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Chapter 36: Problem 9

Why is there no spin-orbit splitting in hydrogen's ground state?

Short Answer

Expert verified
There is no spin-orbit splitting for Hydrogen's ground state because spin-orbit interaction only occurs significantly in atoms/ions with multiple electrons. Hydrogen only has one. Furthermore, Hydrogen's ground state (1s orbital) has zero orbital angular momentum, a key requirement for the existence of spin-orbit interaction.

Step by step solution

01

Understanding Spin-Orbit Interaction

The spin鈥搊rbit interaction (SOI) is a magnetic interaction of a particle's spin with its motion. Specifically, it describes the energy changes due to magnetic interaction between the angular momentum and the spin of the electron. However, this interaction is only significant in atoms/ions with multiple electrons.
02

Knowing Hydrogen's Ground State

The ground state of a hydrogen atom is where the atom's electron is in the lowest possible energy level. This state is known as a 1s orbital. The 's' in 1s means that state has zero orbital angular momentum.
03

Combining the Facts

Given the above observations, in a Hydrogen atom's ground state (1s), there is no spin orbit coupling because the orbital has zero angular momentum. Furthermore, Hydrogen is a one-electron ion, and the central field in such ions is spherically symmetric. Thus, breaking the field's spherical symmetry by the motion of an electron around the nucleus, the basis of spin-orbit interaction, does not occur. Hence, there is no spin-orbit splitting in Hydrogen's ground state.

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Key Concepts

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

Understanding the Hydrogen Ground State
The hydrogen atom's ground state is a fundamental concept in quantum mechanics. In this state, the electron occupies the lowest energy level possible. The orbital in which this condition occurs is called the 1s orbital. In quantum numbers, "1" represents the principal quantum number, indicating the energy level, and "s" refers to the orbital shape, which in this case is spherical.
This spherical shape corresponds to an s-orbital with zero angular momentum.
  • The electron in the ground state is closest to the nucleus.
  • It possesses the minimum energy that it can have without being removed from the atom.
  • This is the most stable state of a hydrogen atom.
Understanding these characteristics is critical because they underlie why certain interactions, like spin-orbit splitting, do not occur in this state.
Basics of Angular Momentum in Quantum Mechanics
Angular momentum in quantum mechanics is a measure of an object's rotation, relevant for both macroscopic objects and particles like electrons. For electrons, angular momentum plays a significant role in the atomic structure and spectral lines of atoms. Angular momentum in atomic systems is quantized, meaning electrons can only possess certain discrete values determined by quantum numbers.
  • In orbital angular momentum, each type of orbital (s, p, d, f) has a characteristic angular momentum.
  • The 's' orbitals have an angular momentum value of zero, hence why electrons in them do not contribute to the orbital motion about the nucleus.
Since the hydrogen ground state is 1s, it has no orbital angular momentum. This is crucial for understanding why spin-orbit interaction is absent in this state.
Spherical Symmetry in Atomic Systems
Spherical symmetry is an essential concept in atomic physics and refers to the uniformity of a field or function when all points on a sphere are considered. In a spherically symmetric field, every point is equidistant from a central point, such as how electrons in an s-orbital spread around the nucleus.
Spherical symmetry simplifies the hydrogen atom's modeling because the field experienced by the electron is independent of direction.
  • In simple terms, a sphere looks the same from any angle.
  • S-orbitals display spherical symmetry as they have no directional preference.
  • This symmetry in hydrogen鈥檚 ground state remains unbroken by electron motion.
Since spin-orbit interaction relies on changes in symmetry, the perfect spherical symmetry of hydrogen's ground state explains the lack of spin-orbit splitting.

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Most popular questions from this chapter

How many quantum numbers are required to specify fully the state of a hydrogen atom?

With sufficient energy, it's possible to eject an electron from an inner atomic orbital. A higher-energy electron will then drop into the unoccupied state, emitting a photon with energy equal to the difference between the two levels. For inner-shell electrons, photon energies are in the keV range, putting them in the X-ray region of the spectrum. These characteristic X rays are labeled with the letter indicating the shell to which the electron drops, followed by a Greek letter indicating the higher level from which it drops; thus \(K \alpha\) designates a transition from the \(L\) shell to the \(K\) shell. Characteristic X rays provide scientists and physicians with an important diagnostic tool. Environmental scientists bombard pollution samples with high- energy electrons, knocking out inner-shell electrons and thus producing X-ray spectra that help identify contaminants (Fig. \(36.20 a\) ). Geologists do the same with rocks. Medical radiologists reverse the process, exploiting the fact that X rays cause inner-shell transitions as well as complete ejection of inner-shell electrons. In particular, radiologists use the element barium in this way to produce high-contrast X-ray images of the intestinal tract \((\text { Fig. } 36.20 b)\)(GRAPH CANNOT COPY) (a) An \(\mathrm{X}\) -ray spectrum from air pollutants trapped on a filter. The labeled peaks show the presence of lead (Pb) and arsenic (As), as evidenced by \(K \alpha, K \beta, L \alpha,\) and \(L \beta\) characteristic X rays. (b) \(\mathrm{X}\) -ray of an intestinal tract, made by coating the intestinal wall with X-ray-opaque barium Molybdenum's X-ray spectrum has its \(K \alpha\) peak at 17.4 keV. The corresponding X-ray wavelength is closest to a. \(1 \mathrm{pm}\) b. \(100 \mathrm{pm}\) c. \(1 \mathrm{nm}\) d. \(100 \mathrm{nm}\)

What are the possible \(j\) values for a hydrogen atom in the \(3 D\) state?

A selection rule for the infinite square well allows only those transitions in which \(n\) changes by an odd number. Suppose an infinite square well of width \(0.200 \mathrm{nm}\) contains an electron in the \(n=4\) state. (a) Draw an energy-level diagram showing all allowed transitions that could occur as this electron drops toward the ground state, including transitions from lower levels that could be reached from \(n=4 .\) (b) Find all the possible photon energies emitted in these transitions.

Differentiate the radial probability density for the hydrogen ground state, and set the result to zero to show that the electron is most likely to be found at one Bohr radius.
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