In a hydrogen atom, an electron carrying charge `'e'` revolves in an orbit of radius `'r'` with speed `'v'`. Obtain an expression for the magnitude of magnetic moment of a revolving electron.

Explanation

In the Bohr model of a hydrogen atom, the electron of charge- `e` performs a uniform circular motion around the positively charged nucleus. Let `r`, `v` and `T` be the orbital radius, speed and period of motion of the electron. Then,
`T=("Circumference")/("Speed")=(2pir)/(v)`......`(i)`
The orbital motion of the electron consituties a loop of coventional current `I` in the opposite sense of its revolution.
Using equation `(i)`, `I=(e )/(T)=(ev)/(2pir)`..........`(ii)`
Therefore, the magnetic dipole moment associated with this electric current loop has a magnitude,
`M_(0)=` Current `xx` Area of the loop
Using equation `(ii)`, `=1(pir^(2))=(ev)/(2pir)xxpir^(2)=(1)/(2)evr`
Multiplying and dividing the right hand side of the above expression by the electron mass `m_(e)`.
`M_(0)=(e)/(2m_(e))(m_(e)vr)=(e)/(2m_(e))L_(0)`
Where `L_(0)=m_(e)vr` is the magnitude of the orbital angular momentum of the electron. `vecM_(0)` is opposite to `vecL_(0)`.
`:. vecM_(0)=-(e)/(2m_(e))vecL_(0)`
Direction of magnetic moment is into the plane. Negatively charged electron is moving in-anti-clockwise direction, leading to closkwise current.