# How can one determine at which distance the Lennard-Jones potential reaches a given value?

My question is fairly simple, but I do need clarification on how to get the inverse of the Lennard-Jones potential V(x).

I am working with the following expression: $$V(x) = e\times[(R/x)^{12} -2\times(R/x)^6]$$

So given a value $V$, how can I find $x(V)$ ?

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The rather arbitrary choice of the exponent 12 gives you a hint. Define $y = 1/x^6$ and you got yourself a simple quadratic equation. – Lagerbaer Feb 1 at 0:31

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– Ϛѓăʑɏ βµԂԃϔ Feb 1 at 1:07
@CrazyBuddy I forget the definition of homework tag, I should give one less step in my answer ;). @ MeloMCR Homework tag does not only apply to the homework, but for the people to not give full ansewr, please dont be angry on it – hwlau Feb 1 at 1:13

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@CrazyBuddy I'm not entirely sure I agree with the homework tag designation in this case even given the "definition". Surely you agree that "any question where it is preferable to guide the asker to the answer rather than giving it away outright" is quite subjective? It sounds like MeloMCR's use for the answer is rather research/industry-oriented, does the phrase "question of primarily educational value" really apply here? Isn't that a rather vague phrase anyway? Couldn't every question on here be considered of primarily educational value? – joshphysics Feb 1 at 1:20

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– David Zaslavsky♦ Feb 2 at 5:18
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## 2 Answers

What you are asking is to take the inverse of the function $V(x)$. Using the $z=(R/x)^6$ as suggested, you will get the quadratic equation:

$$z^2-2z-V/e=0$$

and the solution are

$$z_{\pm} = 1 \pm \sqrt{1+V/e}$$

and so

$$x_{\pm}(V) = \frac{R}{(1\pm\sqrt{1+V/e})^{1/6}} \tag{1}$$

Note that there are at most two real solutions (not twelve) as shown in the figure. The equation (1) needs the condition $z_{\pm}>0$ hold which means $x_{+}$ exists for $V>-e$ and $x_{-}$ exists for $-e>V>0$

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Hey hwlau. I still have some suspicion that the question would probably get closed. Because, an overview simply asks for "How can I find X"..? - which makes me suspicious ;-) – Ϛѓăʑɏ βµԂԃϔ Feb 1 at 1:18
@CrazyBuddy Probably, you may flag it if you want. It is just a mathematical question so it might also be off topic here. After writing few lines to get solution, I cant resist to type the outline here :) – hwlau Feb 1 at 1:22

Substitute $u=\left(\frac{R}{x}\right)^6$:

$V=e(u^2-2u)$

$u^2-2u+\frac V e=0$

Solve the quadratic equation:

$u_{1,2}=1\pm \sqrt{1+\frac V e}$

Insert into the substitution:

$x_{1,2}=R u^{-\frac 1 6}=R \left( 1\pm \sqrt{1+\frac V e}\right)^{-\frac 1 6}$

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