For both $2p$ and $3s$ orbitals,the value of $(n + l)$ is equal to $3$. Which of these has lower energy and why?

(A) The $2p$ orbital has lower energy. According to the $(n + l)$ rule,if the value of $(n + l)$ is the same for two orbitals,the orbital with the lower value of $n$ (principal quantum number) has lower energy. For $2p$,$n = 2$,and for $3s$,$n = 3$. Since $2 < 3$,the $2p$ orbital has lower energy.