Equilibrium peeling force of an adsorbed Gaussian chain ($f=\sqrt{6k_{B}T\mu }/b$)

Let an adsorbed Gaussian chain gain free energy $\mu >0$ per bound monomer. A pulled segment of $n$ Kuhn length steps spanning height $z$ has elastic free energy $3k_{B}T{z}^2/(2n{b}^2)$, giving $F=-(N-n)\mu +3k_{B}T{z}^2/(2n{b}^2)$. Minimizing over $n$ yields $n=(z/b)\sqrt{3k_{B}T/(2\mu )}$ and $F_{\min }=-N\mu +z\sqrt{6k_{B}T\mu }/b$. Its height derivative is the constant peeling force displayed above. The plateau lasts while $0<n<N$ and the Gaussian chain approximation remains valid. In the slab model $F_{\mathrm{ads}}=N{k}_{B}T[(b/\xi {)}^2-ϵb/\xi ]$, optimization gives $\mu =k_{B}T{ϵ}^2/4$ and $f=\sqrt{3/2}ϵk_{B}T/b$.