Critical separation at which molecule breaks

Question

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?

Answer 1

... 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".

Answer 2

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.....