• It is joules per atom. Or, if you multiply by Avogadro's number then you will get joules per mole.

    焦耳每个原子,或者,如果乘以,阿伏伽德罗常数你会得到焦耳数每摩尔。

    麻省理工公开课 - 固态化学导论课程节选

  • But I know when I multiply a vector by a number, I get a vector in the same direction.

    但是我所知道的是当矢量乘以一个常数,我会得到一个同方向的矢量

    耶鲁公开课 - 基础物理课程节选

  • And the relationship that he put forth is that the momentum is equal to Planck's constant times nu divided by the speed of light, or it's often more useful for us to think about it in terms of wavelength.

    爱因斯坦提出的关系式是,动量等于普朗克常数,乘以υ除以光速,或者用波长来表示,通常更容易让我们想明白。

    麻省理工公开课 - 化学原理课程节选

  • We can make some substitutions here using some of the derivation on the previous board which will give us the Planck constant divided by 2 pi mass of the electron times the Bohr radius.

    在这里我们也可以,用我以前在黑板上写过的一些词来取代它,得到的是普朗克常数除以2π电子质量,再乘以波尔半径。

    麻省理工公开课 - 固态化学导论课程节选

  • And that implies that since the quantity we want is given by this quantity, which is zero times a constant, the quantity we want is also zero.

    因为我们需要的量,是由这个量乘以一个常数,因为这个量是零,因此我们需要的量也是零。

    麻省理工公开课 - 热力学与动力学课程节选

  • Then it's very clear the way you patch it up is you multiply it by this constant and now we're all set.

    很明显,弥补的方法就是,乘以这个常数就全部搞定了

    耶鲁公开课 - 基础物理课程节选

  • And what I am going to do is say start with this ion, add up the energy associated with the interactions between that ion and everybody else in the row and then multiply it by Avogadro's number, because that is the number of atoms there are in a row.

    接下来我要从这一离子开始,加上相互作用的能量,也就是这一离子,和其它所有在这一行的离子之间的能量,再乘以阿伏加德罗常数,因为这是在一行的原子的数量。

    麻省理工公开课 - 固态化学导论课程节选

  • What Einstein then clarified for us was that we could also be talking about energies, and he described the relationship between frequency and energy that they're proportional, if you want to know the energy, you just multiply the frequency by Planck's constant.

    爱因斯坦阐述的是我们,也可以从能量的角度来谈论,他描述频率和能量之间的关系,是成比例的,如果希望知道能量值,你用普朗克常数乘以频率就可以了。

    麻省理工公开课 - 化学原理课程节选

  • For the 2 s orbital, at 2 a nought, a0 so it's just 2 times that constant a nought, which is the Bohr radius.

    也就是,乘以常数,玻尔半径,对于3s轨道。

    麻省理工公开课 - 化学原理课程节选

  • R is going to be a fix number R times You don't know what it is right now.

    等于一个常数 R 乘以,你现在还不知道它是多少

    耶鲁公开课 - 基础物理课程节选

  • I'm going to just put that in, and it's the cosine of the number times i plus sine of the number times j times R.

    我要在式子加入这个量,这个式子就等于这个值的余弦乘以 i,加上这个值的正弦乘以 j 再乘以常数 R

    耶鲁公开课 - 基础物理课程节选

  • What I am going to do now is I am going to multiply by N Avogadro and then add Born repulsion.

    我接下来要做的是,将其乘以阿伏加德罗常数,再加入Born的排斥作用。

    麻省理工公开课 - 固态化学导论课程节选

  • So, we can get from these energy differences to frequency h by frequency is equal to r sub h over Planck's constant 1 times 1 over n final squared minus 1 over n initial squared.

    所以我们通过不同能量,得到不同频率,频率等于R下标,除以普朗克常数乘以1除以n末的平方减去。

    麻省理工公开课 - 化学原理课程节选

  • And I use the term photon here, and that's because he also concluded that light must be made up of these energy packets, and each packet has that h, that Planck's constant's worth of energy in it, so that's why you have to multiply Planck's constant times the frequency.

    我这里用光子这个词,是因为他还总结出光,必须由这些能量包组成,每个能量,包有这个h,普朗克常数代表,里面的能量,所以这就是为什么你们,要用普朗克常数乘以频率。

    麻省理工公开课 - 化学原理课程节选

  • And the mathematics of that equation involved a double derivative in time of x 0 plus some constant times x equals zero with some constraints on it.

    那个数学方程式,包括了x对时间的二阶导数,加上常数乘以x等于,还有一些限制条件。

    麻省理工公开课 - 固态化学导论课程节选

  • 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

    麻省理工公开课 - 固态化学导论课程节选

  • n So the velocity is given by this product of the quantum number n Planck constant 2 pi mass of the electron time the radius of the orbit, which itself is a function of n.

    速度是量子数,普朗克常数乘以轨道半径的值,它自身也是n的函数。

    麻省理工公开课 - 固态化学导论课程节选

  • So let's just multiply that by Avogadro's number.

    让我们乘以N,阿伏加德罗常数

    麻省理工公开课 - 固态化学导论课程节选

  • So, if we just rearrange this equation, what we find is that z effective is equal to n squared times the ionization energy, IE all over the Rydberg constant and the square root of this.

    我们可以发现有效的z等于n的平凡,乘以电离能除以里德堡常数,这些所有再开方,所以等于n乘以,除以RH整体的平方根。

    麻省理工公开课 - 化学原理课程节选

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