So that means this node 1 will have an index of 1.

所以这意味着,这个节点索引值将会为。

麻省理工公开课 - 计算机科学及编程导论课程节选

And in terms of radial nodes, we have 2 minus 1 minus 0, so what we have is one radial node.

对于径向节点，我们有２减去１减去０,所以有一个径向节点。

麻省理工公开课 - 化学原理课程节选

Yup, so one total node, 2 minus 1 is 1, and that means since l is equal to 1, we have one angular nodes, and that leaves us with how many radial nodes?

一个节点，２减去１等于１,因为ｌ等于１,我们有一个角向节点，那剩下径向节点有多少个呢？

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

Yup, I heard one, so 2 minus 1, one total node.

嗯，我听到有人说１个，２减去１，１个节点。

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