Problem 18 - University Physics with Modern Physics

Question

A hydrogen atom in the \(n=1, m_{s}=-\frac{1}{2}\) state is placed in a magnetic field with a magnitude of 0.480 \(\mathrm{T}\) in the \(+z\) -direction. (a) Find the magnetic interaction energy (in electron volts) of the electron with the field. (b) Is there any orbital magnetic dipole moment interaction for this state? Explain. Can there be an orbital magnetic dipole moment interaction for \(n \neq 1 ?\)

Step-by-Step Solution

Answer

(a) The interaction energy is approximately \(-2.78 \times 10^{-5}\) eV. (b) No orbital interaction for \(n=1\), but possible for \(n \neq 1\).

Step 1: Understand the Magnetic Interaction Energy Formula

The magnetic interaction energy for an electron in a magnetic field is given by\[ U = - \vec{\mu} \cdot \vec{B} \]where \( \vec{\mu} \) is the magnetic moment of the electron, and \( \vec{B} \) is the magnetic field. For a single electron, the magnetic moment \( \mu = -g \frac{e}{2m_e} S \), where \( g \approx 2 \), \( e \) is the elementary charge, \( m_e \) is the electron mass, and \( S \) is the spin quantum number.

Step 2: Determine the Electron's Magnetic Moment

For electron spin, the z-component of the magnetic dipole moment is given by:\[ \mu_z = -g \frac{e\hbar}{2m_e} m_s \]Substitute the values: \( g = 2 \), \( e = 1.602 \times 10^{-19} \) C, \( \hbar = 1.055 \times 10^{-34} \) J·s, \( m_e = 9.109 \times 10^{-31} \) kg, and \( m_s = -\frac{1}{2} \) into the equation.

Step 3: Calculate \( \mu_z \) using Known Constants

Insert the known values into the formula:\[ \mu_z = -2 \times \frac{1.602 \times 10^{-19} \times 1.055 \times 10^{-34}}{2 \times 9.109 \times 10^{-31}} \times \left(-\frac{1}{2}\right) \]Solving, we find:\[ \mu_z \approx 9.274 \times 10^{-24} \text{ J/T} \], which is also known as the Bohr magneton \( \mu_B \).

Step 4: Calculate the Magnetic Interaction Energy

Using \( \mu_z \) and the relation \( U = -\mu_z B \), where \( B = 0.480 \) T:\[ U = -(9.274 \times 10^{-24} \text{ J/T})(0.480 \text{ T}) \]\[ U \approx -4.451 \times 10^{-24} \text{ J} \]Convert to electron volts by using 1 eV = \( 1.602 \times 10^{-19} \) J:\[ U \approx -4.451 \times 10^{-24} \text{ J} \times \frac{1 \, \text{eV}}{1.602 \times 10^{-19} \, \text{J}} = -2.78 \times 10^{-5} \text{ eV} \].

Step 5: Assess Orbital Magnetic Dipole Moment Interaction

In the \( n=1 \) state, the electron has \( l=0 \), meaning no orbital angular momentum. The orbital magnetic moment is related to \( l \), thus it's zero in this state. Therefore, there's no interaction from the orbital magnetic moment. For \( n eq 1 \), \( l\) can be greater than 0, allowing potential orbital magnetic dipole moment interactions.

Key Concepts

Bohr magneton

The Bohr magneton, symbolized as \( \mu_B \), is a physical constant and the quantum of the magnetic dipole moment. It provides a measure for the magnetic dipole moment of an electron. The Bohr magneton is crucial when discussing magnetic interactions in atomic physics. Its value in SI units is approximately \( 9.274 \times 10^{-24} \text{ J/T} \).

The formula for the Bohr magneton is:

• \( \mu_B = \frac{e \hbar}{2m_e} \)

where:

• \( e \) is the elementary charge, approximately \( 1.602 \times 10^{-19} \text{ C} \)
• \( \hbar \) is the reduced Planck's constant, about \( 1.055 \times 10^{-34} \text{ J}\cdot \text{s} \)
• \( m_e \) is the electron mass, roughly \( 9.109 \times 10^{-31} \text{ kg} \)

The Bohr magneton is a key factor in determining the magnetic interaction energy of electrons within a magnetic field. It represents the intrinsic magnetic moment attributed to the angular momentum of an electron, especially due to its spin.

Electron spin

Electron spin is a fundamental property of electrons, indicative of their intrinsic angular momentum. Unlike classical spinning objects, electron spin does not have an equivalent in macroscopic physics. It's a quantum property and is essential to the magnetic behavior of atoms.

Electrons can have a spin of \( +\frac{1}{2} \) or \( -\frac{1}{2} \), typically referred to as "spin-up" or "spin-down." This spin quantum number, \( m_s \), determines the orientation of an electron's magnetic moment along a magnetic field direction.

The magnetic dipole moment due to electron spin is given by:

• \( \mu_z = -g \frac{e \hbar}{2m_e} m_s \)

where \( g \) is the electron g-factor (approximately 2). The spin of an electron significantly influences the total magnetic interaction energy in a magnetic field. This interaction is crucial in understanding phenomena like hyperfine splitting in atomic spectra.

Orbital magnetic dipole moment

The orbital magnetic dipole moment is a quantum mechanical property associated with an electron's motion around the nucleus of an atom. Unlike spin, which relates to internal angular momentum, the orbital magnetic moment results from an electron's orbit.

This moment is directly tied to the orbital angular momentum quantum number \( l \). For an electron in an orbital, the magnetic moment can be expressed as:

• \( \mu_{orbital} = -\frac{e}{2m_e} L \)

where \( L \) is the orbital angular momentum \( (L = \hbar\sqrt{l(l+1)})\).

In the ground state of hydrogen, \( n=1 \), the orbital angular momentum quantum number \( l \) is 0. This means there is no orbital magnetic dipole moment in this state, hence no interaction with a magnetic field. However, for states where \( n eq 1 \), \( l \) can be greater than zero, allowing for potential interactions between the orbital magnetic dipole moment and an external magnetic field.