91Ó°ÊÓ

Angular momentum is defined as the product of the angular velocity and its moment of inertia.

The state of the hydrogen atom is 2p.

We already know that the probability density does not depend on the azimuthal angleφ. In the absence of information about the z−component of angular momentum it is equally likely that the electron would have any of the allowed values. Adding the angular probability densities with equal weights, we have

$\begin{array}{rcl}{\mathrm{Θ}}_{1.0}\left(\mathrm{θ}\right)+{\mathrm{Θ}}_{1.+1}\left(\mathrm{θ}\right)+{\mathrm{Θ}}_{1.-1}\left(\mathrm{θ}\right)& =& {\left(\sqrt{\frac{3}{4\mathrm{π}}\cos \mathrm{θ}}\right)}^2+{\left(\sqrt{\frac{3}{8\mathrm{π}}\sin \mathrm{θ}}\right)}^2+{\left(\sqrt{\frac{3}{8\mathrm{π}}\sin \mathrm{θ}}\right)}^2\\ & =& \frac{3}{4\mathrm{π}}{\cos }^2\mathrm{θ}+2\frac{3}{8\mathrm{π}}{\sin }^2\mathrm{θ}\\ & =& \frac{3}{4\mathrm{π}}\\ & & \end{array}$

Where,$\mathrm{θ}$ = colatitude angle,$\mathrm{ϕ}$ = azimuth angle

This has no dependence on either $\mathrm{ϕ}\text{or}\mathrm{θ}$

This is clear that the above-mentioned expression is independent of $\mathrm{θ}$and$\mathrm{ϕ}$

Hence, the probability densities for these different possible states are independent of $\mathrm{ϕ}$and $\mathrm{θ}$