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(B) The second excited state corresponds to the principal quantum number $n = 3$.The angular momentum of an electron in an orbit is given by $l = \fra

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

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(B) The second excited state corresponds to the principal quantum number $n = 3$.The angular momentum of an electron in an orbit is given by $l = \frac{nh}{2\pi}$. Since both the hydrogen atom and the $Li^{2+}$ ion are in the same state $(n = 3)$,their angular momenta are equal: $l_{H} = l_{Li}$.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 $Li^{2+}$ $(Z_{Li} = 3)$,the magnitudes of energy are $|E| \propto Z^2$ (since $n$ is constant).Thus,$|E_{Li}| = Z_{Li}^2 |E_H| = 3^2 |E_H| = 9 |E_H|$.Therefore,$|E_H| < |E_{Li}|$.

In a hydrogen atom and a $Li^{2+}$ ion,the electron is in the second excited state. If $l_{H}$ and $l_{Li}$ are the angular momenta of the electrons and $E_H$ and $E_{Li}$ are their respective energies,then:

Similar Questions

An atom of hydrogen gas is in an excited state $n_2$. It absorbs a photon of some energy and jumps to a higher energy state $n_1$. Then,it returns to the ground state after emitting six different wavelengths in the emission spectrum. Determine the values of $n_1$ and $n_2$.

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