There are no correct answers listed.
If n = 4, then l may be 3,2,1,0 so a is out. b,c,d are allowed from n alone.
Look at b. If l = 2, then m(l) may be -2,-1,0,1,2 so b is out since m(l) of 3 is not allowed.
c is out for the reason you point out. Check out d.
n of 4 allows l = 0
l of 0 allows m(l)(ell) of +l(ell) to -l(ell) including 0 so that is ok.
m(s) is -1/2 is ok.
So it must be d.
Which of the following is a legitimate set of n, l, m(l), and m(s) quantum numbers?
a. 4, -2, -1, 1/2
b. 4, 2, 3, 1/2
c. 4, 3, 0, 1
d. 4, 0, 0, -1/2
I don't think it's c because m(s) can only have +1/2 or -1/2 but I'm not sure how to figure out the others. Could you please explain why it's correct?
2 answers
Note that I started with there is no correct answer but show the correct answer as D. I just didn't go back and remove that first sentence.