Atomic Structure Question 242
Question: The four quantum numbers of the outermost orbital of $ K $ (atomic no. =19) are
Options:
A) $ n=2,,l=0,,m=0,,s=+\frac{1}{2} $
B) $ n=4,,l=0,,m=0,,s=+\frac{1}{2} $
C) $ n=3,,l=1,,m=1,,s=+\frac{1}{2} $
D) $ n=4,,l=2,,m=-1,,s=+\frac{1}{2} $
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Answer:
Correct Answer: B
Solution:
$ K_{19}=1s^{2},,2s^{2},2p^{6},,3s^{2},3p^{6},,4s^{1} $ for $ 4s^{1} $ electrons. $ n=4,,l=0,,m=0 $ and $ s=+\frac{1}{2}. $
