An electron in the ground state of hydrogen atom is revolving in anti-clockwise direction in a circular orbit of radius R.

(i) The expression for the orbital magnetic moment of the electrton is `(eh)/(n pi m)`. Find n `square`
(ii) The atom is placed in a uniform magnetic induction B such that the normal to the plane of electron’s orbit makes an angle of `30^(@)` with the magnetic induction. The torque experienced by the orbiting electron is `(ehB)/(n pi m)`. Find n `square`

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.