Explain Magnetic Dipole Moment of a Revolving Electron - QS Study

The magnetic dipole moment of a revolving electron

According to Neil Bohr’s atom model, the negatively charged electron is revolving around a positively charged nucleus in a circular orbit of radius r. The revolving electron in a closed path constitutes an electric current. The motion of the electron in anticlockwise direction produces conventional current in clockwise direction.

Current, i = e/T; where T is the period of revolution of the electron.

If v is the orbital velocity of the electron, then

T = 2πr/v

so, i = ev/2πr

Due to the orbital motion of the electron, there will be orbital magnetic moment μl

μl = i A, where A is the area of the orbit

μl = ev/2πr . πr2

μl = evr/2

If m is the mass of the electron

μl = e/2m (mvr)

mvr is the angular momentum (l) of the electron about the central nucleus.

μl = e/2m.l … … (1)

μ/l = e/2m; is called gyromagnetic ratio and is a constant. Its value is 8.8 × 1010 C kg-1. Bohr hypothesised that the angular momentum has only discrete set of values given by the equation.

l = nh/2π … … (2) where n is a natural number and h is the Planck’s constant = 6.626 × 10-34 Js.

Substituting equation (2) in equation (1)

μl = e/2m . nh/2π = neh/4πm

The minimum value of magnetic moment is

l)min = eh/4πm, n = 1

The value of eh/4πm is called Bohr magneton

By substituting the values of e, h and m, the value of Bohr magneton is found to be 9.27 × 10-24 Am2.

In addition to the magnetic moment due to its orbital motion, the electron possesses magnetic moment due to its spin. Hence the resultant magnetic moment of an electron is the vector sum of its orbital magnetic moment and its spin magnetic moment.