• And what we end up with for the energy then is 2.84 times 10 to the -19 joules.

    我们算出的能量是,2,84乘以10的-19次方焦耳。

    麻省理工公开课 - 化学原理课程节选

  • And we can actually better visualize this if we plot how that energy changes as a function of internuclear distance.

    而我们就能更清楚地看到这些,如果我们画,能量随核间距的变化曲线。

    麻省理工公开课 - 化学原理课程节选

  • So if you demonstrate something by writing an intelligent answer just by outlining it and saying, well, what I would do is I would equate the energy lambda and then solve for lambda, I can see that you know what is going on.

    如果你想证明一些东西,通过写下一些很天才的答案,仅仅大致说一下,好的,我想做的是能量相等,然后解,我能看到你知道怎么做。

    麻省理工公开课 - 固态化学导论课程节选

  • So if I worry about energy I have a pretty good chance of extracting out these heat capacities, right, and I don't have to worry about exactly which path and I can really mix things up.

    能量与热容有密切的关系,从能量就可以就算这两个热容,我不需要考虑到底是哪一条路径,可以随意计算。

    麻省理工公开课 - 热力学与动力学课程节选

  • It's a particle of mass m and momentum p has this energy so you solve for the energy, you get two answers.

    一个质量为m,动量为p的粒子,有这么大的能量,你解能量,得到两个解

    耶鲁公开课 - 基础物理课程节选

  • 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,发射一个电子所需要的最低能量的情况。

    麻省理工公开课 - 化学原理课程节选

  • 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来的,它的能级是最高的,因为这样的话发射它,只需要消耗最少的能量

    麻省理工公开课 - 化学原理课程节选

  • So that means we're going to need to figure out what is the energy per photon that's emitted by that UV light.

    所以那意味着我们将需要,计算从紫外光源发射出的,每个光子的能量

    麻省理工公开课 - 化学原理课程节选

  • So hopefully if some of you have your calculators with you, you can confirm the answer that I got, which is that the energy is 7.82 times 10 to the -19 joules.

    所以如果你们带了计算器,希望你们也能确认一下,我们算出的答案,能量等于,7,82乘以10的-19次方焦耳。

    麻省理工公开课 - 化学原理课程节选

  • The total energy of the system, which we are going to get from postulate number four, which says the energy of the electron, which is the energy of the system, is the sum of the kinetic and the potential energy.

    这个系统的总能量,也就是我们将从第四个假设中算出的能量,也就是电子运动产生的能量,也就是整个系统的能量,是动能和位能的总和。

    麻省理工公开课 - 固态化学导论课程节选

  • And, we could easily calculate the energies.

    我们就能很容易的算能量

    麻省理工公开课 - 固态化学导论课程节选

  • 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,激发一个电子,那这个电子一定是,2,p,轨道上的。

    麻省理工公开课 - 化学原理课程节选

  • 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,我们可以激发其中所有的电子。

    麻省理工公开课 - 化学原理课程节选

  • 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电子,那么我们可以得到很多的剩余能量

    麻省理工公开课 - 化学原理课程节选

  • 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,就是发射一个电子所需要的能量关联起来。

    麻省理工公开课 - 化学原理课程节选

  • So this means that we can go directly from the energy between two levels to the frequency of the photon that's emitted when you go between those levels.

    这意味着我们可以直接,从两个能级的能量得到它们之间,跃迁发射光子的频率。

    麻省理工公开课 - 化学原理课程节选

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