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Q. In $\Psi_{321}$ the sum of angular momentum, spherical nodes and angular node is :-

NTA AbhyasNTA Abhyas 2020

Solution:

$\psi_{321}\Rightarrow n=3,\ell =2,m=1$
angular momentum $=\frac{h}{2 \pi }\sqrt{\ell \left(\right. \ell + 1 \left.\right)}=\frac{\sqrt{6} h}{2 \pi }$
spherical node $=n-\ell -1$
$=3-2-1=0$
Angular node $=\ell =2$
sum $\Rightarrow \frac{\sqrt{6} h}{2 \pi }+0+2$
$\Rightarrow \frac{\sqrt{6} h + 4 \pi }{2 \pi }$