## Atom as a Magnetic Dipole:

In the case of an atom, when an electron revolves in an anticlockwise direction, the current is clockwise. Therefore, the upper face of the electron loop act as a south pole and the lower face acts as the North pole. Hence an atom behaves as a magnetic dipole.

If ‘e’ be the charge of an electron revolving in an orbit of radius ‘r’ with uniform angular velocity ω.

Magnetic moment of the atom is given by-

M = IA ……….(i) But A = πr ^{2} ……….(ii)and I = e/T I = e/(2π/ω) (where T = 2π/ω) I = ωe/2π ……….(iii) Put the value of I and A from (ii) and (iii) in equation (i), M = (ωe/2π) (πr ^{2})M = (1/2) (eωr ^{2}) ……….(iv)According to Bohr’s theory, mνr = nh/2π But ν = rω ⇒ m (rω) r = nh/2π ⇒ mr ^{2}ω = nh/2π ⇒ ωr ^{2} = nh/2πmPut this value of ωr ^{2} in equation (iv),⇒ M = (1/2) e (nh/2πm) ⇒ M = n (eh/4πm) ⇒ M = n . µ _{B} ……….(v)Where, µ _{B} = eh/4πm ……….(vi)From equation (iv), it is clear that magnetic dipole moment is quantized. The minimum value of the magnetic dipole moment of the atom is given by the expression- µ _{B} = eh/4πm, This is called Bohr Magneton.Also, µ _{B} = eh/4πm = (1.6 x 10^{-19} x 6.6 x 10^{-34})/(4 x 3.14 x 9.1 x 10^{-31}) = 9.27 x 10^{-24} Amp.-metre^{2}Hence we may define Bohr magneton as the magnetic dipole moment associated with an atom due to the motion of the electron in the first orbit of the hydrogen atom. |