So if we square sigma 1 s star, we flip the amplitude so it's all positive now, but again we still have this node right in the middle.

如果我们平方1s星，我们把振幅翻转所以现在都是正的，但同样在中间有个节点。

麻省理工公开课 - 化学原理课程节选

But what we're saying is there's a node here, so that there's no probability of finding an electron between those two points.

但我们说在节点这里，这两点是,不可能发现电子的。

麻省理工公开课 - 化学原理课程节选

So that's why we saw, for example, in the p orbitals we had one angular node in each p orbital, because l is equal to 1 there.

这就是为什么在ｐ轨道中,每个轨道节点数都是１,因为这里ｌ等于１．

麻省理工公开课 - 化学原理课程节选

So that's why we have this zero point here, and just to point out again and again and again, it's not a radial node, it's just a point where we're starting our graph, because we're multiplying it by r equals zero.

这就是为什么在这里有个零点，我需要再三强调，这不是径向零点，他只是我们画图的起始处，因为我们用ｒ等于０乘以它。

麻省理工公开课 - 化学原理课程节选

So this is the 1 s star, sigma 1 s star orbital, and what you have in the center here is a node, right in the center between the two nuclei.

这是1s星，sigma1s星轨道，中间这个是节点，它在两个原子核中间。

麻省理工公开课 - 化学原理课程节选

So what we end up with is one radial node for the 2 s orbital of hydrogen, and we can apply that for argon or any other multi-electron atom here, we also have one radial node for the 2 s orbital of argon.

那意味着它们都是径向节点,所以我们得出的结论是，氢的2s轨道是1个径向节点,我们可以将它应用，到氩或者任意一个多电子原子,对于氩的2s轨道。

麻省理工公开课 - 化学原理课程节选