Influences of ${Z=100}$ and ${N=152}$ deformed shells on ${ K^{\pi}=8^{-} }$ isomers and rotational bands in ${N = 150}$ isotones

TL;DR

This paper investigates $K^{ ext{pi}}=8^{-}$ isomers and rotational bands in $N=150$ isotones using the cranked shell model with pairing correlations, revealing the influence of deformed shells at $Z=100$ and $N=152$ on nuclear structure.

Contribution

It demonstrates the role of deformed neutron and proton shells at $N=152$ and $Z=100$ in shaping isomeric states and rotational band behaviors in $N=150$ isotones, using advanced PNC-CSM calculations.

Findings

Reproduces experimental bandhead energies and moments of inertia accurately.

Identifies the lowest $8^{-}$ state as a two-neutron configuration at $N=152$.

Explains irregular MOI behavior through configuration mixing and proton alignment effects.

Abstract

The $K^{\pi}=8^{-}$ isomeric states and rotational bands in the even-even $N = 150$ isotones with $94 \leqslant Z \leqslant 104$ are investigated by the cranked shell model (CSM) with pairing correlations treated by the particle-number-conserving (PNC) method. The experimental bandhead energies and kinematic moments of inertia (MOIs) are reproduced quite well by the PNC-CSM calculation. The two-neutron state with configuration $\nu 9/2^{-}[734] \otimes \nu 7/2^{+}[624]$ is the lowest $8^{-}$ state for these isomers. This is a demonstration of the deformed neutron shell at $N=152$. Low-lying two proton $\pi^{2}8^{-}$($\pi 9/2^{+}[624] \otimes \pi 7/2^{-}[514]$) configuration state is predicted only for $^{252}$No and $^{254}$Rf due to the deformed proton shell at $Z=100$. A distinct upbending is observed for the $\nu^{2}8^{-}$ bands in the lighter isotones while it is absent for bands in the heavier ones. The upbending of the $\nu^{2}8^{-}$ band at frequency $\hbar\omega\approx 0.20$ MeV in $^{244}$Pu attributes to the sudden proton alignment of the interference term $j_x(\pi5/2^{+}[642]\otimes\pi7/2^{+}[633])$. The irregularity of MOI observed in the $K^{\pi}=8^{-}$ band of $^{252}$No can be explained by the mixing of the $\nu^{2}8^{-}$($\nu 9/2^{-}[734] \otimes \nu 7/2^{+}[624]$) and $\pi^{2}8^{-}$($\pi 9/2^{+}[624] \otimes \pi 7/2^{-}[514]$) configurations. The $20\%-30\%$ increase of the bandhead $J^{(1)}$ for the $8^{-}$ bands comparing to the ground-state band is attributed to the $\sim 5\%$ pairing gap reduction of the two-neutron $\nu 7/2^{+}[624] \otimes \nu 9/2^{-}[734]$ configuration state comparing to the ground-state band.