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Yup. So it's just that ionization energy 94×10-18J that we have experimentally measured, 3 . 9 4 times 10 to the negative 18 joules.
它就是我们通过实验测得的,电离能3。,我们用所有的这些。
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So now we can just take the negative of that binding energy here, and I've just rounded up here or 1 . 4 times 10 to the negative 19 joules.
等于4是第三激发态,现在我们可以取它结合能的负值,也就是1。4乘以10的负19次方。
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Now a float is not 1 byte, it's 4 bytes, or 4 times 8, 32 bits.
而float就不只1字节了，它占据4个字节，或者说32比特。
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So it'll be 1 plus 1/4 times 4, 1/4 times 4 is 1, so 1 plus 1 is 2, so Player II's best response in that case will be 2.
这回就是1+4/4,即1+1=2,所以此时参与人II的最佳对策是2
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+4 Because this could be three plus four, 3 times five, or it could be three, plus four times five.
因为这可能是，然后乘以5，也可能是，加上4乘以5。
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That's equal to seven minus six minus two 3 times 4.19, or 4.19 plus 2.3.
算出来是7-6-2*4。19的平方,或者4。19加2。
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During the section times I'll be available if you feel like you want to read Chapters 2 and 4 and then come and ask questions, sort of a tutorial on these topics of chemistry and biochemistry, then I'll be available to talk about that during that time.
在小组讨论的时间里,你们可以来找我,如果你们学习完第二章和第四章之后,想问一些关于,化学和生化方面的问题,可以来找我,我会在那个时候给你们进行讲解
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Our step three is to figure out how many bonding electrons that we need, or excuse me, how many total electrons that we need to fill up our octets, so that's just going to be 4 times 8, which is 32.
我们的第三步是判断,我们需要多少个成键电子，不好意思，是我们总共需要多少个电子,才能填满所有“八隅体“,那么这就应该是四乘上八，也就是三十二。
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it's going to be: 1 minus the probability they'll choose Right times 1, plus the probability that they choose Right times 4.
方程是,1减对手选右的概率再乘以1,加上对手选右的概率乘以4
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4 So this is 16 times 4 equals 64 and though this is not proof by any means, it's not a formal proof because here is one example that happens to prove my point.
所以是16*4，即,尽管如此，这并不是严格的证明,因为是这个例子碰巧证明了我的观点。
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Anyone? How many times can you tear a 4 billion page phone book in half?
谁知道？,40亿页纸每次对半撕，你能撕多少次？
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4 So 2 times 24 plus 16 gives me 64.
*24再加16等于。
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And that factor is 4 pi times epsilon zero.
这个常数是4乘π再乘ε
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So now, 2 times 2 is 4 plus 4, that's 8.
*2是4，再加4就是。
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Three plus four times five.
加4乘以。
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4 Seven times two is 14.
乘以2等于。
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it's an easy calculation -- we're just taking the negative of the binding energy, again that makes sense, because it's this difference in energy here. So what we get is that the binding energy, when it's negative, the ionization energy is 5 . 4 5 times 10 to the negative 19 joules.
这个计算很简单-我们,只需要取结合能的负值,同样这很容易理解，因为这就是这的能量差,所以我们得到的就是结合能,当它取负值，电离能就是5。45乘以。
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So, we can do that by using this equation, which is for s orbitals is going to be equal to dr 4 pi r squared times the wave function squared, d r.
用这个方程,对于s轨道，径向概率分布,4πr的平方,乘以波函数的平方,这很容易理解。
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So this is 3, plus 2 times 3, plus 4 t minus 2.
也就是3+2*3+4t，好。
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And in this case, we go from 8 to 4 to 2 to 1 three times and then on each iteration of this algorithm, each pass across the board I'm touching N numbers, so that means I'm doing N things, log N times.
在这个例子中，我们从8得到4，到2，再到1，是3次,在这个算法的每次迭代中，每一趟我都会操作N个数，也就是所我每次要做N步操作，一共要做，log，N，次。
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And we wrote something that looks, the energy is equal to minus the Madelung constant times Avogadro's number, 0R0 q1 q2 over 4 pi epsilon zero R zero.
我们写下了，晶格能等于负的马德隆常数,乘以阿伏伽德罗常数，乘以q1q2除以4πε
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e The charge on the anion times minus e, so there is the minus e squared, 0R0 and divided by 4 pi epsilon zero r naught, because now I am evaluating this function at r naught, one minus one over n where n is the Born exponent.
阴离子的电荷乘以，因此会有-e的频繁，除以4πε,因为现在我用r圈评估这个函数，1-1/n，n是波恩指数。
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It's times 4, plus PR times 2.
即4+Pr*2
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