I can't seem to understand what this question is asking. I am not sure really what the second and third terms actually represent in the equation.

Near the equilibrium position $r_0$, the interatomic potential energy curve can be written as $$U(x) = -U_0+\frac{1}{2!}\cdot(x-r_0)^2\cdot\left(\frac{d^2U}{dx^2}\right)_{r_0}+\frac{1}{3!}\cdot(x-r_0)^3\cdot\left(\frac{d^3U}{dx^3}\right)_{r_0}+...$$ How can we determine $\left(\frac{d^2U}{dx^2}\right)_{r_0}$ and $\left(\frac{d^3U}{dx^3}\right)_{r_0}$ experimentally? Briefly describe possible experiments, and what they can tell us about the behaviour of the potential energy curve near the equilibrium position.


closed as off-topic by Emilio Pisanty, peterh, Yashas, Kyle Kanos, Jon Custer Jun 8 '17 at 12:43

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  • $\begingroup$ I don't understand the downvotes. Jason is not asking the solution to the problem, he simply didn't understand the question. It's not trivial if you haven't heard of Taylor series before. $\endgroup$ – Renan Nobuyuki Hirayama Jun 7 '17 at 20:16
  • $\begingroup$ @RenanNobuyukiHirayama On my first glance, it's not to clear what he's asking. He also has most of this in an uncropped screenshot of a PDF on his browser, so you have to zoom in to read it on some screens. The PDF part was fixed as I wrote this; but I assume those were the reasons. $\endgroup$ – JMac Jun 7 '17 at 20:19
  • $\begingroup$ In the future, Jason, please type out the text in images (use MathJax when necessary) - not only does it make it easier to read, it also helps search engines (and therefore people who may answer or read your question) find your question. $\endgroup$ – heather Jun 7 '17 at 20:20

Have you studied Taylor expansion yet? Basically, if you have a smooth (continuously differentiable) function $f:\mathbb{R}\rightarrow\mathbb{R}$, you can expand it in terms of its derivatives computed in a certain point $x_0$: $$ f(x)=f(x_0)+ f'(x_0)(x-x_0)+\frac{1}{2!}f''(x_0)(x-x_0)^2+\ ... $$ where the prime denotes a derivative.

Then in the question, since $x_0$ is an equilibrium point, you know $U'(r_0) = 0$. Therefore the potential energy can be expressed as $$ U(x)=-U_0 +\frac{1}{2!}U''(r_0)(x-r_0)^2+\frac{1}{3!}U'''(r_0)(x-r_0)^3 $$

where $U(r_0)\equiv-U_0$.


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