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(A) The energy of an orbital is determined by the $(n+l)$ rule. According to this rule, the orbital with a higher $(n+l)$ value has higher energy.If t

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$A > C > B > D$

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(A) The energy of an orbital is determined by the $(n+l)$ rule. According to this rule, the orbital with a higher $(n+l)$ value has higher energy.If the $(n+l)$ values are the same, the orbital with the higher value of $n$ has higher energy.Calculating $(n+l)$ values for each:$A: n=5, l=2 \implies (n+l) = 5+2 = 7$$B: n=5, l=0 \implies (n+l) = 5+0 = 5$$C: n=4, l=3 \implies (n+l) = 4+3 = 7$$D: n=4, l=1 \implies (n+l) = 4+1 = 5$Comparing the values:For $A$ and $C$, $(n+l) = 7$. Since $A$ has a higher $n$ $(5 > 4)$, $A > C$.For $B$ and $D$, $(n+l) = 5$. Since $B$ has a higher $n$ $(5 > 4)$, $B > D$.Therefore, the decreasing order of energy is $A > C > B > D$.

The correct order of decreasing energy for the electrons whose quantum numbers $n$ and $l$ are given below, is
$A$. $n=5, l=2$
$B$. $n=5, l=0$
$C$. $n=4, l=3$
$D$. $n=4, l=1$

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The correct order of decreasing energy for the electrons whose quantum numbers $n$ and $l$ are given below, is$A$. $n=5, l=2$$B$. $n=5, l=0$$C$. $n=4, l=3$$D$. $n=4, l=1$