Transverse Polarization of the K -Conversion Electrons following the Beta Decay of Hg 203
1963; American Institute of Physics; Volume: 132; Issue: 1 Linguagem: Inglês
10.1103/physrev.132.346
ISSN1536-6065
AutoresRalph Alvin Llewellyn, R. M. Steffen,
Tópico(s)Nuclear physics research studies
ResumoThe transverse polarization of the $K$-conversion electrons following the $\ensuremath{\beta}$ decay of ${\mathrm{Hg}}^{203}$ has been investigated. A lens spectrometer was used to focus the conversion electrons on a Mott-scattering foil and the right-left asymmetry of the scattered conversion electrons in coincidence with the preceding $\ensuremath{\beta}$ particles was measured. A detailed discussion of the instrumental corrections is presented. The degree of transverse polarization of the ${\mathrm{Hg}}^{203}$ conversion electrons in the plane defined by the momenta ${\mathrm{p}}_{\ensuremath{\beta}}=m\mathrm{v}$ of the $\ensuremath{\beta}$ particles and ${\mathrm{p}}_{c}$ of the conversion electrons is described by ${P}_{\mathrm{II}}(\ensuremath{\theta})=K(\frac{v}{c})sin\ensuremath{\theta}$, where $\ensuremath{\theta}$ is the angle between ${\mathrm{p}}_{\ensuremath{\beta}}$ and ${\mathrm{p}}_{c}$. The experiments yielded the value of the polarization constant $K=+0.62\ifmmode\pm\else\textpm\fi{}0.08$. The positive sign of $K$ implies that the spin of the conversion electrons is pointing in the same direction as the $\ensuremath{\beta}$-particle momentum ${\mathrm{p}}_{\ensuremath{\beta}}$. A comparison with the theory of Becker and Rose shows that the experimental value of $K$ is compatible with the spin assignments ${I}_{i}=\frac{1}{2}$ and ${I}_{i}=\frac{3}{2}$ to the ground state of ${\mathrm{Hg}}^{203}$. The spin ${I}_{i}=\frac{5}{2}$ is excluded. The most probable spin assignment is ${I}_{i}=\frac{3}{2}$, which gives good agreement with the observed polarization if the ratio ${y}^{\ensuremath{'}}$ of the scalar and the vector-type beta matrix elements is ${y}^{\ensuremath{'}}=\ensuremath{-}1.25\ifmmode\pm\else\textpm\fi{}0.15$.
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