An electron in the ground state of hydrogen atom is revolving in a circular orbit of radius R. If m is the mass of the electron, e its charge, and h Planck's constant, the orbital magnetic dipole moment of the electron is

Options

A. `h/(2pimR)`
B. `(eh)/(4pi^2R^(2m))`
C. `(eh)/(4pim)`
D. zero

Answer

C

Explanation

From the Bohr quantum condition, we have
mvR = `(nh)/(2pi)`
where v is the speed of the electron.
For ground state, n = 1. So `v= (h)/(2pimR)`
Current around the orbit is
`I=e/T=e/(2piR//v)=(ev)/(2piR) = (eh)/(4pi^2R^2m)`
Magnetic dipole moment is IA `= I xx piR^(2) = (eh)/(4pi^(2)R^(2)m) xxpiR^(2) = (eh)/(4pim)`