A) \[\sqrt{6}h\]
B) \[\sqrt{2}h\]
C) \[2\sqrt{3}h\]
D) \[0h\]
Correct Answer: A
Solution :
| Angular momentum of electron in d-orbital is |
| \[=\sqrt{1+(1+1)}\frac{h}{2\pi }\] for d-orbital, \[I=2\] |
| \[=\sqrt{2(2+1)}\Rightarrow h=\sqrt{6}h\] \[\left( \because h=\frac{h}{2\pi } \right)\] |
You need to login to perform this action.
You will be redirected in
3 sec
