You take on a research assistantship with a molecular...

You take on a research assistantship with a molecular physicist. She is studying the vibrations of diatomic molecules. In these vibrations, the two atoms in the molecule move back and forth along the line connecting them (see Figure $20.5 \mathrm{c}$ ). As an introduction to her research, she asks you to familiarize yourself with the Lennard-Jones potential (see Example 7.9 ), which describes the potential energy function for a diatomic molecule. She asks you to determine the effective spring constant, in terms of the parameters $\sigma$ and $\epsilon,$ for the bond holding the atoms together in the molecule for small vibrations around the equilibrium separation $r_{\mathrm{cq}} .$ After being stumped for a while, you ask her for a hint. She responds, "Example 7.9 provides the derivative of the potential energy function. Compare that to Equation 7.29 to find the force between the atoms. You want to show that $F$ is of the form $-k x$, and find $k .$ Let the separation distance $r=r_{\mathrm{eq}}+x,$ where $x$ is small and take advantage of the series approximations in Appendix Section B.5." Wow, that's several hints! You sit down and get to work.

Transcript

00:01 So this problem talks about something called the leonard jones potential, which has this, this right here is the actual leonard jones potential.

00:18 And this is a graph of that same potential.

00:30 And it tells us that this is the derivative of that potential.

00:35 If we go look it up, this is example 7 .9 provides the derivative.

00:41 This right here is that derivative, and that's the simplified version there.

00:57 And we're supposed to compare that to equation 7 .29 to find the force between atoms, and it's saying we want to show that f is a form of negative kx.

01:08 So i think equation 729 is just that f equals negative d -u -d -x, if it's a conservative potential, which we want to show that that is equal to negative k delta x.

01:35 So we could just make both those positive and say, oh, well, we're trying to find this value k.

01:44 Say if dudx is equal to k delta x, we can find this value k.

01:51 And we have dudx.

01:53 It's right here.

01:55 And if we look at it, we can see, well, this.

01:57 Does not this here does not look anything at all like a linear function right we've got some one over x to the 13th minus uh one over x to the seventh and so that doesn't seem like it looks anything at all like this and that's why i put the graph here because uh it says what we're looking at is some x small x is small and we're going to go around something called this equilibrium point r equals our equilibrium plus x well this would be that equilibrium point right there it's that point where we have minimized the energy where the force would actually be zero the tangent line right here right that would have a zero tangent tangent line so at that point the force would be zero and it'd be in the bottom of the well and we can see that uh down here we can see that it sort of approximates a quadratic, right? i mean, it's not normally a quadratic, but right down here, at the base here, we can see that it is sort of approximating a quadratic.

03:20 And a quadratic potential has a linear force.

03:24 We know that this k -delta -x, if we integrate that, that's one -half kx squared, which would have this quadratic potential.

03:33 So that's the whole idea here.

03:35 So what we see is, ah, there's a point where this function does sort of behave like a linear function, right? there's a, right? we see that, ah, down here, it's derivative is close, it's going to be the same as some linear function here, this d -u -d -x.

03:57 So that's what we're trying to basically find.

04:03 So, the whole idea here is can we make this approximation here? now, one of the things we want to know is what this eq, this r eq value is, right? when we say we're going to be taking these x's, all these x values, and we're going to turn them into some r, where r is equal to r equilibrium, plus some x, where x is a small variation, a small deviation off of r, eq.

04:39 So what we need to know is what req is first.

04:43 And req is when the derivative equals zero.

04:49 So i'll do it in a different color here.

04:52 We want to say, okay, well, 4 epsilon times negative 12 sigma to the 12 over.

05:04 Over x to the 13th minus minus six sigma to the sixth over x to the seventh.

05:17 And what we can just rename these r's, right? we're trying to find that equilibrium position here equals zero.

05:27 So we can just cancel that.

05:30 All right.

05:30 And why don't we make that a positive then? and then we can just move that that term over here.

05:37 So we can say that this is 12 sigma to the 12th over r to the 13th equals 6 sigma to the 6th over r to the 6th.

05:56 Over r to the 7th.

05:59 And then we can cancel 12 and 6 and that just becomes 2.

06:05 And we can cancel this sigma to the 6th and that becomes just sigma to the 6th.

06:11 And we can cancel this r to the seventh, and that becomes r to the 6 as well.

06:23 So that equals 1.

06:26 So what have we got here? that is 2 sigma to the 6 over r to the 6th.

06:37 So therefore r to the 6 equals 2 sigma to the 6 equals 2 sigma to the 6th.

06:46 The sixth, therefore this equilibrium, the r equilibrium is 2 to the one -sixth or the sixth root times sigma.

07:03 So now we have, we know what r equilibrium is and it's just some constant here.

07:11 So you might wonder why do we even need to do that.

07:13 Hopefully it'll become clear later why it was important.

07:17 I think it'll become clear in a moment.

07:22 So now what we want to do is look at this derivative here.

07:32 Look at this thing and we want to do that substitution, r equals r equilibrium plus x.

07:40 So we're going to say, all right, that we know that f equals this negative d -u -d -x, which is going to be negative 4 -epsilon times negative 12 sigma to the 12 over r -eq plus x to the 13th minus minus six sigma to the 6 over r plus x to the seventh.

08:44 And what we can do here is factor out this req.

08:50 So negative 4 epsilon.

08:53 You might say can't we just substitute in this value 2 to the 1 6th sigma? we certainly could just go ahead and substitute that in.

09:03 It just can get a little bit messy.

09:08 So for now, we'll just keep it r .e.

09:11 But we will resubstitute in later.

09:14 You could, it just gets a little bit clunky.

09:26 So we've got the r .eq times one plus x over req and those are both to the 13th minus minus six sigma to the six over and we can pull out that req times one plus x over our eq and those are all to the seventh.

10:14 And now, hopefully what you see here, we got this 1 plus x here.

10:22 In fact, actually we'll go ahead and do one more step here.

10:27 So we can make that negative 4 epsilon, negative 12, sigma to the 12, over r .e .q.

10:48 To the 13th times 1 plus x over r .e .q.

10:55 To the negative 13th and minus 6 sigma to the 6 over r .e .q.

11:11 To the 7th times 1 plus x over x over r .e .q.

11:13 To the 7th times 1 plus x over r .e .q.

11:21 Negative seventh.

11:23 We just took these values here and we made them this this one plus x over req to the 13th that's on the bottom and we just made it on the top to the negative power.

11:34 And now hopefully what you see is we have these these one plus x over req to some power.

11:41 We can use now this tailor expansion here and we just expand it to first order because x is very, very small...

Question

Please give Ace some feedback