论文标题
$^{6,8} $的物质半径和皮肤来自质子+$^{6,8} $的反应横截面，基于Love-Franey $ t $ -matrix模型（出版在物理学的结果中）
Matter radii and skins of $^{6,8}$He from reaction cross section of proton+$^{6,8}$He scattering based on the Love-Franey $t$-matrix model (published in Results in Physics)
论文作者
论文摘要
对于$^{4,6,8} $ He，Tanihata等。确定物质半径$ r_ {m}（σ_{\ rm i}）= 1.57（4），2.48（3），2.52（3），2.52（3）$ 〜fm，来自交互交叉部分$σ_{\ rm i} $ for $^{\ rm I} $ for $^{4,6,8} $ for be cal al targes，c al cal al targes n n n n n n n n n n n n n n cal al arn/n c. al n n cars上。 Lu等。测量了$^{4,6,8} $ He的原子同位素移位（AIS），并确定了proton radii $ r_ {p}（{\ rm ais}）$ for $^{4,6,8} $ HE。至于p+$^{4,6,8} $散射，反应横截面$σ_ {\ rm rm r}（{\ rm exp}）$在700〜MEV处可用，以高度准确。我们的目的是确定$^{\ rm rm rm rm rm}（\ rm rm r}（{\ rm exp}）$ $^{6,8} $ He for $^{6,8} $ He for $^{\ rm exp}）$ radi $ r_ {m} $ r_ {m} $和skins $ r _ {\ rm skin} $。 {\ bf方法：}我们的模型是Love-Franey $ t $ -matrix折叠模型，因为该模型优于Glauber模型的光学限制。我们的$^{6,8} $的结果是$ r_ {m}（{\ rm exp}）= 2.48（3），2.53（2）$ 〜fm和$ r _ {\ rm skin} = $ 0.78（3），0.82（3），0.82（2）〜fm。对于$^{6,8} $，我们的结果$ r_ {m}（σ_{\ rm rm}）$与tanihata {\ it等人}的$一致。对于$^{8} $ HE，$^{4} $之间的距离和价值中心四个中子为2.367〜fm。
For $^{4,6,8}$He, Tanihata et al. determined matter radii $r_{m}(σ_{\rm I})=1.57(4), 2.48(3), 2.52(3)$~fm from interaction cross sections $σ_{\rm I}$ for $^{4,6,8}$He scattering on Be, C Al targets at 790~MeV/nucleon. Lu et al. measured the atomic isotope shifts (AIS) for $^{4,6,8}$He and determined proton radii $r_{p}({\rm AIS})$ for $^{4,6,8}$He. As for p+$^{4,6,8}$He scattering, reaction cross sections $σ_{\rm R}({\rm exp})$ are available at 700~MeV with high accuracy. Our aim is to determine matter radii $r_{m}$ and skins $r_{\rm skin}$ for $^{6,8}$He from the $σ_{\rm R}({\rm exp})$ and the $r_{p}({\rm AIS})$. {\bf Method:} Our model is the Love-Franey $t$-matrix folding model, since the model is better than the optical limit of Glauber model. Our results for $^{6,8}$He are $r_{m}({\rm exp})=2.48(3), 2.53(2)$~fm and $r_{\rm skin}=$0.78(3), 0.82(2)~fm. For $^{6,8}$He, our results $r_{m}(σ_{\rm R})$ agree with those of Tanihata {\it et al.}. For $^{8}$He, the distance between $^{4}$He and the center of mass of valence four neutrons is 2.367~fm.
