Question

{"given":"Both hydrogen (H) and $\\mathrm{Li}^{2+}$ are in the second excited state, so $n = 3$ for both. $Z_H = 1$, $Z_{Li} = 3$.","key_observation":"In Bohr's model, angular momentum $l = n\\hbar$ depends only on $n$, while energy $E_n = -13.6\\,\\dfrac{Z^2}{n^2}$ eV depends on both $Z$ and $n$.","option_analysis":[{"label":"(A)","text":"$l_H > l_{Li}$ and $|E_H| > |E_{Li}|$","verdict":"incorrect","explanation":"Since both atoms are at $n = 3$, $l_H = l_{Li} = 3\\hbar$, so $l_H > l_{Li}$ is false. Also, $|E_{Li}| = 9|E_H|$, so $|E_H| > |E_{Li}|$ is false."},{"label":"(B)","text":"$l_H = l_{Li}$ and $|E_H| < |E_{Li}|$","verdict":"correct","explanation":"Both atoms at $n = 3$ have $l = 3\\hbar$, so $l_H = l_{Li}$. Since $|E| \\propto Z^2$, $|E_{Li}| = 9|E_H|$, confirming $|E_H| < |E_{Li}|$."},{"label":"(C)","text":"$l_H = l_{Li}$ and $|E_H| > |E_{Li}|$","verdict":"incorrect","explanation":"While $l_H = l_{Li}$ is correct, $|E_H| > |E_{Li}|$ is false because $|E_{Li}| = Z_{Li}^2 \\times |E_H| / Z_H^2 = 9|E_H|$."},{"label":"(D)","text":"$l_H < l_{Li}$ and $|E_H| < |E_{Li}|$","verdict":"incorrect","explanation":"The second part $|E_H| < |E_{Li}|$ is correct, but $l_H < l_{Li}$ is false since both equal $3\\hbar$ at $n = 3$."}],"answer":"(B)","formula_steps":[]}