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So, actually I want you to go ahead in your notes and circle that zero point and write "not a node."
在你们笔记上把这个零点圈出来,在旁边写上“不是节点“,它不是节点“
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You can go ahead and use that equation, or you could figure it out every time, because if you know the total number of nodes, and you know the angular node number, then you know how many nodes you're going to have left.
你们可以直接用这个方程，或者每次都自己算出来，因为如果你们知道了总的节点数，又知道角向节点数，就知道剩下的节点数是多少。
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And then we'll construct our tree as follows: each node, well, let me put an example here.
然后我们如下建立我们的决策树：,每一个节点，好的，让我们在这里举一个例子。
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And that zero point is the node.
这个零平面就是节点。
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And, in fact, these are the only two types of nodes that we're going to be describing, so we can actually calculate both the total number of notes and the number of each type of node we should expect to see in any type of orbital.
事实上，我们只,描述这两种节点，所以我们可以,计算任何轨道中的,总结点数以及各种节点数。
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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？
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5 0 The first node will be the to-pull 2, 5 and 0.
第一个节点是可获取的物品。
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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。
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We have one node here, and we can again define that most probable radius.
在这里有个一节点，另外我们可以定义最可能半径。
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And we ask, what do we get with this node?
然后我们问，这个节点我们获得的是什么？
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Yeah. And what does this node look like?
太好了，那这个节点应该是怎样的呢？
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So what we should expect to see is one radial node, and that is what we see here 3s in the probability density plot.
个节点,这就是我们，在这概率密度图上所看到的,如果我们考虑。
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And also that we know that the zero does not count as a node, if per se I actually had managed to hit zero in drawing that, so the correct answer would be the bottom one there.
另外你们要知道零点不是节点，假设说我确实把零点画成０了，那正确的结果就是底下这个。
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And in terms of radial nodes, we expect to see one node.
对于径向节点，应该有１个。
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And what is this node going to look like?
这个节点应该是什么样子呢？
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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。
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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。
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You can also have angular notes, and when we talk about an anglar node, what we're talking about is values of theta or values of phi at which the wave function, and therefore, the wave function squared, or the probability density are going to be equal to zero.
我们也可以有角向节点，当我们说道一个角向节点时，我们指的是在某个theta的值,或者phi的值的地方，波函数以及波函数的平方，或者概率密度等于零。
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We can look at the 2 p, which should have one radial node, and we just figured it out for the, excuse me, 3d for the 3 p has one radial node, and for the 3 d here, we should have zero radial nodes, we just calculated that.
我们看2p，它有一个节点，我们已经知道对于，不好意思，是3p有一个节点，对于，它应该没有径向节点，我们刚刚算过这个。
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We'll introduce in the next course angular nodes, but today we're just going to be talking about radial nodes, psi and a radial node is a value for r at which psi, and therefore, 0 also the probability psi squared is going to be equal to zero.
将会介绍角节点,但我们今天讲的是，径向节点，径向节点就是指，对于某个r的值,当然，也包括psi的平方，等于,当我们说到s轨道时。
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We call that a node, r and a node, more specifically, is any value of either r, the radius, or the two angles for 0 which the wave function, and that also means the wave function 0 squared or the probability density, is going to be equal to zero.
节点就是指对，于任何半径，或者，两个角度,波函数等于，这也意味着波函数的平方或者概率密度,等于，我们可以看到在1s轨道里。
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sb So in this case we would have 1 s a and 1 s b, and instead we would subtract one from the other, and what we would see is that instead of having additional, more wave function in the middle here, we've actually cancelled out the wave function and we end up with a node.
在这种情况下我们有1sa和，两者相减，我们可以看到不是在，中间有了额外更多的波函数，而是我们消掉了,波函数得到一个节点。
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