Beta-Decay Matrix Elements inSb122

Abstract

Dynamic nuclear orientation has been used to study the ${2}^{\ensuremath{-}}$ \ensuremath{\leftrightarrow} ${2}^{+}$ 1.42-Mev beta ray in the decay of ${\mathrm{Sb}}^{122}$. The ${\mathrm{Sb}}^{122}$, which was a substitutional donor atom in a silicon crystal, was oriented by saturating each of the four $\ensuremath{\Delta}({m}_{I}+{m}_{J})=0$ forbidden transitions. The angular distribution of the gamma ray following the beta ray was measured with two scintillation counters. The nuclear and electron relaxation times were determined by the rate of growth and decay of the nuclear orientation. The electron ($\ensuremath{\Delta}{m}_{J}=\ifmmode\pm\else\textpm\fi{}1$, $\ensuremath{\Delta}{m}_{I}=0$) relaxation time was found to be (4.9\ifmmode\pm\else\textpm\fi{}1.2) min. The nuclear relaxation can be represented as due to a combination of the modulation of the isotropic hyperfine interaction and nuclear quadrupole relaxation. For the dipole mechanism, $50 min\ensuremath{\le}{T}_{N}\ensuremath{\le}100 min$ and for the quadrupole mechanism, $150 min\ensuremath{\le}{T}_{N}\ensuremath{\le}1700 min$. An analog computer was used to correct the initial orientation parameters for the effects of nuclear relaxation. From these, data restrictions can be placed upon the relative amounts of angular momentum carried off by the 1.42-Mev $\ensuremath{\beta}$ ray. The modified ${B}_{\mathrm{ij}}$ approximation was then used to analyze this result in conjunction with the beta-gamma angular correlation. There are three sets of matrix elements which can explain the observed data. One set implies that all the antimony atoms are in the simple donor sites; the other two sets imply that only 40% of the antimony atoms are in the donor sites. The first set gives $V=\ensuremath{-}0.5\ifmmode\pm\else\textpm\fi{}0.1$, $Y=\ensuremath{-}0.5\ifmmode\pm\else\textpm\fi{}0.1$; the second set, $V=\ensuremath{-}4.2\ifmmode\pm\else\textpm\fi{}2.0$, $Y=\ensuremath{-}1.4\ifmmode\pm\else\textpm\fi{}0.5$; the third set, $V=\ensuremath{-}6.3\ifmmode\pm\else\textpm\fi{}1.0$, $Y=+1.8\ifmmode\pm\else\textpm\fi{}1.5$.