So, we end up with a total of six electrons that are possible that have that 2 p orbital value.

所以我们最后,总共得到了6个电子,在所有可能的“2p“轨道值中。

麻省理工公开课 - 化学原理课程节选

Let's look now at a p orbital, so how many total nodes do we have here?

让我们来看看ｐ轨道，它有多少个节点呢？

麻省理工公开课 - 化学原理课程节选

So total quantity demand is 1 -P where P is the lower of the two prices.

总需求量是1-P,而P是两个公司的价格中较低那个

耶鲁公开课 - 博弈论课程节选

V Minus p2, V2. Minus p delta V. So the total work is the work from the left hand side plus the work on the right hand side, p1 V1- p2 V2 which is p1 V1 minus p2 V2.

负p2V2【w=-p2v2】，即负pΔ,所以全部的功就是,左边的功加上右边的功，等于。

麻省理工公开课 - 热力学与动力学课程节选

So when we talk about p orbitals, it's similar to talking about s orbitals, and the difference lies, and now we have a different value for l, so l equals 1 for a p orbital, and we know if we have l equal 1, we can have three different total orbitals that have sub-shell of l equalling 1.

当我们考虑ｐ轨道时，这和ｓ轨道的情形和相似，不同之处在于ｌ的值不一样，对于ｐ轨道，ｌ等于１,我们知道如果ｌ等于１,我们有３个,不同的轨道。

麻省理工公开课 - 化学原理课程节选

So we'll assume just to make life simple, that the total quantity produced is 1 -P.

为了简化,我们假设,总产量是1-P

耶鲁公开课 - 博弈论课程节选

So if we're talking about oxygen, 2s2 we would say that it's 1 s 2, then we have 2 s 2, 2p and then we have 2 p, and our total number of electrons 4 in the p orbitals are four.

所以如果我们在讨论氧，我们会说它是1s2然后我们有，然后，我们在p轨道的,电子总数是。

麻省理工公开课 - 化学原理课程节选