So if we draw the 2 p orbital, what we just figured out was there should be zero radial nodes, so that's what we see here.
如果我们画一个2p轨道,我们刚才知道了是没有径向节点的,我们在这可以看到。
So, let's change our graph where we now have this zero point set as the two individuals hydrogen atoms, and then we see that our h 2 molecule is at the negative of the dissociation energy, or the negative what that bond strength is.
那么让我们把曲线图中的零点能改到,两个分离的氢原子处,那我们就会看到,氢分子就是负的离解能,或者负的键的强度。
Is the strength length greater than two or greater than zero?
字符串长度大于2或者大于0么?
And so, in this case, l could equal zero, 2 l could equal one or l could equal two.
所以这样的话,l可以等于,也可以等于1或。
Interest rates have hit zero 0.2%; that's twenty basis points above zero.
利率已经接近零了0。2%;,这只有零点以上的二十个基本点。
So what we do is we pick an origin, call it zero, we put some markers there to measure distance, and we say this guy is sitting at 1,2,3,4,5.
我们现在选择一个原点,称其为零点,我们做一些标记来测量距离,这个点在五个单位长度的地方
Yup, zero radial nodes. So, for a 2 p orbital, all the nodes actually turn out to be angular nodes.
没有,对于2p轨道,所有的节点都是角向节点。
I don't exactly know where they are in RAM, but I know if I've got 2 gigabytes of RAM, zero is here, byte 1 is here, byte 2 is here, dot, dot, dot so maybe byte 123 is over here and 456 is here.
我不知道他们在内存中哪个地方,但是我知道如果我有2G内存,零在这里,字节1在这里,字节2在这里,等等等,所以可能123字节在这里,456字节在这里。
And a bit, B-I-T, just a shorthand for binary digit, 2 so a digit of zero over a one, bi means two, you only have two digits.
比特,B-I-T,是binary,digit的简写,从0到1的数字,而bi表示,所以你就只有两个数字。
We put the magnesium 2 plus here, and then we put the oxygen here, O 2 minus, and we know that this gives us our R zero, right?
我们把Mg2+放这,然后O2-放这,我们都知道这给我们的是R0,对么?
I know the energy in this first pair would equal -e^2 That is just going to equal minus e squared over 4 pi epsilon zero r naught.
我们明白第一对的能量将会等于,等于,/4πε0,R圈。
It's gonna be zero if N is less than 2, right.
如果N小于2,T应该是。
Well, this question mark becomes an O, this question mark becomes an O, and then the loop terminates 0 1 2 because it's iterating from zero to N so that's zero, 1, 2 and the length of the string is 3 so the loop terminates, but I remember that I needed to have this special sentinel value so I'm just going to put it there manually.
嗯,这个问号变成了,这个问号变成了0,然后循环结束了,以为迭代从0到N,那就是,字符串的长度是3,然后循环就结束了,但是我记得我需要这个特殊的标记值,所以需要手动的加上它。
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