The deuteron's binding energy is held fixed while the assumed square-well range is varied

PHYSICS · Multi-part · Fermi University Physics · Nuclear Physics · The nuclear force and nuclear models · The nuclear force and nuclear models

The Nucleus Resolved, Ch The deuteron and what two nucleons do to each other

Problem

The deuteron's binding energy B=2.2246 MeVB = 2.2246\,\mathrm{MeV} is held fixed while the assumed square-well range RwR_{\rm w} is varied. Take μredc2=469.46 MeV\mu_{\rm red}c^2 = 469.46\,\mathrm{MeV} and ℏc=197.327 MeV fm\hbar c = 197.327\,\mathrm{MeV\,fm}, and use the matching condition kincot⁡(kinRw)=−κk_{\rm in}\cot(k_{\rm in}R_{\rm w}) = -\kappa with V0−B=ℏ2kin2/2μredV_0 - B = \hbar^2k_{\rm in}^2/2\mu_{\rm red}. ((a)) Find the well depth V0V_0 needed for Rw=1.7 fmR_{\rm w} = 1.7\,\mathrm{fm}. ((b)) Find the well depth V0V_0 needed for Rw=2.5 fmR_{\rm w} = 2.5\,\mathrm{fm}. ((c)) Someone claims the depth must scale as V0∝1/Rw2V_0 \propto 1/R_{\rm w}^2, since that is how the just-bound threshold depth scales. Using the fitted value V0=33.7 MeVV_0 = 33.7\,\mathrm{MeV} at Rw=2.1 fmR_{\rm w} = 2.1\,\mathrm{fm} as a reference, check whether this simple scaling reproduces your answers to (a) and (b).

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