The ratio (in SI unit) of magnetic dipole moment to that of the angular momentum of an electron of mass `m kg` and charge `e` coulomb in Bohr's orbit of hydrogen atom is

Options

A. `(e)/(m)`
B. `(m)/(e)`
C. `(2m)/(e)`
D. `(e)/(2m)`

Answer

D

Explanation

Angular momentum,
`L = m_(e) vr " " … (i)`
Therefore , orbital motion of electron is equivalent to a current .
`I = e(1// T)`
Period of revolution of electron is given by
`T = (2pi r)/(v)`
`therefore I = e((1)/(2pi r //v)) = (ev)/(2 pi r)`
Area of electron orbit , `A = pi r^(2)`
Magnetic dipole moment of the atom
`M = (ev)/( 2pi r) xx pi r^(2) = evr `
Using Eq. (i) , we have
`M = ((e)/(2m_(e))) L`