A compound nucleus of mass number is formed at excitation energy

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

The Nucleus Resolved, Ch Reaction mechanisms: resonances, the compound nucleus, direct reactions and the optical model

Problem

A compound nucleus of mass number A=60A = 60 is formed at excitation energy E∗=12.0 MeVE^* = 12.0\,\mathrm{MeV}. Take the Fermi-gas level-density parameter ald=A/8=7.50 MeV−1a_{\rm ld} = A/8 = 7.50\,\mathrm{MeV^{-1}} and a spin cut-off parameter σJ=3.0\sigma_J = 3.0. Consider compound levels of spin and parity Jπ=1−J^\pi = 1^-. ((a)) Using ω(E∗)=π12ald−1/4E∗−5/4exp⁡ ⁣(2aldE∗)\omega(E^*) = \dfrac{\sqrt\pi}{12}a_{\rm ld}^{-1/4}E^{*-5/4}\exp\!\left(2\sqrt{a_{\rm ld}E^*}\right), compute the total state density ω(E∗)\omega(E^*). ((b)) Compute the level density of all spins, ρlev=ω/(2π σJ)\rho_{\rm lev} = \omega/(\sqrt{2\pi}\,\sigma_J), then the density of 1−1^- levels, ρ(Jπ)=12ρlev 2J+12σJ2exp⁡ ⁣[−(J+12)22σJ2]\rho(J^\pi) = \tfrac{1}{2}\rho_{\rm lev}\,\dfrac{2J+1}{2\sigma_J^2}\exp\!\left[-\dfrac{(J+\frac{1}{2})^2}{2\sigma_J^2}\right], and the mean spacing D=1/ρ(1−)D = 1/\rho(1^-) of those levels. ((c)) If the summed transmission coefficients of all open exit channels at this excitation are ∑cTc=4.5\sum_c\mathcal{T}_c = 4.5, estimate the total width Γ=(D/2π)∑cTc\Gamma = (D/2\pi)\sum_c\mathcal{T}_c and decide whether the statistical (compound-nucleus) treatment is fully justified here, using the criterion that levels overlap once ∑cTc>2π\sum_c\mathcal{T}_c \gt 2\pi.

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