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(B) The angular momentum of an electron in an orbit is given by $L = mvr$.The magnetic dipole moment $M$ associated with the orbital motion of the ele

Vedclass · Accepted Answer

$\frac{e}{2m}$

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(B) The angular momentum of an electron in an orbit is given by $L = mvr$.The magnetic dipole moment $M$ associated with the orbital motion of the electron is given by $M = IA$,where $I$ is the current and $A$ is the area of the orbit.Since $I = \frac{e}{T} = \frac{ev}{2 \pi r}$ and $A = \pi r^2$,we have $M = \left( \frac{ev}{2 \pi r} ight) 	imes (\pi r^2) = \frac{1}{2} evr$.Taking the ratio of the magnetic dipole moment to the angular momentum:$\frac{M}{L} = \frac{\frac{1}{2} evr}{mvr} = \frac{e}{2m}$.Thus,the ratio is $\frac{e}{2m}$.

The ratio of the magnetic dipole moment to the angular momentum of the electron in the $1^{st}$ orbit of a hydrogen atom is

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