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(B) For electron $A$,$n=3$ and $l=2$,which corresponds to the $3d$ subshell.For electron $B$,$n=3$ and $l=0$,which corresponds to the $3s$ subshell.Ac

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$A$ has more energy than $B$

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(B) For electron $A$,$n=3$ and $l=2$,which corresponds to the $3d$ subshell.For electron $B$,$n=3$ and $l=0$,which corresponds to the $3s$ subshell.According to the $(n+l)$ rule,the energy of an orbital increases with the $(n+l)$ value.For $A$,$(n+l) = 3+2 = 5$.For $B$,$(n+l) = 3+0 = 3$.Since the $(n+l)$ value for $A$ is greater than that for $B$,electron $A$ has higher energy than electron $B$.

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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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 $B$?