I have the following question.

The potential energy of two atoms, a distance $r$ apart, is: $$U = -Ar^{-2} + Br^{-10}$$ Given that the atoms form a stable molecule at a separation of $\pu{0.3 nm}$ with an energy of $\pu{-4 eV}$, calculate $A$ and $B$. Also find the force required to break the molecule, and the critical separation at which the molecule breaks.

Molecule is stable at the distance when the first derivative is zero. But how to find the critical separation at which the molecule breaks?

  • $\begingroup$ Please don't deface the question, it could be useful to someone in the future $\endgroup$
    – jonsca
    Commented Dec 7, 2013 at 10:54
  • $\begingroup$ Have you found the solution to your own question? If so, we'd appreciate if you could write up an answer. $\endgroup$ Commented Dec 7, 2013 at 11:28

2 Answers 2


... As far as I know, in the absence of an external influence, the molecule never actually breaks. I mean, most people wouldn't call two atoms a metre apart a molecule as such, but there's no negative curvature stationary point in a potential like this at which you could say "if $r > x$, this molecule is broken and will not reform".

  • $\begingroup$ The potential energy curve is rather unusual though, so maybe it was created to "fit the problem" rather than to fit reality. $\endgroup$ Commented Dec 6, 2013 at 22:37
  • $\begingroup$ @NicolauSakerNeto: Well... It's kind of unusual, but it's a lot like a Lennard-Jones potential in effect. $\endgroup$
    – Aesin
    Commented Dec 6, 2013 at 22:40
  • $\begingroup$ I thought that maybe changing the powers would have somehow changed the result even though I couldn't see it, but after graphing it in Wolfram Alpha, evidently not. $\endgroup$ Commented Dec 6, 2013 at 22:45
  • $\begingroup$ But there is always an external influence. Thermal energy from the environment matters as other molecules will randomly buffet the two atoms in question meaning that, in reality, the atoms will separate. $\endgroup$
    – matt_black
    Commented Jan 1, 2014 at 21:04
  • $\begingroup$ @matt_black: While this is true, in the absence of stated external conditions the statistics of that process are undefined. $\endgroup$
    – Aesin
    Commented Jan 3, 2014 at 22:02

I start from the expression $$\ce{U = - Ar^{-2} + Br^{-10}} ............. (1)$$ If this expression ($1$) is derived with respect to r, it gives ($2$) $$\ce{0 = -A(-2)·r^{-3} + B(-10)·r^{-11}} ... ........(2) $$ And this can be rearranged to give ($3$) : $$\ce{A = 5B·r^{-8}} ... ........(3) $$ If A from ($3$) is introduced into ($1$), the following expression is obtained : $$\ce{U = - \frac{5 B r^{-8}}{r^{2}} + B r^{-10} = - 4B·(r)^{-10}} ............ (4)$$

Now, introducing U = $4$ eV, and r = $0.3$ nm, the following expression ($5$) is obtained : $$\ce{4 eV = - \frac{5 B (0.3 nm)^{-8}}{0.3 (nm)^{-2}} + B (0.3nm)^{-10} = - 4B·(0.3 nm)^{-10}} ............ (5)$$ This can be simplified and yield the final value of B in eV ($6$) : $$\ce{B = (0.3 nm)^{10} ...}..........(6) $$ Then from ($3$) and ($6$) the final A value in eV is ($7$) : $$\ce{A = 5 B·r^{-8} = 5 (0.3 nm)^{10}·(0.3 nm)^{-8} = 5(0.3 nm)^{2}}.........(7)$$ I have the strange impression that this solution is too simple.....


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