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A type of interaction between atoms in a molecule is called a Van der Waals interaction. This can be described by the potential energy function; $$U= U_{0}\left(\frac{R_{0}}{r}\right)^{12}-2\left(\frac{R_{0}}{r}\right)^{6}$$

Also, the two atoms are distance $R_0$ from one another when in equilibrium. Let $x$ be the displacement from this equilibrium position so that, $x=r-{R_{0}} , r={R_{0}}+x$ Where the constant $U_{0}>0$.

So, we can find the $\text{force}(F)$ on the second atom by finding the negative derivative of $U$.

Show that the negative derivative of $U$ is

$$F= 12 \frac{U_{0}}{R_{0}}\left[\left(\frac{R_{0}}{r}\right)^{13}-\left(\frac{R_{0}}{r}\right)^{7}\right].$$

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So you have to take a derivative here. What do you know about derivatives? –  mixedmath Sep 30 '12 at 6:10
    
In particular, this would appear to be a (negative) derivative with respect to $r$. –  Gerry Myerson Sep 30 '12 at 7:22
    
@John: $F = - \dfrac{dU}{dr}$ is the relationship between potential energy and force here. The minus sign in front of the derivative is why your problem states "negative derivative." So the problem's just asking you to compute $- \dfrac{dU}{dr}$ and show you end up with the expression for $F$ that is given. Are there parentheses missing from the definition of $U$? –  Jonas Meyer Oct 5 '12 at 6:39
    
No, thats how it was given. –  John Oct 5 '12 at 12:47

4 Answers 4

U can be rewritten as $$ U = U_0R_0^{12} \left( 1\over r\right)^{12} - 2R_0^{6} \left( 1\over r\right)^{6} $$ Now, $$ F = - \frac{dU}{dr} $$ $$ = -\frac{d}{dr}\left( U_0R_0^{12} \left( 1\over r\right)^{12} - 2R_0^{6} \left( 1\over r\right)^{6} \right) $$ $$ = -\left( U_0R_0^{12} \frac{d}{dr}r^{-12} - 2R_0^{6} \frac{d}{dr}r^{-6} \right) $$ $$ = -\left(U_0R_0^{12} (-12r^{-13}) - 2R_0^{6} (-6r^{-7}) \right) $$ Multiple and divide by $R_0$ $$ = -\frac1{R_0}\left(U_0R_0^{13} (-12r^{-13}) - 2R_0^{7} (-6r^{-7}) \right) $$ $$ = \frac{12}{R_0}\left(U_0R_0^{13} r^{-13} - R_0^{7} (r^{-7}) \right) $$ This equals to $$ F = \frac{12}{R_0}\left[U_0\left(\frac{R_0}{r} \right)^{13} - \left(\frac{R_0}{r} \right)^{7}\right] $$ instead of $$ F = 12\frac{U_0}{R_0}\left[\left(\frac{R_0}{r} \right)^{13} - \left(\frac{R_0}{r} \right)^{7}\right] $$ $U_0$ is not the coefficient of the second term.

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ok Thanks Shashwat, but i dont see how the second last equation equals to the last one. Could you clear that up? –  John Oct 5 '12 at 6:26
    
thanks mate i really appreciate it. –  John Oct 5 '12 at 6:46
    
But it doesn't equals to the one which you gave. –  Shashwat Oct 5 '12 at 7:49
    
But it must right? –  John Oct 5 '12 at 12:27
    
Then the answer can be wrong. You can see the $U_0$ is not the coefficient of the second term, so it can't be taken out as common as in the expression on $F$. –  Shashwat Oct 8 '12 at 4:40

Hint: $\left( \frac c x\right)^n=c^n\cdot x^{-n}$

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sorry its been so long since answering, I lost this site. embarrassing, I know. I still dont know what is meant by negative derative and this hint doesnt help me much. Sorry, but could someone help me with the steps? –  John Oct 5 '12 at 5:17

$$F={\bf \color{red}-}\frac{dU}{dr}$$$ { } $ $ { } $$ { } $

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Here's where understanding the physical content of a problem is just as important as understanding it mathematically. The Van Der Waals interaction can be thought of as the net electrostatic attraction or repulsion between 2 atoms caused solely by the minor but constantly relative charge of each atom due to the rotating of each nucleii's "electron cloud". For example, say we have 2 iodine atoms. Since iodine has a large and relatively diffuse electron cloud (why is complicated,don't worry about it) .as the cloud rotates around it's nuclei, it's not symmetrical. As a result, the nuclei is partially exposed when the cloud "bulges" to one side and this causes a momentary net positive charge on this side,while the "bulge" side has a net negative charge. Now imagine these kinds of transitory charge imbalances happen throughout all the iodine atoms in a given sample of iodine gas.Then at any given moment, the "average" of these charges may be either net positive or net negative. Positive charges will attract negative ones, resulting in a net attractive force. Like charges will repel, so likewise a net repulsive force results in this case. The force of the net charge effect at any given moment is the Van der Waals interaction. It turns out to have immense importance in chemistry.For example, Van Der Walls interaction is one of the reasons why gases in the real world don't obey the ideal gas law.

In any event, like any net force, the VDW interaction has both a potential and kinetic energy function. Notice the potential energy function is an inverse function of the radius r of the nuclei to the electron cloud (the greater r is,the further away from the nuclei the electrons are and therefore the Van Der Waals interaction potential energy diminishes). That's why we need the negative derivative to obtain the force of the interaction. Remember from basic physics force is a vector and direction matters!

The important thing to realize in solving this problem is that the potential energy function can be expressed in terms of either $x$ or $r$ depending on how you choose to express it from the relation $x = r - R_0$. In either case, note that $U$ is a composition of functions. $U$ is a function of $x$ which in turn is a function of $r$.

Remember-we got rules around here for that kind of thing! (HINT!)

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Can someone please explain what thier problem with this answer is? I provided the physical intutition behind the solution rather then simply mapping out a solution like everyone else.How does this deserve a downvote bombardment?I'm simply going to delete it if I don't get a response. This is completely absurd! –  Mathemagician1234 Jan 3 '13 at 6:21

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