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And we can simplify this expression as saying negative e squared over 4 pi, epsilon nought r squared. Epsilon nought is a constant, it's something you might see in physics as well.
也会遇到它,在这里，你可以就把它,理解为一个转换系数,我们需要做的。
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Once you've got that, you can do minus 7 times a vector Just take the vector, multiply it by Pi and flip it over.
明白这点之后,你就可以计算-7乘以矢量,只需用 π 去乘以那个矢量,然后将其方向调转
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Bohr said that the angular momentum, mvr where n is this integer counter h over 2 pi.
波尔提及到角动量，是被量化了的，mvr，is，quantized,这里的n等于一个整数乘以h除以2π
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Remember, we didn't hybridize the 2 p y orbital, so that's what we have left over to form these pi bonds.
记住，我们并没有杂化2py轨道，这是我们剩下的那个行成了π键。
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r=nh/mv That will give us 2 pi r goes as nh over mv.
2π
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n*pi/L It turns out it is n pi over l.
结果是。
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q1*q2/ That's simply q1, q2 over 4 pi epsilon zero R.
那只是简单的。
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And now the force, in its most general term / is q1q2 over 4 pi epsilon zero, which is the conversion factor r squared.
库仑力的最基本形式,就是，其中r是一个变量。
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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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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。
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When we look at this angular part, we see that it's always the square root of 1 over 4 pi, it doesn't matter what the angle is, it's not dependent on the angle.
当我们看这角向部分，可以看到它总是等于１除以４pai开根号，这和是什么角度没有关系，它和角度无关。
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