How many unique combinations of the quantum numbers l and ml are there when (b) n = 5?

Video transcript

Hello everyone in this video. We're given the principle quantum number that we're trying to see how many unique combinations of quantum numbers of L and ml there will be So let's not forget that our quantum numbers include L ML and MS. Alright. So since we're given n equals to four So I guess I can say this is my part one So n equals to four then R. L is dependent on R. N. And that can range from 02, N -1. We said their end value is equal to four. So a new range is going to be from zero through 4 -1 is three. So listing out Oliver Values that sees a part two then so we have zero or when L equals to zero, L equals to one. Well he goes to two and L equals to three. So I'm just listing out all the L values that come from this range here. So next will be our M. L. Value. So we found the N. L. And now M L. So M L is dependent on our L. Value and we have them here. So go ahead, do the end pink. So just a little kind of side note here that M M L value is always going to be ranging from negative L through positive L. So since we have our L value being zero the only ml values possible is going to be zero. The next is going to be ranging from negative one to positive one. So that range includes numbers -1, 0 and positive one. Next with the L B. Going to Our ML values is going to be in the range of positive two or negative to positive two. So the values are included within that range Are negative 2 -10 one and two. The same thing for our last L value ranging from negative three to positive three. It's going to be -3 -2. No one sarah one two three. And so the number of possible Combinations from the 1st 1 is one. The 2nd 1 would be three, then we have bye and last night seven. So adding the numbers here that we just wrote in purple which is one plus three plus five seven. There's going to be 16 total. And so from lee running out our answer. So there are 16 unique combinations. Alright let's go and highlight that as well. Alright, so that is going to be our final answer for this problem.