A synthetic heavy nucleus

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Nuclear Physics · Fission, fusion and nuclear energy

The Nucleus Resolved, Ch Fission: the double-humped barrier and the shells the drop cannot see

Problem

A synthetic heavy nucleus, provisionally labelled X, has the following drop energy and Strutinsky shell-plus-pairing correction along its fission path. Data in MeV, as β2\beta_2: ELDME_{\rm LDM}, δUshell+δPpair\delta U_{\rm shell}+\delta P_{\rm pair}: 00: 00, +1.8+1.8; 0.250.25: 0.90.9, −2.7-2.7; 0.450.45: 2.62.6, +1.2+1.2; 0.600.60: 3.33.3, −2.6-2.6; 0.800.80: 4.04.0, −0.8-0.8. ((a)) Sum the rows and give the inner barrier height EAE_A, the second-minimum excitation energy EIIE_{\rm II}, and the outer barrier height EBE_B, all measured from the deformed ground state. ((b)) A fission isomer occupies the lowest state of the second well. Treating the outer hump as a parabola of curvature ℏωB\hbar\omega_B, estimate its fission half-life by WKB using the assault-frequency estimate λf=ℏωII2πℏexp⁡ ⁣(−2π(EB−EII)ℏωB)\lambda_f = \dfrac{\hbar\omega_{\rm II}}{2\pi\hbar}\exp\!\left(-\dfrac{2\pi(E_B - E_{\rm II})}{\hbar\omega_B}\right) with ℏωII=1.0 MeV\hbar\omega_{\rm II} = 1.0\,\mathrm{MeV} and ℏωB=0.60 MeV\hbar\omega_B = 0.60\,\mathrm{MeV}. Repeat with ℏωB=0.55 MeV\hbar\omega_B = 0.55\,\mathrm{MeV} and 0.65 MeV0.65\,\mathrm{MeV} to show how sensitive the estimate is to the barrier curvature. ((c)) Higher up, this same nucleus is formed as a compound system that must evaporate four neutrons before reaching a stable product, and at each step the level-density ratio gives Γn/Γf=4×10−3\Gamma_n/\Gamma_f = 4\times10^{-3}. Estimate the overall probability that the nucleus survives all four steps without fissioning, taking the per-step survival probability as Γn/Γf\Gamma_n/\Gamma_f (valid because Γn≪Γf\Gamma_n \ll \Gamma_f).

Related theory

Browse the problem library