Question

In ${\Psi }_{321}$ the sum of angular momentum, spherical nodes and angular mode is:

• A. $\frac{\sqrt{6h}+4\pi }{2\pi }$
• B. $\frac{\sqrt{6h}}{2\pi }+3$
• C. $\frac{\sqrt{6h}+2\pi }{2\pi }$
• D. $\frac{\sqrt{6h}+8\pi }{2\pi }$

Answer 1

Answer: (A)

$\frac{\sqrt{6h}+4\pi }{2\pi }$

${\Psi }_{321}⟶{\Psi }_{l,m,n}$

$n=3,l=2,m=1$

Angular Momentum=$\sqrt{l(l+1)}\cdot \frac{h}{2\pi }=\sqrt{2(2+1)}\cdot \frac{h}{2\pi }=\frac{\sqrt{6}\cdot h}{2\pi }$

Radial node or spherical node $=n-l-1=3-2-1=0$

Angular node $=l=2$

Thus, sum=$\frac{\sqrt{6}\cdot h}{2\pi }+2=\frac{\sqrt{6}\cdot h+4\pi }{2\pi }$