How to determine the spring constant in a Lennard-Jones potential

[happyparticle]

Homework Statement

Estimate the effective spring constant (k) for this system

Relevant Equations

F= -kx
$F = -\frac{du}{dr}$

Hi,
First of all I hope it doesn't bother if I ask too much question.I found the values of $u1,u2$ for 2 differents posistions $(r1,r2 )$ and I now have to determine the spring constant (k).I'm thinking about using$$
F= -kx
$$
with $F = -\frac{du}{dr}$ then

$$
U = \int -kr \cdot dr =-k\frac{r^2}{2}
$$

I'm wondering if I can use $r=r2$ and $U=U2$ or I'm completely wrong by using $F=−kx$

Otherwise, I found $k=2\varepsilon\Big(\frac{n}{r_0}\Big)^2$ Here, but I'm not sure how to get this equation from $U_{LJ}(r)=\varepsilon\Big[\Big(\frac{r_0}{r}\Big)^{2n}-2\Big(\frac{r_0}{r}\Big)^n\Big]$

 

[haruspex]

Homework Statement:: Estimate the effective spring constant (k) for this system

What system?

 

[etotheipi]

For a general potential $V(r)$, write $r = r_0 + \eta$, where $V'(r_0) = 0$, i.e.$$V(r) = V(r_0) + \eta V'(r_0) + \frac{1}{2} \eta^2 V''(r_0) + \mathcal{O}(\eta^3) = V(r_0) + \frac{1}{2} \eta^2 V''(r_0) + \mathcal{O}(\eta^3)$$Write $\tilde{V}(\eta) = V(r)$, and then$$F = -\tilde{V}'(\eta) = - V''(r_0) \eta + \mathcal{O}(\eta^3)$$you identify the "effective spring constant" with the term $V''(r_0)$. So for the LJ potential, you just need to twice differentiate$$V(r) = 4\varepsilon \left[ \left(\frac{r_0}{r}\right)^{12} - \left(\frac{r_0}{r}\right)^6 \right]$$and evaluate at $r_0$.

 

[happyparticle]

I found k = -42373736. I don't know what the spring constant should looks like for 2 atoms of argon.

I can't find any value of K to give me an idea.

I had $U_1$ and $U_2$

 

[haruspex]

Otherwise, I found $k=2\varepsilon\Big(\frac{n}{r_0}\Big)^2$ Here, but I'm not sure how to get this equation from $U_{LJ}(r)=\varepsilon\Big[\Big(\frac{r_0}{r}\Big)^{2n}-2\Big(\frac{r_0}{r}\Big)^n\Big]$

Where are you up to with this? I assume that using @etotheipi's explanation you now know how to get $k=2\varepsilon\Big(\frac{n}{r_0}\Big)^2$ from $U_{LJ}(r)=\varepsilon\Big[\Big(\frac{r_0}{r}\Big)^{2n}-2\Big(\frac{r_0}{r}\Big)^n\Big]$. I also assume you are taking n=6.
You wrote that you have values for U at two values for r. Did you use those to get a value for $\varepsilon$, and hence that value for k?

 

[happyparticle]

Where are you up to with this? I assume that using @etotheipi's explanation you now know how to get $k=2\varepsilon\Big(\frac{n}{r_0}\Big)^2$ from $U_{LJ}(r)=\varepsilon\Big[\Big(\frac{r_0}{r}\Big)^{2n}-2\Big(\frac{r_0}{r}\Big)^n\Big]$. I also assume you are taking n=6.
You wrote that you have values for U at two values for r. Did you use those to get a value for $\varepsilon$, and hence that value for k?

I found U by using
$U_{LJ}(r)=\varepsilon\Big[\Big(\frac{r_0}{r}\Big)^{2n}-2\Big(\frac{r_0}{r}\Big)^n\Big]$. and $\epsilon = 1.654 \cdot 10^{-21}$

 

[haruspex]

I found U by using
$U_{LJ}(r)=\varepsilon\Big[\Big(\frac{r_0}{r}\Big)^{2n}-2\Big(\frac{r_0}{r}\Big)^n\Big]$. and $\epsilon = 1.654 \cdot 10^21$

In that case, where are you getting values for $\epsilon$ and $r_0$ from?

 

[happyparticle]

In that case, where are you getting values for $\epsilon$ and $r_0$ from?

Sorry, $\epsilon$ is given in the question and I found $r$ by using $\frac{du}{dr}$ which is the max. and $\sigma$ if given as well.

$U(r) = 4\epsilon[(\frac{\sigma}{r})^{12}- (\frac{\sigma}{r})^6]$

I made a mistake. Now I found $k = 6.82\cdot 10^{-12}$

 

[haruspex]

Sorry, $\epsilon$ is given in the question and I found $r$ by using $\frac{du}{dr}$ which is the max. and $\sigma$ if given as well.

