I've got two states here, three states here, two here, I need four, and if I can come up with these bonds, four, by the Hund rule I'd fill them like this.
我们已经有2个状态在这里,3个在这里,2个在这里,我需要4个,如果我能想到这些键,四个,通过洪特规则,我们就能像这样排布。
- But there's one potential gotcha here -- and this is where you have to be kind -- of piecing together the little clues along the way -- what have I done that's interesting here that had I not, I would have had a mathematical error?
但是这里有一个可能性-,这就是你需要把,一些线索拼凑在一起-,我这里所做的是很有趣的,如果我不这样做,我可能已经犯了个数学错误?
s1 Here is 1s atomic. But lithium has 2s1, so I need a 2s atomic orbital here and likewise over here.
这是1s原子,但锂有两层,所以我还需要在这里添加2s轨道,就像那样。
It says they're not the same, and boy, I need help on this one, John, John pre-defined eq in there.
我并没有对它进行定义,它说它们并不相等,噢同学们,在这里我需要帮助了,在这里并没有,it’s,not,,there’s,no,提前定义的等于啊。
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。
I want to do this carefully and slowly, and before I even start, let me point out in passing that calculus is not going to help us here.
我要慢点仔细讲这部分,在我开始之前,我需要说明微积分,在这里是用不着的
Another way of saying it is, anything that uses get float doesn't care what the details are inside or shouldn't, and if I change that definition, I don't have to change anything elsewhere in my code, whereas if I just have the raw code in there, I have to go off and do it.
换种说法就是,任何用到获取,输入这个功能的人不用担心具体的实现细节,如果我改变了这里的实现,我并不需要去改变我的代码,因为我的最底处的源码就在这里,我去改这里就可以了。
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