_0\right )^{12} - \left ( {1/x_0} \over 1+ x/x_0\right )^6 & = \\ {1\over 4}\left (1-12\;x/x_0 +144 \;(x/x_0)^2 ...\right ) - {1\over 2}\left (1-6\, x/x_0 +36 \,(x/x_0)^2 ...\right ) & = \\ -{1\over 4} + 18 \,(x/x_0)^2 ... = -{1\over 4} + 9/ \sqrt[3]2 \;x^2 ... & \approx -{1\over 4} + 7.143\; x^2 ... \end{align*} (my plot was a guess with ##9x^2##)

##\ ##

Last edited:
kuruman said:
Are we talking about the same thing?
The question asks how the force behaves, not the potential.
BvU said:
First order of force is linear.
Yes, but in general the first nonzero term could have been any odd power. Whatever, it is not ##\Delta x^6##.

haruspex said:
The question asks how the force behaves, not the potential.
Right. But the first two terms in the expansion are $$U(x)= U(x_m)+\frac{1}{2}U''(x_m)(x-x_m)^2$$ That's a harmonic oscillator potential with restoring force $$F=-k(x-x_m)~;~~~~~~k=U''(x_m).$$ Any even potential with a local minimum is harmonic about that minimum.

kuruman said:
restoring force $$F=-k(x-x_m)$$
which is linear in ##\Delta x=x-x_m##, as I wrote.

haruspex said:
which is linear in ##\Delta x=x-x_m##, as I wrote.
Yes, we are talking about the same thing.

In unison. Hope the OP is happy too

##\ ##

I_Try_Math

How does the force between atoms in a diatomic molecule change as the distance increases?

As the distance between atoms in a diatomic molecule increases, the force initially becomes more attractive, reaching a minimum at the equilibrium bond length. Beyond this point, the force becomes repulsive as the atoms move further apart, following an inverse-square law at large distances.

What is the equilibrium bond length in a diatomic molecule?

The equilibrium bond length is the distance between the two atoms at which the force is zero. At this distance, the potential energy of the molecule is at its minimum, and the attractive and repulsive forces between the atoms are balanced.

How do attractive and repulsive forces contribute to the force on an atom in a diatomic molecule?

Attractive forces, primarily due to electrostatic interactions between the nuclei and electrons, dominate at longer distances. Repulsive forces, mainly due to electron-electron repulsion and the Pauli exclusion principle, become significant at shorter distances. The net force is a combination of these, leading to a potential energy curve with a minimum at the equilibrium bond length.

What is the Lennard-Jones potential and how does it describe the force on atoms in a diatomic molecule?

The Lennard-Jones potential is a mathematical model that describes the potential energy of interaction between two atoms. It includes a term for attractive forces that varies with the inverse sixth power of the distance and a term for repulsive forces that varies with the inverse twelfth power of the distance. The derivative of this potential with respect to distance gives the force between the atoms.

How does quantum mechanics influence the force on atoms in a diatomic molecule?

Quantum mechanics introduces the concept of quantized energy levels and wavefunctions for electrons in a diatomic molecule. The force on the atoms is influenced by the distribution of electron density, which is determined by solving the Schrödinger equation for the molecule. Quantum mechanical effects lead to phenomena such as bond formation, bond length variation, and vibrational modes.

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