8.2 Orbital magnetic dipole moment of the electron (Page 3/8)

University physics volume 3 Page 3 / 8

U (θ) = − μ B c o s θ = − μ_{z} B = − (− μ_{B} m) B = m μ_{B} B,

where $μ_{B}$ is the Bohr magneton and m is the angular momentum projection quantum number (or magnetic orbital quantum number ), which has the values

m = − l, − l + 1,..., 0,..., l - 1, l .

For example, in the $l = 1$ electron state, the total energy of the electron is split into three distinct energy levels corresponding to $U = − μ_{B} B, 0, μ_{B} B .$

The splitting of energy levels by an external magnetic field is called the Zeeman effect . Ignoring the effects of electron spin, transitions from the $l = 1$ state to a common lower energy state produce three closely spaced spectral lines ( [link] , left column). Likewise, transitions from the $l = 2$ state produce five closely spaced spectral lines (right column). The separation of these lines is proportional to the strength of the external magnetic field. This effect has many applications. For example, the splitting of lines in the hydrogen spectrum of the Sun is used to determine the strength of the Sun’s magnetic field. Many such magnetic field measurements can be used to make a map of the magnetic activity at the Sun’s surface called a magnetogram ( [link] ).

The figure shows the effect of magnetic field, B sub ext, on two different spectral lines, corresponding to the l=1 to l=0 transition on the left and the l=2 to l=0 transition on the right. The spectra are shown for no external field, for a non zero external field and for a large external field. With no external field, both transitions appear as single lines. In the second case, when magnetic field is applied, the spectral lines split into several lines; the line on the left splits into three lines. The line on the right splits into five. In the third case, the magnetic field is large. The left line is again split into three lines and the right into five, but the split lines are farther apart than they are when the external magnetic field is not as strong. — The Zeeman effect refers to the splitting of spectral lines by an external magnetic field. In the left column, the energy splitting occurs due to transitions from the state $(n = 2, l = 1$ ) to a lower energy state; and in the right column, energy splitting occurs due to transitions from the state $(n = 2, l = 2)$ to a lower-energy state. The separation of these lines is proportional to the strength of the external magnetic field.

A magnetogram of the sun, which appears as a gray disc against a black background, with white and black spots scattered on it. Most of the spots are concentrated in the center right part of the image. — A magnetogram of the Sun. The bright and dark spots show significant magnetic activity at the surface of the Sun.

Summary

A hydrogen atom has magnetic properties because the motion of the electron acts as a current loop.
The energy levels of a hydrogen atom associated with orbital angular momentum are split by an external magnetic field because the orbital angular magnetic moment interacts with the field.
The quantum numbers of an electron in a hydrogen atom can be used to calculate the magnitude and direction of the orbital magnetic dipole moment of the atom.

Conceptual questions

Explain why spectral lines of the hydrogen atom are split by an external magnetic field. What determines the number and spacing of these lines?

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A hydrogen atom is placed in a magnetic field. Which of the following quantities are affected? (a) total energy; (b) angular momentum; (c) z-component of angular momentum; (d) polar angle.

a, c, d; The total energy is changed (Zeeman splitting). The work done on the hydrogen atom rotates the atom, so the z -component of angular momentum and polar angle are affected. However, the angular momentum is not affected.

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On what factors does the orbital magnetic dipole moment of an electron depend?

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Problems

Find the magnitude of the orbital magnetic dipole moment of the electron in in the 3 p state. (Express your answer in terms of $μ_{B} .$ )

The 3 p state corresponds to $n = 3$ , $l = 2$ . Therefore, $μ = μ_{B} \sqrt{6}$

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A current of $I = 2 A$ flows through a square-shaped wire with 2-cm side lengths. What is the magnetic moment of the wire?

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Estimate the ratio of the electron magnetic moment to the muon magnetic moment for the same state of orbital angular momentum. ( Hint: $m_{μ} = 105.7 MeV / c^{2})$

The ratio of their masses is 1/207, so the ratio of their magnetic moments is 207. The electron’s magnetic moment is more than 200 times larger than the muon.

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Find the magnitude of the orbital magnetic dipole moment of the electron in in the 4 d state. (Express your answer in terms of $μ_{B} .$ )

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For a 3 d electron in an external magnetic field of $2.50 \times 10^{− 3} T$ , find (a) the current associated with the orbital angular momentum, and (b) the maximum torque.

a. The 3 d state corresponds to $n = 3$ , $l = 2$ . So,
$I = 4.43 \times 10^{−7} A .$
b. The maximum torque occurs when the magnetic moment and external magnetic field vectors are at right angles ( $sin θ = 1)$ . In this case:
$| \vec{τ} | = μ B .$
$τ = 5.70 \times 10^{−26} N \cdot m .$

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An electron in a hydrogen atom is in the $n = 5$ , $l = 4$ state. Find the smallest angle the magnetic moment makes with the z -axis. (Express your answer in terms of $μ_{B .}$ )

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Find the minimum torque magnitude $| \vec{τ}$ | that acts on the orbital magnetic dipole of a 3 p electron in an external magnetic field of $2.50 \times 10^{- 3} T$ .

A 3 p electron is in the state $n = 3$ and $l = 1$ . The minimum torque magnitude occurs when the magnetic moment and external magnetic field vectors are most parallel (antiparallel). This occurs when $m = \pm 1$ .The torque magnitude is given by
$| \vec{τ} | = μ B sin θ,$
Where
$μ = (1.31 \times 10^{−24} J/T) .$
For $m = \pm 1,$ we have:
$| \vec{τ} | = 2.32 \times 10^{21} N \cdot m .$

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An electron in a hydrogen atom is in 3 p state. Find the smallest angle the magnetic moment makes with the z -axis. (Express your answer in terms of $μ_{B} .$ )

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Show that $U = − \vec{μ} \cdot \vec{B}$ .

( Hint : An infinitesimal amount of work is done to align the magnetic moment with the external field. This work rotates the magnetic moment vector through an angle $- d θ$ (toward the positive z -direction), where $d θ$ is a positive angle change.)

An infinitesimal work dW done by a magnetic torque $τ$ to rotate the magnetic moment through an angle $- d θ$ :
$d W = τ (− d θ)$ ,
where $τ = | \vec{μ} \times \vec{B} |$ . Work done is interpreted as a drop in potential energy U , so
$d W = − d U .$
The total energy change is determined by summing over infinitesimal changes in the potential energy:
$U = − μ B c o s θ$
$U = − \vec{μ} \cdot \vec{B} .$

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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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