$U(r) = 4\epsilon[(\frac{\sigma}{r})^{12}- (\frac{\sigma}{r})^6]$

I made a mistake. Now I found $k = 6.82\cdot 10^{-12}$

Ok, but now I'm confused about something else.
In post #1 you had

$U_{LJ}(r)=\varepsilon\Big[\Big(\frac{r_0}{r}\Big)^{2n}-2\Big(\frac{r_0}{r}\Big)^n\Big]$

What happened to that factor of 2? Or is $r_0=2^{\frac 16}\sigma$?

 

[kuruman]

Ok, but now I'm confused about something else.
In post #1 you had

What happened to that factor of 2? Or is $r_0=2^{\frac 16}\sigma$?

It seems there is a typo in post #3 by @etotheipi. The expression for the LJ potential should be $$V(r) = 4\varepsilon \left[ \left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^6 \right].$$$r_0$ is obtained from this.

 

[etotheipi]

Oh whoops, yeah I always remember that potential wrong! Yes I just checked wikipedia and it's the same as what @kuruman wrote down in #10.

To me, so long as it looks something like $Ar^{-12} + Br^{-6}$ then I don't really care too much about what the exact forms of the pre-factors are

 

[haruspex]

It seems there is a typo in post #3 by @etotheipi. The expression for the LJ potential should be $$V(r) = 4\varepsilon \left[ \left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^6 \right].$$$r_0$ is obtained from this.

So you are confirming

$r_0=2^{\frac 16}\sigma$

?

 

[kuruman]

So you are confirming
$r_0=2^{\frac 16}\sigma$
?

Yes.

Oh whoops, yeah I always remember that potential wrong! Yes I just checked wikipedia and it's the same as what @kuruman wrote down in #10.

To me, so long as it looks something like $Ar^{-12} + Br^{-6}$ then I don't really care too much about what the exact forms of the pre-factors are

Perhaps you should care, at least a little bit, because the casual observer might be confused when $r_0$, the conventional symbol for the value of $r$ at which the LJ potential is minimum, is used to denote the scaling parameter.

 

[etotheipi]

Life's too short to worry about things we got wrong

I think, everybody mis-remembers numerical factors sometimes. But the rest of my explanation is correct!

 

[happyparticle]

So basically, am I completely wrong or that's not too bad?

 

[kuruman]

So basically, am I completely wrong or that's not too bad?

In post #8 you say

Sorry, $\epsilon$ is given in the question and I found $r$ by using $\frac{du}{dr}$ which is the max. and $\sigma$ if given as well.

$U(r) = 4\epsilon[(\frac{\sigma}{r})^{12}- (\frac{\sigma}{r})^6]$

I made a mistake. Now I found $k = 6.82\cdot 10^{-12}$

Can you show us how you got that number and what numbers you were given? Once you do that, we will be able to tell you if you are right or wrong.

 

[hutchphd]

I will point out that you gave a number for k that had no units and you did not show your work. What are we supposed to check?
I believe $72\frac {\epsilon} {r_0^2}$ is correct but I never trust my algebra

 

[kuruman]

I will point out that you gave a number for k that had no units and you did not show your work. What are we supposed to check?
I believe 72ϵr02 is correct but I never trust my algebra

I get your expression divided by $2^{\frac{1}{3}}$. I don't trust my algebra either so I used Mathematica.

 

[hutchphd]

I don't see how the cube root gets there but I presume the OP can do Algebra and so he can tell us! I am truly incompetent unless terrified.

 

[kuruman]

I don't see how the cube root gets there but I presume the OP can do Algebra and so he can tell us! I am truly incompetent unless terrified.

The second derivative has terms like $\dfrac{\sigma^6}{r^8}$ and $\dfrac{\sigma^{12}}{r^{14}}$. At $r=2^{\frac{1}{6}}\sigma$, you don't get integer exponents for the $2$ in the denominators.

 

[haruspex]

I get your expression divided by $2^{\frac{1}{3}}$. I don't trust my algebra either so I used Mathematica.

I get @hutchphd's result, which is the same as $72\frac {\epsilon} {2^{\frac 13}\sigma^2}$.
Is it more of the r₀/σ confusion?

 

[hutchphd]

I notice from the OP:

Otherwise, I found $k=2\varepsilon\Big(\frac{n}{r_0}\Big)^2$ Here, but I'm not sure how to get this equation from $U_{LJ}(r)=\varepsilon\Big[\Big(\frac{r_0}{r}\Big)^{2n}-2\Big(\frac{r_0}{r}\Big)^n\Big]$

And no $\sigma$ or roots of $r_0$ need appear.

 

[kuruman]

Is it more of the r0/σ confusion?

Yes. We are good and reconciled.

 

[happyparticle]

$\Delta E = \frac{1}{2} k \Delta r = > -1.6199\cdot 10^{-21} - (-1.654 \cdot 10^{-21}) = \frac{1}{2} k (3.92198 \cdot 10^{-10} - 3.82198\cdot 10^{-10})$

This is how I got k

 

[hutchphd]

That is not the energy of a spring.