Which of the following set of quantum numbers | vedclass

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(C) The energy of an orbital is determined by the $(n + l)$ rule.For $A$: $n = 3, l = 0$,so $(n + l) = 3 + 0 = 3$ ($3s$ orbital).For $B$: $n = 3, l =

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$n = 3, l = 2, m = 1, s = +1/2$

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(C) The energy of an orbital is determined by the $(n + l)$ rule.For $A$: $n = 3, l = 0$,so $(n + l) = 3 + 0 = 3$ ($3s$ orbital).For $B$: $n = 3, l = 1$,so $(n + l) = 3 + 1 = 4$ ($3p$ orbital).For $C$: $n = 3, l = 2$,so $(n + l) = 3 + 2 = 5$ ($3d$ orbital).For $D$: $n = 4, l = 0$,so $(n + l) = 4 + 0 = 4$ ($4s$ orbital).Comparing the $(n + l)$ values,$5$ is the highest value.Therefore,the set of quantum numbers $n = 3, l = 2, m = 1, s = +1/2$ represents the highest energy.

Column-$I$	Column-$II$
$(a)$ $4s$	$(p)$ Circular orbit around nucleus
$(b)$ $4p$	$(q)$ Non-directional orbital
$(c)$ $1s$	$(r)$ Angular momentum $= 2h/\pi$
$(d)$ $3d$	$(s)$ Radial node is zero

Which of the following set of quantum numbers represents the highest energy of an atom?

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Column-$I$ Column-$II$

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$(b)$ $4p$ $(q)$ Non-directional orbital

$(c)$ $1s$ $(r)$ Angular momentum $= 2h/\pi$

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Two electrons $A$ and $B$ in an atom have the following set of quantum numbers:
$A: n=3, l=2, m_l=-2, m_s=+1/2$
$B: n=3, l=0, m_l=0, m_s=+1/2$
What is true for $A$ and $B$?

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