Discussion
The Lennard-Jones (LJ) potential of interaction between two molecules as a function of distance \((r)\), is given by \(V_{LJ}(r) = 4\epsilon[(\frac{\sigma}{r})^{12} - (\frac{\sigma}{r})^{6}]\), where \(\epsilon\) and \(\sigma\) are constants. The expression of \(r\) at which \(V_{LJ}(r)\) reaches minimum is
Answer: B
The potential is lowest at \(r = 2^{1/6}\sigma\).
Set the derivative of the potential to zero:
\[ \begin{aligned} \frac{dV_{LJ}}{dr} &= 4\epsilon\left[-\frac{12\sigma^{12}}{r^{13}} + \frac{6\sigma^{6}}{r^{7}}\right] = 0 \\ \frac{12\sigma^{12}}{r^{13}} &= \frac{6\sigma^{6}}{r^{7}} \\ r^{6} &= 2\sigma^{6} \\ r &= 2^{1/6}\sigma \approx 1.12\,\sigma \end{aligned} \]
The option \(\sigma\) is where the potential crosses zero, not where it is lowest: at \(r = \sigma\) the two bracketed terms are equal. The other two expressions cannot be a distance at all: \(\sigma^{3/2}\) has units of length to the power 3/2, and \(\epsilon^{1/6}\) is an energy term with no \(\sigma\) in it.
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