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Derive an expression for the magnetic moment vec(mu) of an electron revolving around the nucleus in terms of its angular momentum vecl. What is the direction of the magnetic moment of the electron with respect to its angular momentum? |
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Answer» Solution :We know that a revolving electron constitutes an electric current given by `I = e/T = (ev)/(2 pi r)` where `T = (2 pi r)/v` = peroid of revolution of electron revolving in a CIRCULAR orbit of radius r with a speed v. `:.` Magnetic moment `(mu_l)` associated with this CIRCULATING current, `mu_l = I cdot A = (ev)/(2pi r) cdot pi r^2 = (evr)/(2)` As an electron revolving is an anticlockwise direction is equivalent to current in clockwise direction, hence in ACCORDANCE with right hand rule the magnetic moment is in a direction perpendicular to plane of paper (or plane of orbit) directed inward. The above relation may also bewritten as : `mu_l = (e v rm)/(2m ) = e/(2m) cdot l` and in vector notation, `vec(mu)_(l) = -e/(2m) cdot vecl ,` where `l = m v r` = orbital angular momentum of the electron around the nucleus . The -ve sign APPLIED here indicates that the angular momentum of the electorn is opposite in direction to the magnetic moment. 
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