Argon is described by the Lennard-Jones potential V(r) = 4ε[(σ/r)¹² − (σ/r)⁶] with σ = 0.34 nm. Which is closest to the interatomic distance r₀ at which the energy is a minimum?

Answer: B. About 0.38 nm

Setting dV/dr = 0 gives −12σ¹²/r¹³ + 6σ⁶/r⁷ = 0, so r⁶ = 2σ⁶ and r₀ = ⁶√2 × σ ≈ 1.122 × 0.34 ≈ 0.38 nm. The energy there is −ε, and ε represents the bond strength (the depth of the well). About 0.34 nm is σ itself, the distance at which V(r) = 0; there the repulsive and attractive energies merely cancel to give zero, and the force acts in the repulsive direction. About 0.43 nm is ∛2 × σ, the value obtained by mistaking r⁶ = 2σ⁶ for r³ = 2σ³. r₀ is a guide to the nearest-neighbor distance in the solid and is close to the measured nearest-neighbor distance in crystalline argon at low temperature (about 0.376 nm).