Dislocation-assisted diffusion of oxygen in albite
Dislocation-assisted diffusion of oxygen in albite
复制标题
钠长石中氧的位错辅助扩散
DOI：
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发表时间：
1981
期刊：
影响因子：
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通讯作者：
J. Tullis
中科院分区：
文献类型：
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作者：
R. Yund;B. Smith;J. Tullis
摘要 钠长石中的氧扩散已通过积分（本体 18O）法在 750° 至 450°C 之间、PH2O 为 2 kb 时测定。原始材料具有低位错密度（<106 cm−2），其晶格扩散系数（D1）如下所示，与之前的测定结果非常吻合。样品在高温高压下变形，产生 5 × 109 cm−2 的均匀位错密度。下面给出的这种变形材料的扩散系数 (Da) 分别比 700°C 和 450°C 时的 D1 大约 0.5 和 0.7 个数量级。这种增强被认为是由于沿着位错核心的更快扩散。假设位错核心半径为 4 Å，计算出的管道扩散系数 (Dp)（如下所示）比 D1 大约 5 个数量级。这些结果表明，位错的存在可能仅略微增强变质条件下的体积扩散。
$$\begin{聚集} D_1 = 9.8 \pm 6.9 \times 10^{ - 6} (cm^2 /\sec ) \hfill \\ {\text{ }} \cdot \exp [ - 33.4 \pm 0.6(kcal/mole)/RT] \hfill \\ \end{聚集} $$
$$\begin{聚集} D_a = 7.6 \pm 4.0 \times 10^{ - 6} (cm^2 /\sec ) \hfill \\ {\text{ }} \cdot \exp [ - 30.9 \pm 1.1(kcal/mole)/RT] \hfill \\ \end{聚集} $$
$$\begin{聚集} D_p \约1.2 \times 10^{ - 1} (cm^2 /\sec ) \hfill \\ {\text{ }} \cdot \exp [ - 29.8(kcal/mole)/RT]。 \hfill \\ \end{聚集} $$
AbstractOxygen diffusion in albite has been determined by the integrating (bulk 18O) method between 750° and 450° C, for a PH2O of 2 kb. The original material has a low dislocation density (<106 cm−2), and its lattice diffusion coefficient (D1), given below, agrees well with previous determinations. A sample was deformed at high temperature and pressure to produce a uniform dislocation density of 5 × 109 cm−2. The diffusion coefficient (Da) for this deformed material, given below, is about 0.5 and 0.7 orders of magnitude larger than D1 at 700° and 450° C, respectively. This enhancement is believed due to faster diffusion along the cores of dislocations. Assuming a dislocation core radius of 4 Å, the calculated pipe diffusion coefficient (Dp), given below, is about 5 orders of magnitude larger than D1. These results suggest that volume diffusion at metamorphic conditions may be only slightly enhanced by the presence of dislocations.
$$\begin{gathered} D_1 = 9.8 \pm 6.9 \times 10^{ - 6} (cm^2 /\sec ) \hfill \\ {\text{ }} \cdot \exp [ - 33.4 \pm 0.6(kcal/mole)/RT] \hfill \\ \end{gathered} $$
$$\begin{gathered} D_a = 7.6 \pm 4.0 \times 10^{ - 6} (cm^2 /\sec ) \hfill \\ {\text{ }} \cdot \exp [ - 30.9 \pm 1.1(kcal/mole)/RT] \hfill \\ \end{gathered} $$
$$\begin{gathered} D_p \approx 1.2 \times 10^{ - 1} (cm^2 /\sec ) \hfill \\ {\text{ }} \cdot \exp [ - 29.8(kcal/mole)/RT]. \hfill \\ \end{gathered} $$