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Orbital magnetic dipole moment

What is the magnitude of the orbital dipole magnetic moment μ of an electron in the hydrogen atom in the (a) s state, (b) p state, and (c) d state? (Assume that the spin of the electron is zero.)

Strategy

The magnetic momentum of the electron is related to its orbital angular momentum L . For the hydrogen atom, this quantity is related to the orbital angular quantum number l . The states are given in spectroscopic notation, which relates a letter ( s , p , d , etc.) to a quantum number.

Solution

The magnitude of the magnetic moment is given in [link] :

μ = ( e 2 m e ) L = ( e 2 m e ) l ( l + 1 ) = μ B l ( l + 1 ) .
  1. For the s state, l = 0 so we have μ = 0 and μ z = 0 .
  2. For the p state, l = 1 and we have
    μ = μ B 1 ( 1 + 1 ) = 2 μ B μ z = μ B m , where m = ( 1 , 0 , 1 ) , so μ z = μ B , 0 , μ B .
  3. For the d state, l = 2 and we obtain
    μ = μ B 2 ( 2 + 1 ) = 6 μ B μ z = μ B m , where m = ( 2 , 1 , 0 , 1 , 2 ) , so μ z = 2 μ B , μ B , 0 , μ B , −2 μ B .

Significance

In the s state, there is no orbital angular momentum and therefore no magnetic moment. This does not mean that the electron is at rest, just that the overall motion of the electron does not produce a magnetic field. In the p state, the electron has a magnetic moment with three possible values for the z -component of this magnetic moment; this means that magnetic moment can point in three different polar directions—each antiparallel to the orbital angular momentum vector. In the d state, the electron has a magnetic moment with five possible values for the z -component of this magnetic moment. In this case, the magnetic moment can point in five different polar directions.

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A hydrogen atom has a magnetic field, so we expect the hydrogen atom to interact with an external magnetic field—such as the push and pull between two bar magnets. From Force and Torque on a Current Loop , we know that when a current loop interacts with an external magnetic field B , it experiences a torque given by

τ = I ( A × B ) = μ × B ,

where I is the current, A is the area of the loop, μ is the magnetic moment, and B is the external magnetic field. This torque acts to rotate the magnetic moment vector of the hydrogen atom to align with the external magnetic field. Because mechanical work is done by the external magnetic field on the hydrogen atom, we can talk about energy transformations in the atom. The potential energy of the hydrogen atom associated with this magnetic interaction is given by [link] :

U = μ · B .

If the magnetic moment is antiparallel to the external magnetic field, the potential energy is large, but if the magnetic moment is parallel to the field, the potential energy is small. Work done on the hydrogen atom to rotate the atom’s magnetic moment vector in the direction of the external magnetic field is therefore associated with a drop in potential energy. The energy of the system is conserved, however, because a drop in potential energy produces radiation (the emission of a photon). These energy transitions are quantized because the magnetic moment can point in only certain directions.

If the external magnetic field points in the positive z -direction, the potential energy associated with the orbital magnetic dipole moment is

Practice Key Terms 5

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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