Past exam of the soft condensed matter course of the University of Cambridge / 2022 / Section B / 4 / e

During slow peeling in atomic force microscopy, $n$ adjusts to minimize the Helmholtz free energy at each imposed height. Define the positive binding gain $\mu =k_{B}T[ϵb/\xi -(b/\xi {)}^2]$ for the adsorbed segment. The stationary point satisfiesSince ${\partial }_n^{2}F=3k_{B}T{z}^2/(b^2{n}^3)>0$, this is a minimum while $0<n_{\ast }<N$. The upward pulling force is the height derivative of the minimized free energy. The implicit $n_{\ast }(z)$ term vanishes because ${\partial }_{n}F=0$, givingUsing $\xi =2b/ϵ$,It is independent of $z$: extra pulling height releases more monomers while leaving the fractional extension of the stretched part unchanged. Equivalently $F_{\min }=-N\mu +f_{\mathrm{peel}}z$. This is the equilibrium peeling force of an adsorbed Gaussian chain.

The plateau ends when $n_{\ast }=N$, atThereafter all monomers are detached and the Gaussian chain force is $f=3k_{B}Tz/(N{b}^2)$, agreeing continuously with the plateau at $z_{\det }$. Very near the start and final release, finite-chain and junction effects lie outside the blob approximation.