| 查看: 276 | 回复: 3 | |||
| 当前主题已经存档。 | |||
| 【有奖交流】积极回复本帖子，参与交流，就有机会分得作者 li_qun 的 15 个金币 | |||
[交流]
【求助】请帮我解释一段文献，不长
|
|||
|
第一段作者对图1做了解释，第二段作者提到了一个发光模型并提出质疑，他认为只有电子完全逃逸到导带才可以说有空穴产生，但被激发到5d态的电子仍然被Eu2+束缚所以不能说有空穴产生，按作者的意思电子并未进入导带，但最后一段中，作者自己提出的模型中又包含了电子进入导带的内容，这不是矛盾吗，还是我理解有问题？ Fig. 1 shows the level scheme for the lanthanides in SrAl2O4 [5]. The curves connecting the lowest levels of the 4f configurations and that of the 5d configurations are shown as function of the number n of electrons in the 4f configuration of the trivalent lanthanides. The characteristic pattern with n can be recognized clearly. Note that the energy of the 5d state is relatively constant with n. The energy difference between the location of the 5d state of trivalent lanthanides (curve 2) and divalent lanthanides (curve 4) is about 0.8 eV. The ground state energy of Ce3+, Pr3+, Tb3+, andDy3+ is above the top of the valence band defined as the zero of energy. This means that these ions are able to trap a hole from the valence band. The 5d level of Eu2+ is about 0.4 eV below the conduction band at 7.4 eV. This is large enough for 5d–4f luminescence to occur, however, the proximity of the conduction band bottom makes the emission prone to thermal quenching by means of the ionization of the 5d electron. The excellent persistent luminescence properties ofSrAl2O4 doped with Eu2+ and Dy3+ were discovered in1996 by Matsuzawa et al. [6]. The following mechanismwas proposed that is still widely used today. Excitation ofEu2+ by daylight excites Eu2+ to its 5d state. The hole left behind in the 4f state of Eu is thermally excited to the valence band and subsequently trapped by Dy. A slow release of the hole trapped at Dy followed by recombination at Eu then leads to the luminescence that may persist throughout the night. Actually there is no such thing as a hole left behind in the 4f state because the electron excited to the 5d state is still bonded to Eu2+. One may only speak of a hole after the electron has completely escaped to the conduction band. The scheme in Fig. 1 shows that the Matsuzawa model cannot be correct. Illuminating the phosphor by shorter than 390nm wavelength light, i.e. daylight, brings the electron from Eu2+ inside the conduction band of SrAl2O4. The electron will escape to conduction band states and Eu2+ is converted to Eu3+. The escaped electron is subsequently trapped by Dy3+ to create Dy2+. Fig. 1 shows that the Dy2+ level is about 1 eV below the conduction band. A slow thermal release of the electron followed by re-trapping on Eu and thus creating 5d–4f emission is a much more likely mechanism for the persistent luminescence. The scheme reveals [ Last edited by li_qun on 2009-5-9 at 11:15 ] |
回复此楼» 猜你喜欢
290求调剂
已经有3人回复
-
求调剂
已经有3人回复
-
327求调剂
已经有3人回复
-
085600 286分 材料求调剂
已经有11人回复
-
一志愿河北工业大学0817化工278分求调剂
已经有15人回复
-
085601材料工程找调剂
已经有15人回复
-
材料与化工272求调剂
已经有15人回复
-
南京大学化学调剂
已经有14人回复
-
一志愿北京工业大学，324分求调剂
已经有6人回复
-
求调剂
已经有5人回复
2楼2009-05-09 11:16:30
3楼2009-05-10 22:28:26
4楼2009-05-12 10:38:20
10