Question

The $H$-like species $L{i}^{2+}$ is in spherically symmetrical state $S_1$ and one radial node. Upon absorbing light, the ion undergoes transition to state $S_2$. The state $S_2$ has one radial node and it has the same energy as the ground state energy of a $hydrogen$ atom.

The orbital angular momentum quantum number of state $S_2$ is:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

The spherically symmetric state $S_1$ of $L{i}^{2+}$ with one radial node is $2s$. Upon absorbing light the ion gets excited to state $S_2$, which also has one radial node. The energy of electrons in $S_2$ is the same as that of the energy of electrons in the ground state $Hydrogen$ atom.

Since $Z$=3 for $L{i}^{2+}$ and $E_n$ = $E_1$, where $E_1$ is the energy for ground state H-atom.
$\therefore \frac{3^2}{n^2}13.6=13.6$

$\Rightarrow n=3$

Thus the state $S_2$ of $L{i}^{2+}$ having one radial node is $3p$.
Orbital angular momentum quantum number of $3p$ is $1$.