Question

For an electron with $n=3$ has only one radial node. The orbital angular momentum of the electron will be :

• A. 0
• B. $\sqrt{6}\frac{h}{2\pi }$
• C. $\sqrt{2}\frac{h}{2\pi }$
• D. $3\frac{h}{2\pi }$

Answer 1

The correct option is C $\sqrt{2}\frac{h}{2\pi }$

Number of radial nodes $=n-l-1=1$

Since, $n=3.\therefore l=1$

Orbital angular momentum $=\sqrt{l(l+1)}\frac{h}{2\pi }=\sqrt{2}\frac{h}{2\pi }$