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li_qun

铜虫 (著名写手)

应助: 2 (幼儿园)
金币: 61
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帖子: 1240
在线: 419.7小时
虫号: 692320
注册: 2009-01-14

[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 ]

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