Structure of Atom

Concept:In hydrogen-like atoms, orbital energy depends solely on principal quantum number $n$; in multi-electron atoms, energy depends on both $n$ and $l$ due to shielding and penetration effects.Explanation:For hydrogen (single‑electron), all orbitals with same $n$ have equal energy. Therefore $E_{2s}(\mathrm{H}) = E_{2p}(\mathrm{H})$ → option B is correct.For lithium (multi‑electron), the $(n+l)$ rule applies: $2s$ has $n+l=2$, $2p$ has $n+l=3$. Lower $(n+l)$ gives lower energy. Also, the $2s$ orbital penetrates closer to the nucleus, experiencing a higher effective nuclear charge. Thus $E_{2s}(\mathrm{Li}) < E_{2p}(\mathrm{Li})$ → option A is correct.Comparing hydrogen and lithium: Even after shielding by two $1s$ electrons, the effective nuclear charge $Z_{\text{eff}}$ for a $2s$ electron in Li is greater than that for H. This makes $E_{2s}(\mathrm{Li})$ more negative (lower) than $E_{2s}(\mathrm{H})$. Hence $E_{2s}(\mathrm{H}) > E_{2s}(\mathrm{Li})$ → option D is correct.Since $E_{2p}(\mathrm{H}) = E_{2s}(\mathrm{H})$ and $E_{2s}(\mathrm{Li}) < E_{2s}(\mathrm{H})$, we have $E_{2p}(\mathrm{H}) > E_{2s}(\mathrm{Li})$; therefore option C is incorrect.Answer:Options A, B, and D are correct.