A) \[\frac{h}{\pi }\]
B) \[2\pi h\]
C) \[\frac{2h}{\pi }\]
D) \[\frac{\pi }{h}\]
Correct Answer: A
Solution :
According to Bohr's model of atom, electrons in an atom can revolve in those orbits (as suggested by classical theory) in which its angular momentum about the nucleus is an integer multiple of\[\frac{h}{2\pi }\], where \[h\] is Planck's constant\[(=6.6\times {{10}^{-34}}J\text{-}s)\] Thus,\[I\omega =mvr=\frac{n\,\,h}{2\pi }\] For 2nd orbit,\[n=2\] \[\therefore \] \[I\omega =\frac{2h}{2\pi }=\frac{h}{\pi }\]
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