All the energies are negative because it is a bound system. I start up here with n equals one.

所有能量级都是负数，因为它是一个束缚系统,在这里我从n等于1讲起。

麻省理工公开课 - 固态化学导论课程节选

So if I start off with a list of length n, how many times can I divide it by 2, until I get to something no more than two left?

我能够除以多少次2呢？,直到我得到的长度不超过2么？,对数次，对吧？就像刚才那位同学说的那样？

麻省理工公开课 - 计算机科学及编程导论课程节选

Absorption is just the opposite of emission, so instead of starting at a high energy level and dropping down, when we absorb light we start low and we absorb energy to bring ourselves up to an n final that's higher.

吸收就是发射的逆过程,与从一个高能量到低能量不同，当吸收光时，我们从低能量开始,吸收能量到一个更高的能量。

麻省理工公开课 - 化学原理课程节选

So when we count, generally, again, we start from zero, we go to N minus 1.

当我们计数时，一般的，再次强调，我们从0开始，到N-1结束。

哈佛公开课 - 计算机科学课程节选

And unlike n, l can start all the way down at 0, and it increases by integer value, so we go 1, 2, 3, and all the way up.

不像n，l可以从0开始取，然后每次增加一个整数，所以我们可以去1，2，3，一直下去。

麻省理工公开课 - 化学原理课程节选

Log n Log n, because at each stage I'm cutting the problem in half. So I start off with n then it's n n/2 n/4 n/8 over two n over four n over eight.

因为总共有多少层？,因为在每一层，我都是把问题分解成两半,因此以n开始，然后是。

麻省理工公开课 - 计算机科学及编程导论课程节选

So what we'll do is this problem here, which is let's calculate out what the wavelength of radiation n would be emitted from a hydrogen atom if we start at the n equals 3 level and we go down to the n equals 2 level.

我们来做这个问题,让我们来计算一下,从n等于3到,等于2能级氢原子辐射的波长是多少。

麻省理工公开课 - 化学原理课程节选

So this is our final equation, and this is actually called the Balmer series, which was named after Balmer, and this tells us the frequency of any of the lights where we start with an electron in some higher energy level and we drop down to an n final that's equal to 2.

把2代入到这里，所以得到1除以,这就是我们最终的方程，这叫做Balmer系，以Balmer名字命名的,它告诉我们从高能级掉到n等于2的,最终能级所发出光的频率。

麻省理工公开课 - 化学原理课程节选