The correct set of four quantum numbers for the valence electron of rubidium atom $(Z=37)$ is:
(1) $5,0,0,+\frac{1}{2}$
(2) $5,0,1,+\frac{1}{2}$
(3) $5,1,0,+\frac{1}{2}$
(4) $5,1,1,+\frac{1}{2}$
$Rb=[Kr] 5 s^{1}$
$n=5$
$l=0$
$m=0$
$s=+1 / 2$ or $-1 / 2$
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