 Okay, I'm going to do a homework problem. I just picked one out. So this one gives you a mole of nickel and for a wire with a mass and a density. And so you put a mass on the end of it and it stretches. And so the point is it says, what's the interatomic spring consum? Okay. So let's just do this generically first. Okay. So here's a wire like that. And so this is broken into atoms that are like a box big, right? The little boxes, okay. And so this has a side of length D and this is, let's say this has a cross-section in the bottom of A and a length of L. Okay. So if I take, if I put some mass on the end of here, what was it, in this case it was 53 kilograms, then that would be a force of 53 times 9.8 Newton per kilogram to pull it down and the whole thing would stretch, it would stretch a mountain delta L. Okay. So if that were a spring, you know, I know the force from a spring is KS. So if I know S, which would be delta L in this case, and the force I can find, I can find the spring concept for that wire, but I don't want that. I want the spring concept for the interatomic springs, okay. So here this is going to be K wire, it's going to be, I'll call this big M, G, that's this mass, divided by delta L, right, that's what that is. That's the spring concept for the whole thing. Well, if we think of this as a whole bunch of springs in between the atoms, we have some in series and we have some in parallel, like that, okay. So what, if they're all the same spring, how can I find the spring concept of those? Well, the first thing to think about is if I take a spring and I add another one right here on the end of it in series, then if I stretch it, it's going to stretch twice as long. So that makes the spring concept effectively half as much, if there were two of them, right? Yep. Okay, and then if I take two springs, we'll call this series, and if I take two in parallel, then when I add a load to the bottom of it, each one only has to stretch half as much in order to support the same load. So it's going to make an effective spring concept that's stronger, okay. So how can I relate this to the inner atomic spring concept? Well, let's see. If I have the spring concept for everyone that's in series, it's going to make it weaker, right? So if I had three in series, it would be a third as strong, yep. So I'm going to write this, that Ks is the inner atomic spring concept, this is the number of springs in series, and for every spring in parallel, it makes it stronger. So I have this number of springs in parallel divided by the number of springs in series for this particular thing. So how do I find that? How do I find the number of springs parallel? Well if I look at this head on, this is some area A, and if I know the, this is A, if I know the inner atomic diameter, D, then I can find that. I can say Np is just that total area A divided by D squared, right? Because each one of those squares has a size of D times D, and then if I divide A by D times D, that'll give me the number of squares there, which is the number of springs in parallel. How do you find D? I'm going to leave that as a problem for you. I think you can do that, because you just need the density and the atomic mass, right? And you can find it. Okay, so and then what about in series? Again, if I know the length of this, and I know the length of each one of these, I could find it. The number in series is going to be L divided by D. So we put that in here, this is going to be, Np is going to be Aks over D squared, Ns is going to be L over D, so one of those cancels. And I get that, Aks over DL. Okay, so now I want to find the interatomic spring constant. That's what I wanted to. Is that DL? Yeah, that's right, L, not delta L. So let me solve, I have this equal to this, I can solve for Ks. Ks equals mg over delta L, and then I have DL over A. Okay, so this is D, this is F, right? That's F. So let me rewrite this, Ks equals F over A, and then I have over delta L over L, right? I can remove that, make an improper fraction, D. So this is Young's modulus times D. That is the correct answer, right? Oh, Young's modulus over D. Wait, Ks, why is there D on the bottom, A, I made a mistake. Number in parallel is A over D squared, and when you make a mistake in a video, it's like I'm going to have to delete this video and start over, and I don't want to do that. How much time have I used so far? Six minutes. Okay, let's see, I guess these are backwards. If I have the spring constant, if I want to get the real spring, if that's, I think I have that back there, MP is A over D squared, A over D squared, L over D is in S. So if I have a spring constant of 1 and I have 3 in parallel, it would have a spring constant of 3. That's right, help! What did I do wrong? Okay, so A over D squared, L over D, you know, once you make an error, it's like, you're screwed, you're never going to find it, A, L, D, S, K, S, D, K, S, what the hell? Oh, that's Jung's modulus. Okay, now that's right. I was right the whole damn time, okay. So that's right, I didn't make a mistake, okay, so that's right. So now, Jung's modulus are all things that you can determine by actual macroscopic properties. But this gets down to the, in the book, it says Y equals KS over D, duh, it's the same thing. Hello? I feel so dumb now. Okay, so I think that should help you with those homework problems. You know, you've got to think of it making these connections between atomic and macroscopic properties.