**V = – 4 ε [(σ/r) ^{6} – (σ/r)^{12}]**

**Question 1:** The variables in the Lennard-Jones equation are V and r, Firstly, find the derivatives of V with respect to r dV/dr. Secondly, differentiate the first derivative with respect to r to find the second derivative d ^{2}V/dr^{2} for the interaction between N_{2} molecules.

**Question 2:** Use the solutions of Question 2 to find the separation at which the Lennard-Jones potential V is at a minimum, that is the distance between two N_{2} molecules at which the potential energy is a minimum.

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**Solution 1:**

Bring the denominators of the pair of inner brackets up to the numerators before differentiating.

**V = – 4 ε [(σ/r) ^{6} – (σ/r)^{12}];**

**V = – 4 ε [(σ ^{6}r^{-6} – σ^{12} r^{-12}]**

**(dV/dr) = – 4 ε [- σ ^{6}r^{-6} – (- 12) σ^{12} r^{-13}]**

**Solution 2:**

To find the separation at which the potential V is at a minimum we use the first derivative. Leaving a in angstroms will give r in angstroms. For clarity I have not shown the units in this evaluations of the equation.

**(dV/dr) = – 4 ε [(6 σ ^{6}/r^{7}) + (12 σ^{12} / r^{13})]**

Either 4ε must equal zero, which it cannot, or the term in brackets equals zero giving the following,

R_{min} = 4.40 A is the separation of two N2 molecules at the minimum potential energy (see below Fig) and is larger than the 3.7 A “hard sphere” intermolecular interactions diameter of N_{2}.

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