The hyperfine interaction in a hydrogen atom between the...

The hyperfine interaction in a hydrogen atom between the magnetic dipole moment of the proton and the spin magnetic dipole moment of the electron splits the ground level into two levels separated by $5.9 \times 10^{-6} eV$. (a) Calculate the wavelength and frequency of the photon emitted when the atom makes a transition between these states, and compare your answer to the value given at the end of Section 41.5. In what part of the electromagnetic spectrum does this lie? Such photons are emitted by cold hydrogen clouds in interstellar space; by detecting these photons, astronomers can learn about the number and density of such clouds. (b) Calculate the effective magnetic field experienced by the electron in these states (see Fig. 41.18). Compare your result to the effective magnetic field due to the spin-orbit coupling calculated in Example 41.7.

Key Concepts

Spin-Orbit Coupling

Spin-orbit coupling is the interaction between an electron’s spin and its orbital motion around the nucleus, leading to further splitting of atomic energy levels. Comparing the effective magnetic fields due to hyperfine interactions and spin-orbit coupling helps in understanding the relative significance of these effects in atomic structure.

Effective Magnetic Field

In the context of atomic interactions, the effective magnetic field is a conceptual tool that describes the net magnetic field experienced by an electron due to nearby sources, including other magnetic moments of the atom. It provides insight into the interactions that split the energy levels and affects the electron’s behavior.

Magnetic Dipole Moment

The magnetic dipole moment, associated with both the nucleus and the electron, is a measure of the strength and orientation of their magnetism. Its interaction in an atom is crucial for understanding both hyperfine splitting and other magnetic interactions such as spin-orbit coupling.

Energy Transition and Photon Emission

Energy transitions between closely spaced hyperfine levels result in the emission or absorption of photons. The energy difference between the initial and final states determines the energy of the photon, which can be directly related to its frequency and wavelength via Planck’s relation.

Photon Wavelength and Frequency Calculation

Calculating the wavelength and frequency of an emitted photon involves using the energy difference together with fundamental constants (Planck’s constant and the speed of light). This process, typically done by the equations ? = hc/E and f = E/h, links the microscopic energy changes in atoms to measurable electromagnetic radiation properties.

Hyperfine Interaction

This is the interaction between the magnetic moments of the nucleus and the electron, which leads to small energy splitting of atomic levels. It arises due to coupling between the nuclear magnetic dipole moment and the magnetic field created by the electron’s spin, and is key to understanding fine details of atomic spectra in systems such as hydrogen.

Electromagnetic Spectrum Region

The region of the electromagnetic spectrum in which a photon is emitted is determined by its wavelength or frequency. In cases of hyperfine transitions, the resulting photons typically lie in the microwave region. This identification is important in applications like astronomical observations where specific wavelength ranges help in studying interstellar phenomena.

Transcript

0:00 Aloha.

00:01 So in this problem, we are considering the hyperfine interaction in the hydrogen atom.

00:06 So that's the magnetic dipole interaction between the magnetic dipole of the proton and the spin magnetic dipole of the electron.

00:15 And the electron is split into two different energy states.

00:19 And something that we're told is the energy difference between these states shall call delta e, which is 5 .9 times 10 to the minus 6 electron volt.

00:30 That we're told and we're in the ground state to energy levels.

00:41 And we want to find the wavelength of the light, which i'll call lambda, of the photon that's emitted when the electron goes between these two states.

00:52 So when it's going from the higher state to the lower state, it'll emit a photon.

00:57 And we can use the equation for the energy of a photon, which is that e is h.

01:05 C over lambda where that plonks constant and c's speed of light.

01:11 So here the, we want to just flip this around to an equation for the wavelength of the light.

01:19 And the energy here is the delta e for this energy transition.

01:24 That's the energy of the photon.

01:26 That photon is basically carrying away the energy.

01:29 The electron makes this transition.

01:32 So then we can just plug in these numbers.

01:34 So we'll have lambda is 6.

01:41 6 .626.

01:42 You can just look up what the value of planck's constant, the textbook somewhere, and the speed of light.

01:52 You just want to make sure that you're using consistent units...

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