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小木虫论坛-学术科研互动平台» 学术交流区» 论文翻译 » 论文翻译求助完结子版 » 求翻译一段话！自己翻译的说表达不到位，希望高手帮助！
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wxlwww123

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1楼 2012-11-21 09:39:43
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james.bond

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【答案】应助回帖

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220chengeng: 金币+2, 感谢应助。 2012-12-13 07:49:37
sltmac: 金币+40 2012-12-28 12:47:26
With the increase of the amount of dopant sodium, the degradation activity of the catalyst gradually increased again. Besides, when the doping amount reaches 15%, the activity will be restored to the considerable level as that of pure ZnO. This is because the doping of Na into  ZnO will produce Nai + and NaZn- defects, and at the initial stage of doping the Fermi level is big so that the formation energy of NaZn- is less than that of Nai + providing priority for generating NaZn – and provided acceptor defect level.Therefore, the cavity concentration in ZnO is enlarged, leading to the decline of n-type light response and decrease of photo-catalytic activity. At the mean time, the previously formed acceptor level is strong competitioner with photoinduced electron, thereby inhibits the yield of photoinduced electron and cavity at the catalyst surface leading to the decrease of photocatalysis activity. However, the Fermi level decreases while the cavity concentration increases, as a result the formation energy of NaZn-can increases and that of Nai + decreases. Consequently, further increase of dopant concentration will lead to the generation of Nai +, resulting in a self-compensating effect and the cavity concentration reduces. The n-type optical response is then restored, and photocatalytic activity will gradually restore as well.

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研究方向：人工单晶,x射线行貌学，X射线多重衍射，x射线精细结构分析，晶体模拟
3楼2012-12-13 03:51:31
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220chengeng

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你把自己的贴出来，毕竟专业术语您还是比其它人熟悉些，不存在不好意思，敢贴出来才能学到东西啊，就像练口语，怕难为情不开口始终都练不好了。
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2楼2012-11-21 09:50:12
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sklyer_123

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sltmac: 金币+10 2012-12-28 12:47:30
However with the sodium doped quantity increase, catalyst degradation activity also increases gradually, when the doping of 15%, active returned to levels comparable with pure ZnO. This is because the Na doped ZnO can generate Nai+ and NaZn- the two impurity defects, dopant in the initial stage, the Fermi level is larger, NaZn- formation can be less than Nai+, will give priority to generate NaZn-, formed acceptor defect levels, resulting in the increase of the concentration of ZnO hole, type N optical response is weakened, photocatalytic activity drop. Simultaneous formation of the acceptor level is the light induced electron contenders, thus inhibiting the catalyst surface of photogenerated electrons and holes of the yield, photocatalytic activity decline. However in the hole density increase while the Fermi level will gradually decrease, leading to the formation of NaZn- can rise, Nai+ can drop formation. So further increased dopant concentration, can lead to the generation of Nai+, self compensation effect, so that the hole concentration decreased. Type N optical response due to recovery, photocatalytic activity came back.
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只要努力就会更加接近成功，加油，亲爱的自己
4楼2012-12-25 13:58:39
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