Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$1$

For, $S_1$ (spherically symmetrical)
node $=1$
$\Rightarrow n-1=1$
$n=2$
For $S_2$, radial node $=1$
$E_{s_2}=\frac{-13.6\times z^2}{n^2}=E_H$ in ground state $=-13.6$
$E=\frac{-13.6\times 9}{n^2}\Rightarrow n=3$
So, state $S_1$ is $2s$ and $S_2$ is $3p$.
Azimuthal quantum number in $S_2$ for p is 1 and hence, the answer is ${}^{\prime }{1}^{\prime }$.