How force on atom in diatomic molecule varies with distance

[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.

 

[haruspex]

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]

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.

 

[haruspex]

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.

 

[BvU]

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.

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

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[haruspex]

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.

 

[kuruman]

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?

 

[BvU]

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.

[edit] 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 $9x^2$)

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[haruspex]

Are we talking about the same thing?

The question asks how the force behaves, not the potential.

First order of force is linear.

Yes, but in general the first nonzero term could have been any odd power. Whatever, it is not $\Delta x^6$.

 

[kuruman]

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.

 

[haruspex]

restoring force $$F=-k(x-x_m)$$

which is linear in $\Delta x=x-x_m$, as I wrote.

 

[kuruman]

which is linear in $\Delta x=x-x_m$, as I wrote.

Yes, we are talking about the same thing.

 

[BvU]

In unison. Hope the OP is happy too

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