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Q. According to Bohr’s theory, the angular momentum for an electron of $3rd$ orbit is

VITEEEVITEEE 2014

Solution:

Angular momentum, mvr
$ = \frac{nh}{2\pi} = \frac{3 \times h}{2 \pi} = \frac{1.5\, h }{\pi} $
$ = 3\hbar \left[ \because \hbar = \frac{h }{2 \pi}\right]$