For a neutron bound by a small separation energy in an orbital outside a compact core

PHYSICS · Multi-part · Fermi University Physics · Nuclear Physics · Nuclear reactions · Nuclear reactions

The Nucleus Resolved, Ch Beams and the edges of the chart: emittance, heavy ions, high spin and the drip lines

Problem

For a neutron bound by a small separation energy SnS_n in an ℓ=0\ell=0 orbital outside a compact core, the radial wavefunction outside the core is well approximated by the pure exponential tail u(r)=2κ e−κru(r)=\sqrt{2\kappa}\,e^{-\kappa r}, normalised so ∫0∞∣u(r)∣2 dr=1\int_0^\infty |u(r)|^2\,\mathrm{d}r=1, where κ=2μredSn/ℏc\kappa=\sqrt{2\mu_{\rm red}S_n}/\hbar c and μred\mu_{\rm red} is the neutron-core reduced mass. Treat the core as pointlike compared with 1/κ1/\kappa, so that all of ⟨r2⟩\langle r^2\rangle comes from the tail. ((a)) Derive ⟨r2⟩=∫0∞r2∣u(r)∣2 dr=1/(2κ2)\langle r^2\rangle=\int_0^\infty r^2|u(r)|^2\,\mathrm{d}r=1/(2\kappa^2), and hence the tail's rms radius 1/(2 κ)1/(\sqrt2\,\kappa). ((b)) The ground state of 15C^{15}\mathrm{C} is 1/2+1/2^+, a single neutron in a 2s1/22s_{1/2} orbital outside a 14C^{14}\mathrm{C} core, with one-neutron separation energy Sn=1.22 MeVS_n=1.22\,\mathrm{MeV}. Using μredc2=1415mnc2\mu_{\rm red}c^2=\tfrac{14}{15}m_nc^2 with mnc2=939.57 MeVm_nc^2=939.57\,\mathrm{MeV} and ℏc=197.327 MeV fm\hbar c = 197.327\,\mathrm{MeV\,fm}, compute κ\kappa, 1/κ1/\kappa and the tail's rms distance from the core. ((c)) A neutron with the same Sn=1.22 MeVS_n=1.22\,\mathrm{MeV} sitting instead in a 1d5/21d_{5/2} orbital (ℓ=2\ell=2) does not produce a comparably extended halo. Explain briefly, in terms of the centrifugal barrier, why the ℓ=0\ell=0 case reaches so much further than an ℓ=2\ell=2 case of the same binding energy.

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