Q.Derive an expression for the magnetic dipole moment of a revolving electron.
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Start your 14-day free trial to unlock the full solution →Treating the orbiting electron as a tiny current loop gives its magnetic dipole moment as mu_l = (e v r)/2 = (e/2m) L, showing that the magnetic moment is directly proportional to the orbital angular momentum.
Consider an electron of charge magnitude e and mass m revolving in a circular orbit of radius r around the nucleus with speed v (as in the Bohr model). Since a moving charge constitutes a current, this orbiting electron is equivalent to a tiny current loop.
Equivalent current: The electron completes one revolution in time T = 2pir / v, so the equivalent current is the charge passing a point per unit time:
I = e / T = e v / (2 pi r)
Magnetic moment of the equivalent current loop: A current loop of current I and area A has a magnetic moment mu = I A. Here, the loop area is A = pi r^2, so:
mu_l = I A = [e v / (2 pi r)] x (pi r^2) = (e v r) / 2
Relation to angular momentum: The orbital angular momentum of the electron about the nucleus is L = m v r. Rewriting mu_l:
mu_l = (e v r)/2 = (e / 2m) x (m v r) = (e/2m) L
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