 Solution

Use a freely rotating chain: fixed bond length $b$, fixed angle $\vartheta$ between forward bond vectors, and independently random torsion angles. Then $⟨{\hat{t}}_0\cdot {\hat{t}}_n⟩=(\cos \vartheta {)}^n=e^{-nb/{\ell }_p}$, giving the orientational persistence length of a freely rotating chainThe equivalent long-chain Kuhn length divided by two, $b(1+\cos \vartheta )/[2(1-\cos \vartheta )]\simeq 3.48\mathrm{nm}$, is nearly equal here but is a distinct discrete-chain convention. A chemical internal bond angle is supplementary to the forward-vector angle used above.

Synthetic examples are relatively stiff poly(p-phenylene terephthalamide) and flexible polydimethylsiloxane. Light scattering and viscometry established rodlike solution conformations for the former; single-chain mechanical measurements support the latter's flexibility. Persistence is solvent- and model-dependent, so no universal numerical value is inferred for either example.^[ref][ref]