A Hydrogen atom and a Li^{++} ion are both in | vedclass

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(B) The second excited state corresponds to the principal quantum number $n = 3$.According to Bohr's quantization condition,the angular momentum $l$ i

Vedclass · Accepted Answer

$l_H = l_{Li}$ and $|E_H| < |E_{Li}|$

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(B) The second excited state corresponds to the principal quantum number $n = 3$.According to Bohr's quantization condition,the angular momentum $l$ is given by $l = n \left( \frac{h}{2\pi} ight)$.Since both are in the same state $n = 3$,their angular momenta are equal: $l_H = l_{Li} = 3 \left( \frac{h}{2\pi} ight)$.The energy of an electron in a hydrogen-like atom is given by $E = -13.6 \frac{Z^2}{n^2} 	ext{ eV}$.For Hydrogen $(Z_H = 1)$ and $Li^{++}$ $(Z_{Li} = 3)$,the energies are $E_H = -13.6 \frac{1^2}{3^2}$ and $E_{Li} = -13.6 \frac{3^2}{3^2}$.Taking the magnitude,$|E_H| = \frac{13.6}{9} 	ext{ eV}$ and $|E_{Li}| = 13.6 	ext{ eV}$.Clearly,$|E_H| < |E_{Li}|$. Thus,option $(b)$ is correct.

$A$ Hydrogen atom and a $Li^{++}$ ion are both in the second excited state. If $l_H$ and $l_{Li}$ are their respective electronic angular momenta,and $E_H$ and $E_{Li}$ are their respective energies,then:

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