00:01
In this problem, we're going to talk about the quantum angular momentum.
00:03
And what we need to remember is that the angular momentum, l of a particle, as a function of its orbital quantum number, little l is equal to the square root of l times l plus 1 times h bar.
00:20
L plus 1 is the angular quantum number, and we need to remember that the possible values of l for a given value of the principal number n ranges from 0 to n minus 1.
00:37
Okay? and our, in our exercise, we have an atom that has an angular momentum of 10 times the square of 50, 7 times hbar.
00:56
And our goal in the question a is to find what is the value of the orbital number.
01:06
So basically, notice that l can be written as a square root of 5 ,700 h -bar, and this is equal to the square root of l times l plus 1, h -bar, which means that l times l plus 1 is equal to 5 ,700.
01:27
So l -squared plus l is equal to 5 ,700.
01:34
Plus l minus 5 ,700 is equal to 0.
01:39
So l is equal to minus 1 plus or minus the square root of 1 plus 4 times 5700 divided by 2.
01:54
One of the root are negative, so i'm not going to care about it.
01:58
And the positive 1 is equal to 75...