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A hydrogen atom is in a d state. In the absence of an external magnetic field the states with different \(m_{l}\) values have (approximately) the same energy. Consider the interaction of the magnetic field with the atom's orbital magnetic dipole moment. (a) Calculate the splitting (in electron volts) of the \(m_{l}\) levels when the atom is put in a \(0.400-\mathrm{T}\) magnetic field that is in the \(+z\) -direction. (b) Which \(m_{l}\) level will have the lowest energy? (c) Draw an energy-level diagram that shows the \(d\) levels with and without the external magnetic field.

Short Answer

Expert verified
The energy splitting is 4.63 x 10鈦烩伒 eV. The lowest energy is at \(m_l = -2\).

Step by step solution

01

Understanding the Zeeman Effect

For a hydrogen atom in a magnetic field, the energy levels split due to the Zeeman effect. Each atomic state characterized by quantum numbers gets energy shifts depending on the magnetic field strength and the magnetic quantum number \( m_l \). The energy shift for a level is given by:\[\Delta E = m_l \mu_B B\]where \( \mu_B \) is the Bohr magneton \( (5.788 \times 10^{-5} \text{ eV/T}) \), \( B \) is the magnetic field strength, and \( m_l \) takes integer values between \( -l \) and \( +l \).
02

Calculating Splitting for m_l Levels

Since we are dealing with a d state, \( l = 2 \). The possible values for \( m_l \) are \(-2, -1, 0, 1, 2\). The magnetic field \( B \) is given as \( 0.400 \text{ T} \).We compute the energy shift \( \Delta E \) for the extremal \( m_l \) values:\[\Delta E = (2)(5.788 \times 10^{-5} \text{ eV/T}) (0.400 \text{ T}) = 4.6304 \times 10^{-5} \text{ eV}\]The difference in energy between consecutive \( m_l \) values is \( \Delta E \). For \( m_l = 卤2 \), \( \Delta E = 卤2 \times 4.6304 \times 10^{-5} \text{ eV} \).
03

Identifying Lowest Energy State

The levels will have energies determined by \( m_l \), meaning higher \( m_l \) will have higher energy under a positive magnetic field. Thus, the state with the lowest energy is \( m_l = -2 \) because it has the most negative shift. Therefore, the lowest energy level is \(-4.6304 \times 10^{-5} \text{ eV}\) relative to the center of the levels.
04

