Select the acceptable sets of quantum numbers in an atom.

Select the acceptable sets of quantum numbers in an atom.
a.) (2, 2, 1,+1/2) b.) (3,2,1,1) c.) (1, 0, 1/2, 1/2) d) (4, 3, -2, +1/2)
e.) (3, 0, 0, +1/2)


The correct answers are d and e. Here’s why: First, you need to know that the four numbers represent n, l, m(l), and m(s). Two things can help you rule out some options right away: 1) the last number m(s) must either be +1/2 or -1/2, and 2) the first three numbers must always be an integer (no fractions!). So, let’s break down the answers one at a time: a) 2, 2, 1, +1/2 –> n=2, if l=n-1, then 2-1 must be 1, so you can rule this answer out. b) 3, 2, 1, 1 –> The last number m(s) must either be +1/2 or -1/2, so this is wrong. c) 1, 0, 1/2, 1/2 –> The third number cannot be a fraction, so this is wrong. d) 4, 3, -2, +1/2 –> n=4, so l=4-1=3, good so far. If l=3, then that means you can have three possible values for l (0, 1, 2). To determine m(l), take the highest possible value for l (in this case, 2) and solve for m(l)=2(l)+1, or 2(2)+1=5. The five means you can have five values for m(l) with 0 in the middle and work your way out so that you have -2, -1, 0, 1, 2 (five total possibilties), so, yes, -2 is a possible value. Still good. Lastly, you have m(s), which must be either +1/2 or -1/2 and in this case it is, so you’re good! This is correct!

e) 3, 0, 0, +1/2 –> n=3, so l=3-1=2, which means you have two possible values for l (0, 1), so the second number checks out. Now for m(l)=2(l)+1=2(1)+1=3, so you have three possibilities for m(l) represented by -1, 0, 1, so the third number is good. And by now, you know the last number is good, so this answer is also correct!

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