An electron in the ground state of hydrogen atom is revolving in anticlockwise direction in circular orbit of radius R. The orbital magnetic dipole moment of the electron will be

Answer

(i) 4 (ii) 8

Explanation

(i)4(ii)8
(i) In ground state `(n=1)` according to Bohr's theory
`mvR=h/(2pi)` or `v=h/(2pi m R)`
Now time period `T=(2piR)/v=(2piR)/(h//2pimR)=(4pi^(2)mR^(2))/h`
Magnetic moment `M=iA`
where `i=("charge")/("time period")=e/((4pi^(2)mR^(2))/h)=(eh)/(4pi^(2)mR^(2))` and `A=piR^(2)`
`:.M=(piR^(2))((eh)/(4pi^(2)mR^(2)))` or `M=(eh)/(4pim)`
Direction of magnetic moment M is perpendicular to the plane of orbit.
(iI) `tau=MxxB,|tau|=MBsin theta`
Where `theta` is the angle betweeen M and B `, theta=30^(@)`
`:.tau=((eh)/(4pim))(B)sin 30^(@),tau=(ehB)/(8pim)`
The direction of `tau` is perpendicular to both M and B.