 Solution

Write the Morse potential relative to its minimum as $V(a+x)=D[e^{-2x/\xi }-2e^{-x/\xi }]$. Expanding one order beyond the harmonic term givesThe negative cubic coefficient means that stretching is softer than compression. Thermal motion therefore has positive mean displacement even though the most probable minimum is at $x=0$.

For thermal expansion of a weakly anharmonic bond, classical equipartition gives $⟨x^2{⟩}_0=k_{B}T/\kappa$. Vanishing mean restoring force, to first anharmonic order, requiresFor a bond free to expand under negligible applied tension, $L=⟨r⟩\simeq a+⟨x⟩$. Its linear thermal expansion coefficient is thereforeKeeping the leading perturbative numerical factor gives $\alpha \simeq 3k_B/(2a\kappa \xi )=5.9\times 10^{-5}{\mathrm{K}}^{-1}$. The question only requires the scale, not that factor. The thermal expansion is positive; a purely harmonic approximation would give zero mean displacement and zero thermal expansion. The local Taylor expansion describes a bound bond and is not a globally stable cubic potential.