Question

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

• A. $l_H\,>\,l_{Li}\,and\,\|E_H\|\,>\,\|E_{Li}\|$
• B. $l_H\,=\,l_{Li}\,and\,\|E_H\|\,<\,\|E_{Li}\|$
• C. $l_H\,=\,l_{Li}\,and\,\|E_H\|\,>\,\|E_{Li}\|$
• D. $l_H\,<\,l_{Li}\,and\,\|E_H\|\,<\,\|E_{Li}\|$

Answer 1

Answer: (B)

According to Bohr, the angular momentum of an electron of hydrogen atom and hydrogen like atom Li$^{+ \, +},$ $l = n \frac{h}{2 \pi}$.
For both, $n$ = 2 $\therefore $ $l_H = l_{Li} = 2 \times \frac{h}{2 \pi} = \frac{h}{\pi}$
Energy of electron in nth orbit of hydrogen atom, $E_H = - \frac{13.6}{n^2}$ eV
For $n$ = 2, we get $E_H = - \frac{13.6}{4} $ - 3.4 eV
Energy of electron in nth orbit of hydrogen like atom (say Li$^{+ \, +}$ ion), $E_{Li} = \frac{Z^2 \times 13.6}{n^2}$ eV
Here $n$ = 2 and for Li, Z = 3 $\therefore $ $E_{Li} = \frac{9 \times 13.6}{4}$ = - 30.6 eV
$i.e.$ $|E_H| < |E_{Li}|$