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No, OK and that's correct, because each photon of light actually has more energy than is needed to eject an electron.
没有，好，对了，因为每一个光子实际上,都有更多的能量去逐出一个电子。
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So we can actually pop an electron or eject an electron from any single orbital that is occupied within the atom.
任何一个被占据轨道,打出一个电子,或者说发射出一个电子。
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Yes. In fact, there is not enough energy in a single photon to go ahead and eject an electron from this zinc surface.
是的，事实上，这里的光子没有足够的能量,去从锌表面逐出一个电子。
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So in this case, we don't have enough energy to eject an electron, so, an electron is not ejected.
所以在这种情况下，我们没有足够的能量,去逐出一个电子，这样没有电子被逐出。
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So what we're saying here is the incident energy, so the energy coming in, is just equal to the minimum energy that's required to eject an electron.
这里我们来讨论一下，入射能量正好等于,发射出一个电子所需要的最低能量的情况。
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The electron's going to come out of that highest occupied atomic orbital, that one that's the highest in energy, because that's going to be the at least amount of energy it needs to eject something.
这个电子应该是从,最高的被占据轨道上出来的，它的能级是最高的，因为这样的话发射出它,只需要消耗最少的能量。
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If we have a higher z effective, it's pulled in tighter, we have to put in more energy in order to eject an electron, so it turns out that that's why case 2 is actually the lowest energy that we need to put in.
而如果有效核电量更高，原子核的束缚也就更紧，我们不得不输入更多的能量来打出一个电子，这就是第二种情况,所需要输入的,能量更少的原因。
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All right. So please raise your hand now if you think there'll be sufficient energy to eject electrons from the metal surface?
好的，如果你们认为,有足够的能量从金属表面,射出电子请举手？
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So, which has the smallest energy that you have to put in in order to eject this electron?
也就是，哪一项为了打出一个电子,所需要注入的能量最少？
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We're only using up a little bit to eject the electron, then we'll have a lot left over.
如果我们只需要用很少一部分来发射出电子，那么我们可以得到很多的剩余能量。
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So we can use an equation to relate the incident energy and the kinetic energy to the ionization energy, or the energy that's required to eject an electron.
因此我们可以用一个公式将入射能量,与动能和电离能，就是发射出一个电子所需要的能量关联起来。
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So, for example, we were talking about a threshold frequency as in a minimum frequency of light that you need in order to eject an electron from a metal surface.
举个例子来说，我们谈论的临界频率是,光从金属表面逐出一个电子,所需的最小频率。
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It makes sense that it's going to come out of the highest occupied atomic orbital, because that's going to be the lowest amount of energy that's required to actually eject an electron.
从最高占据轨道上,去掉一个电子是合理的，因为这样是发射一个电子,所损失的最低能量。
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So if we can figure out the binding energy, we can also figure out how much energy we have to put into our atom in order to a eject or ionize an electron.
所以如果我们可以计算出结合能，我们也可以计算出,我们需要注入多少能量到原子中,去逐出或电离一个电子。
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Does it make any difference at all in terms of whether we can eject an electron?
这究竟与我们能否,逐出一个电子有区别吗？
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So we see that we do not eject electrons in the case of the laser pointer, even if we have this intensity, it is still not related to the energy of an individual photon, so we won't see an effect.
所以我们看到我们用激光笔,还是没有逐出电子，即使我们有这样的强度，它仍然与一个单个的光子能量无关,所以我们不会看到光电效应。
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So if we think about the work function for zinc, and the work function for zinc is 6.9 times 10 to the -19 joules, do we expect that when we shine our UV light on the zinc, we'll be able to eject electrons?
如果我们考虑锌的功函数,它是6，9乘以10的-19次方焦耳，我们是否可以预测当,用紫外灯照射锌时，我们可以射出电子呢？
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putting all those things together, if you looked at this question again we'd get 100% on it, 0 9 that our only option here is 0. 9, and that it's not the negative, it's the positive version, because we're talking about how much energy we have to put into the system in order to eject an electron.
把这些放在一起，你们再看一下题目,大家100%都能选对，我们唯一的选择就是这个，它不是负数，它是正的，因为我们说的，是要,把电子激发出来,需要提供的能量。
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So if we're going to eject an electron using a minimum amount of energy, that's where it's going to come from.
因此，如果我们要用最少的能量,激发出一个电子，那这个电子一定是，2，p，轨道上的。
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Well, we can't guarantee with UV light we'll have enough energy to eject every single electron, so that's why when we use x-rays, they're higher energy, you can pretty much be guaranteed we're going to eject all of those electrons there.
好，我们不能保证紫外光有足够的能量,激发出每一个电子，所以我们要使用，X，射线，它的能量更高，你可以非常确定,我们可以激发出其中所有的电子。
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Remember, because that 1 s orbital is all the way down in terms of if we're thinking about an energy diagram, we're all the way down here, so we have a huge amount of energy we have to put into the system in order to eject an electron.
还记得吧，因为1，s，轨道在能量示意图里,是在最底部的，我们要一直到最下面，所以我们要向这个系统注入非常大的能量,才能打出一个电子。
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So what we'll end up with is boron plus, 1 s 2 2 s 2 1 s 2, 2 s 2, E and what we say is the delta energy or the change in energy as the same thing as saying the energy of the products minus the energy of our reactant here, that's how much energy we have to put into the system to eject an electron.
所以结果应该是正一价的硼，电子排布为，我们说，Δ,也就是，E，的变化量,等于生成物的能量,减去反应物的能量，这就是我们从这个系统中,打出一个电子所需要的能量。
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