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Q. The ratio of the magnetic dipole moment to the angular momentum of the electron in the 1st orbit of hydrogen atom is

KCETKCET 2012Atoms

Solution:

We have, $ L=m v r$
and $ M=IA=\frac{e v}{2 \pi r} \times \pi r^{2}=\frac{1}{2} evr$
$\therefore \frac{M}{L}=\frac{e}{2 m}$