Nuclear Models — Question 9

Section 42-8 · Physics pages 7148–7149

Question 9 Separate proton and neutron Fermi seas set a kinetic-energy scale In a zero-temperature free Fermi-gas model of symmetric nuclear matter, total nucleon density is .160 fm−3. Protons and neutrons each have spin degeneracy two and density half the total. Use kF = (3π2ns)1/3, ℏc = 197.3 MeV fm and mc2 = 939 MeV. protons neutrons separate filled momentum spheres Schematic; labels, not drawing lengths, specify the data. (a) Find Fermi wave number and nonrelativistic Fermi kinetic energy for either species. (b) Derive mean kinetic energy from state density proportional to k2dk and evaluate it.
Show solutionHide solution
Question 9 - Solution Separate proton and neutron Fermi seas set a kinetic-energy scale protons neutrons separate filled momentum spheres (a) Method and calculation ns = .0800 fm−3, kF = (3π2ns)1/3 = 1.333 fm−1 . EF = (ℏc)2k2 F /(2mc2) = 36.83 MeV . (b) Interpretation and further calculation ⟨E⟩= ∫kF 0 (ℏ2k2/2m)k2dk ∫kF 0 k2dk = 3 5EF . ⟨E⟩= 22.10 MeV . Physical check The species density, not total nucleon density, enters each Fermi sphere. Attractive nuclear interactions must offset this kinetic energy to produce a bound system.

Question 9

Separate proton and neutron Fermi seas set a kinetic-energy scale

In a zero-temperature free Fermi-gas model of symmetric nuclear matter, total nucleon density is .160fm−3.160\,\mathrm{fm^{-3}}. Protons and neutrons each have spin degeneracy two and density half the total. Use kF=(3π2ns)1/3k_F=(3\pi^2n_s)^{1/3}, ℏc=197.3MeVfm\hbar c=197.3\,\mathrm{MeV\,fm} and mc2=939MeVmc^2=939\,\mathrm{MeV}.

Diagram shown in the original SVG worksheet.

  1. Find Fermi wave number and nonrelativistic Fermi kinetic energy for either species.

  2. Derive mean kinetic energy from state density proportional to k2dkk^2dk and evaluate it.

Question 9 - Solution

Separate proton and neutron Fermi seas set a kinetic-energy scale

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ns=.0800fm−3,kF=(3π2ns)1/3=1.333fm−1.\boxed{n_s=.0800\,\mathrm{fm^{-3}},\quad k_F=(3\pi^2n_s)^{1/3}=1.333\,\mathrm{fm^{-1}}}.EF=(ℏc)2kF2/(2mc2)=36.83MeV.\boxed{E_F=(\hbar c)^2k_F^2/(2mc^2)=36.83\,\mathrm{MeV}}.

(b) Interpretation and further calculation

⟨E⟩=∫0kF(ℏ2k2/2m)k2dk∫0kFk2dk=35EF.\langle E\rangle=\frac{\int_0^{k_F}(\hbar^2k^2/2m)k^2dk}{\int_0^{k_F}k^2dk}=\frac35E_F.⟨E⟩=22.10MeV.\boxed{\langle E\rangle=22.10\,\mathrm{MeV}}.

Physical check

The species density, not total nucleon density, enters each Fermi sphere. Attractive nuclear interactions must offset this kinetic energy to produce a bound system.