Drawing Energy-Level Diagram

In the absence of a magnetic field, all \( m_l \) levels at \( l = 2 \) have the same energy. With the introduction of the magnetic field, these levels split linearly according to \( \Delta E = m_l \mu_B B \). The energy diagram will have five lines, with the central line at \( m_l = 0 \), and two lines above and two below representing \( m_l = 卤1, 卤2 \). The spacing between each line is equal and represented by \( \Delta E \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Quantum Number
In quantum mechanics, each electron in an atom is identified by a set of quantum numbers, one of which is the magnetic quantum number, denoted as \( m_l \). The magnetic quantum number arises from the electron's orbital angular momentum. It can take on integer values ranging from \( -l \) to \( +l \), where \( l \) is the azimuthal quantum number associated with the electron's orbital type (e.g., s, p, d, f states with \( l = 0, 1, 2, 3 \), respectively).
The significance of \( m_l \) comes into play when an atom is placed in an external magnetic field. This field causes the electron's energy levels to split, a phenomenon known as the Zeeman Effect. The energy shift for a particular state is directly proportional to the value of \( m_l \).
  • Increased value of \( m_l \) corresponds to higher energy shifts in a positive magnetic field.
  • Lower \( m_l \) values correspond to lower energy, making them more stable energetically in a positive magnetic field.
Therefore, the magnetic quantum number plays a crucial role in determining the energy distribution when atoms interact with magnetic fields.
Bohr Magneton
The Bohr magneton is a fundamental physical constant, symbolized as \( \mu_B \), and is crucial for understanding magnetic interactions at the atomic level. It provides a convenient unit for expressing the magnetic dipole moment of an electron resulting from its orbital or spin angular momentum.
The Bohr magneton is defined mathematically as:\[\mu_B = \frac{e\hbar}{2m_e}\]where:
  • \( e \) is the electron charge (\( 1.602 \times 10^{-19} \) C),
  • \( \hbar \) is the reduced Planck鈥檚 constant (\( 1.055 \times 10^{-34} \) Js),
  • \( m_e \) is the electron mass (\( 9.109 \times 10^{-31} \) kg).
In more practical terms, the Bohr magneton is approximately \( 5.788 \times 10^{-5} \) eV/Tesla. It represents the smallest magnetic moment that can be observed in atomic physics.
In scenarios like the Zeeman Effect, the energy shift \( \Delta E \) due to a magnetic field is expressed using \( \mu_B \), such that \( \Delta E = m_l \mu_B B \), where \( B \) is the magnetic field strength. This relationship clearly shows how the Bohr magneton links quantum mechanics with electromagnetic theory.
Orbital Magnetic Dipole Moment
The orbital magnetic dipole moment is a concept critical to understanding how electrons in atoms interact with external magnetic fields. It represents the magnetic moment due to an electron's motion in its orbit, akin to a current loop generating a magnetic dipole.
In simpler terms, when an electron moves in its orbital path, it creates a magnetic field, with a magnetic dipole moment proportional to its angular momentum. This moment is denoted as \( \mu \) and can be typically expressed in terms of the Bohr magneton \( \mu_B \).
For an electron with a given magnetic quantum number \( m_l \) in a particular orbital, the magnetic dipole moment is proportional to \( m_l \) and \( \mu_B \):\[\mu = - m_l \mu_B\]When subject to an external magnetic field, the orientation and magnitude of this dipole moment influence the electron's energy, leading to the splitting of energy levels, which is central to the Zeeman Effect.
  • The direction of the magnetic field determines the alignment or anti-alignment of the dipole moment.
  • Changes in energy levels are directly related to the electron's orbital magnetic dipole moment.
Overall, the orbital magnetic dipole moment provides essential insight into how atomic structures interact with magnetic fields, linking atomic motion to macroscopic magnetic properties.

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Most popular questions from this chapter

(a) Write out the ground-state electron configuration \(\left(1 s^{2}, 2 s^{2}, \ldots\right)\) for the carbon atom. (b) What element of next-larger \(Z\) has chemical properties similar to those of carbon? Give the ground-state electron configuration for this element.

(a) If the value of \(L_{z}\) is known, we cannot know either \(L_{x}\) or \(L_{y}\) precisely. But we can know the value of the quantity \(\sqrt{L_{x}^{2}+L_{y}^{2}}\) . Write an expression for this quantity in terms of \(l\) \(m_{l},\) and \(\hbar .\) (b) What is the meaning of \(\sqrt{L_{\mathrm{r}}^{2}+L_{\mathrm{y}}^{2}} ?(\mathrm{c})\) For a state of nonzero orbital angular momentum, find the maximum and minimum values of \(\sqrt{L_{x}^{2}+L_{y}^{2}}\) . Explain your results.

A hydrogen atom in the 5\(g\) state is placed in a magnetic field of 0.600 T that is in the \(z\) -direction. (a) Into how many levels is this state split by the interaction of the atom's orbital magnetic dipole moment with the magnetic field? (b) What is the energy separation between adjacent levels? (c) What is the energy separation between the level of lowest energy and the level of highest energy?

CP Classical Electron Spin. (a) If you treat an electron as a classical spherical object with a radius of \(1.0 \times 10^{-17} \mathrm{m}\) , what angular speed is necessary to produce a spin angular momentum of magnitude \(\sqrt{\frac{3}{4}} \hbar ?\) (b) Use \(v=r \omega\) and the result of part (a) to calculate the speed \(v\) of a point at the electron's equator. What does your result suggest about the validity of this model?

CALC The wave function for a hydrogen atom in the 2\(s\) state is $$\psi_{2 s}(r)=\frac{1}{\sqrt{32 \pi a^{3}}}\left(2-\frac{r}{a}\right) e^{-r / 2 a}$$ (a) Verify that this function is normalized. (b) In the Bohr model, the distance between the electron and the nucleus in the \(n=2\) state is exactly 4\(a\) . Calculate the probability that an electron in the 2\(s\) state will be found at a distance less than 4\(a\) from the nucleus.

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