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Q.
For a $d$ electron, the orbital angular momentum is
Structure of Atom
Solution:
The orbital angular momentum is: $\frac{h}{2 \pi} \sqrt{l(l+1)}$
The orbital angular momentum for an electron ind orbital $(l=2)$ is
$\frac{h}{2 \pi} \sqrt{l(l+1)}=\frac{h}{2 \pi} \sqrt{2(2+1)}=\frac{h}{2 \pi} \sqrt{6}$