X射线多晶体衍射（XRD）

各位大侠，望赐教: 1.溅射速度的问题。   在我们的样品溅射时，软件打开后显示的溅射速率是3.21nm/s，但是我想知道这个数据是我样品的真实溅射速度吗？因为不同的材料用同样能量的离子束溅射，溅射速率是不一样的，现在知道的是五氧化二钽的溅射速率，是否这个数值是参比五氧化二钽的，而不是我样品真实的溅射速率，这点有疑问？ 2.X射线入射角与光电子出射角度之间有没有什么关系？光电子接收器是在垂直样品的方向么？ 各位大侠，望赐教: 1.溅射速度的问题。   在我们的样品溅射时，软件打开后显示的溅射速率是3.21nm/s，但是我想知道这个数据是我样品的真实溅射速度吗？因为不同的材料用同样能量的离子束溅......

instrument is serving as a "reference" only meaning its never the REAL etch-rate on YOUR SAMPLE. However, its still serving as a very important reference for knowing some thickness information because otherwise you have no idea at all how the thickness relationship to be on your sample. The value you see in your software should be calibrated by the service-engineer and as you said by using Ta2O5 as standard. Ta2O5 (or some use SiO2) are typical samples using for sputtering rate check on surface analysis instruments.



And further to your question because sputtering rate is on Ta2O5 and you do not know what is actual rate on YOUR SAMPLE. In this case, what exactly is your sample material? Sometimes it may be possible to get some reference on the internet about different materials sputtering rate as reference to SiO2 or Ta2O5. So give it a search on Google or Baidu for it. On the other hand, certainly it is impossible to know sputtering rate on ALL materials in the world. So in this case, a very typical way when explaining a depth profile data, one would put into the data stating "sputtering rate is xx nm/min on Ta2O5/SiO2 standard sample". It is widely accepted by the surface analysis field.



Answer 2: Usually no need to consider the angles how photo-e comes out because basically they're coming out at ALL angles. Usually what would affect your data are:

(1) Angle between X-ray and Analyzer

(2) Angle between Analyzer and your sample-flat-surface.

For (1) it is instrumentation so you don't care much.

For (2), it may relate to later part of your question. As continue to guess your system is Thermo 250 or 250xi, YES the analyzer is perpendicular to the sample. Its called as Take-of-angle (TOA) 90-degree. At this angle, the sensitivity would be highest (just imagine because sample is directly facing to analyzer) while the detection depth is deeper. If one tilt the sample stage so making the TOA is smaller like 90 > 45 > 30..., then sensitivity will get lower

谢谢!

1：（第一个问题反馈）

      既然大家都公认溅射深度可以以五氧化二钽或者二氧化硅作为基准的数据作为结果，那就可以接受。因为，有些文章里都是用溅射深度，而不是溅射时间，我很奇怪他们是如何找到样品的溅射速率的，所以严格来讲都不是他们样品的真实溅射深度，只是参照。

2：（第二个问题反馈）仍有小困惑。

    光电子的出射是类似漫反射装的，那么分析器在各个方向接收应该是一样的吧（这时入射X射线与样品的角度不变）？反倒是我觉得如果样品不动，接收器也不动，当我们改变入射X射线角度时（实际上是通过转动样品实现），那么根据Beer-Lambert公式是存在0>30>45>60>90(角度为入射方向与样品表面法线夹角）？我如下的两个表面检测，得到的结果仍有些地方不是很明白。表面1与入射线夹角小，所以测试的合金元素含量较表面2高，但是当溅射后，120,160,200s。。。。。。，两个表面的元素含量几乎一样，但是我想，溅射并未改变表面1与入射线的夹角，这样我是否就可以认为表面1内的元素含量确实本来就是比表面2内的要高，尽管角度也可能是影响因素之一？这里面是否还含有其他影响因素是我没考虑到的？

3：（新问题）关于劈裂峰

      光致电离所形成的内壳层空位便同价轨道未配对自旋电子发生耦合，使体系出现不只一个终态，在XPS谱图上将有一条谱线对应，这叫多重分裂。由于轨道角动量是一个值，自旋有两个值，但是在测试时，电子要么是顺时针，要么是逆时针自旋，怎么会出现不只一个终态，是因为一部分电子是顺时针，一部分电子是逆时针吗？对一些元素的XPS谱为什么会劈裂成两个峰不理解。

谢谢！

第二个问题，现已理解。

劈裂峰哪位大侠能解答下么？

The reason for the doublet peak is due to spin orbital splitting (or called spin orbital coupling). Please see above page for explanation. When spin happen in favorable direction, the binding energy becomes lower (e.g. 2p3/2 peak) while the opposite case binding energy become higher (e.g. 2p1/2 peak). This phenomena will always happen but in some cases the energy difference is so small and as in XPS analysis it may or may not able to separate the 2 peaks.

关于自旋时是顺时针还是逆时针这个很好理解，其实我不理解的就是最终状态是顺时针or逆时针的状态，但是在XPS检测时确是2个状态都有数值，劈裂成了两个峰，一大一小，作为测试的时候，不可能同一个电子即是顺时针状态又是逆时针状态，而只能是两个状态中的一个，如何就出现了两个峰，这时不是or而是and了。

They are not the same electron. In XPS, X-ray keep shining onto the samples and excites the photo electrons out.



Still taking the example of p orbit:

- l = 1

- so J can be 1/2 or 3/2

- For J = 1/2, it can be only 1 state = 1/2 (2 electrons)

- For J = 3/2, it can be 1/2 or 3/2 (4 electrons)

- So this is why commonly we say 2p3/2 and 2p1/2 ration are 4:2 = 2:1 (area of the XPS peaks)



***Continue the concept for d or f orbit to get to the 3:2 and 4:3 area ratio.



So we know a probability of 2:1 hitting ratio (if only considering 2p) between the 3/2 and 1/2 splitting peaks. In some sense, you are right to say that "at just 1 moment" when "1 photo electron" is generated, then it has to be EITHER 2p3 OR 2p1. But the thing is X-ray keep shining and photo e- keeps generating and it will get probability of getting both so in a real XPS spectrum you see 2p3 AND 2p1 at the same time.



I hope it can make some sense now.

谢谢！Thank you very much!