Question

The orbital angular momentum quantum number of the state S${}_2$ is :

• A. $0$
• B. $\sqrt{2}$$\frac{h}{2\pi }$
• C. $1$
• D. $\frac{h}{2\pi }$

Answer 1

Answer: (B)

$\sqrt{2}$$\frac{h}{2\pi }$

Since S1 is spherically symmetric it means it is s orbital. Also,it is given that no.of radial nodes = 1
No.of radial nodes is given by -: $n-l-1$ ; where n = principal quantum number, l = angular quantum number
Here, $n-l-1=1$
For s orbital, l = 0
Thus, n = 2
S${}_1$ is 2s orbital. So, S${}_2$ will be the orbital next to 2s i.e. 2p
The orbital angular momentum is given by $\sqrt{l\times l+1}\times h/2\pi$
Here, for 2p orbital l=1
Orbital angular momentum will be $\sqrt{1\times 1+1}\times h/2\pi$ i.e. $\sqrt{2}\times h/2\pi$