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The 21 cm Line: One of the most important windows to the mysteries of the cosmos is the 21 cm line. With it, astronomers map hydrogen throughout the universe. An important trait is that it involves a highly forbidden transition that is, accordingly, quite long-lived. But it is also an excellent example of the coupling of angular momenta. Hydrogen's ground state has no spin-orbit interaction - for \(\boldsymbol{\theta}=0,\) there is no orbit. However, the proton and electron magnetic moments do interact. Consider the following simple model. (a) The proton sees itself surrounded by a spherically symmetric cloud of Is electron, which has an intrinsic magnetic dipole moment/spin that, of course, has a direction. For the purpose of investigating its effect on the proton, treat this dispersed magnetic moment as behaving effectively like a single loop of current whose radius is \(a_{0}\), then find the magnetic tield at the middle of the loop in tenns of \(e, h, m_{e}, \mu_{0},\) and \(a_{0} .\) (b) The proton sits right in the middle of the electron's magnetic moment. Like the electron. the proton is a spin- \(\frac{1}{2}\) particle, with only two possible orientations in a magnetic field. Noting, however, that its spin and magnetic moment are parallel rather than opposite, would the interaction energy be lower with the proton's spin aligned or antialigned with that of the electron'? (c) For the proton, \(g_{p}\) is \(5.6 .\) Obtain a rough value for the energy difference between the two orientations. (d) What would be the wavelength of a photon that carries away this energy difference?

Short Answer

Expert verified
A step-by-step calculation gives the magnetic field at the center of the 'loop', the orientation for lower energy, the energy difference between the two orientations, and the wavelength of the photon that carries away this energy difference.

Step by step solution

01

Calculate the Magnetic Field

Using the fact that the magnetic field \( B \) at the center of a current loop is given by the formula \( B = \mu_0 I / 2R \), where \( \mu_0 \) is the permeability of free space, \( I \) is the current and \( R \) is the radius of the loop. The current here can be found as \( I = e / T \), where \( e \) is the charge of the electron and \( T \) is the period of revolution. The period can be expressed in terms of the mass \( m_e \) and radius \( a_0 \) as \( T = 2 \pi a_0 /v \), and the speed \( v \) can be found from the centripetal force principle as \( v = \sqrt{e^2 / 4 \pi \epsilon_0 m_e a_0} \). Substituting these in the initial equation we find the magnetic field.
02

Determine the Lower Energy Orientation

The energy of interaction can be modeled as \( U = - \vec{\mu} \cdot \vec{B} \), where \( \vec{\mu} \) is the magnetic moment of the proton and \( \vec{B} \) is the magnetic field of the electron 'loop'. Noting that the spin and thus the magnetic moment of the proton are parallel, we can conclude that the energy will be lower when the proton's spin is aligned with that of the electron, as it yields a negative interaction energy.
03

Compute the Energy Difference

The magnitude of the magnetic moment of the proton is \( \mu_p = g_p \mu_N \), where \( \mu_N \) is the nuclear magneton and \( g_p = 5.6 \) is given. The energy difference between the two orientations obtained in previous step can be calculated with the formula \( \Delta U = 2 |U| = 2 |\mu_p B| \). Substituting for \( \mu_p \) and \( B \) we find the energy difference.
04

Find the Photon Wavelength

The energy of a photon is given by \( E = h c / \lambda \), where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the photon. If a photon carries away the energy difference obtained in the previous step, equating these gives us \( \lambda = h c / \Delta U \), which can be computed to find the wavelength.

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Key Concepts

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

Hydrogen Mapping
The 21 cm line is crucial in the cosmic search for hydrogen. Often dubbed the 'fingerprint of hydrogen,' it allows astronomers to detect and map the distribution of hydrogen gas in galaxies and intergalactic space. - **Why 21 cm?**: The 21 cm line arises from the hyperfine transition in neutral hydrogen, where the spin alignment of the electron shifts relative to the proton. - **Astronomical importance**: Since hydrogen is abundant throughout the universe, this spectral line provides valuable insights into the structure and movement of galaxies. It helps create detailed maps of hydrogen clouds that are otherwise invisible in optical wavelengths.
- **Applications in Cosmology**: By analyzing this line, scientists can trace the large-scale structure of the universe and gain information about cosmic evolution and even the rate of expansion of the universe. This non-visual map of hydrogen can penetrate dust clouds that obscure visible light, offering a clearer view of the universe's architecture.
Angular Momentum Coupling
Angular momentum coupling occurs in atoms when the angular momenta of different particles interact. In our exercise with the hydrogen atom, this concept is crucial in understanding the 21 cm line emission. - **Types of Coupling**: - **Spin-Spin Coupling**: It refers to the interaction between the spins of two particles, such as the electron and the proton in a hydrogen atom. - **Orbital and Spin-Orbit Coupling**: While the ground state of hydrogen has no spin-orbit interaction due to no orbit, coupling can occur in other states.
- **Energy States**: When angular momenta couple, they split the energy states into closely spaced levels. These levels determine the energy difference that emits or absorbs radiation, like in the 21 cm transition.
- **Outcome**: Understanding these interactions helps predict the allowed energy transitions and explains why certain transitions, like the 21 cm line, are relatively rare and long-lived in comparison to other transitions.
Spin-Orbit Interaction
Spin-orbit interaction is a fundamental concept in quantum mechanics, occurring when an electron's spin interacts with its orbital movement around the nucleus. Despite its absence in the hydrogen atom's ground state, understanding this interaction can enrich students' grasp of atomic behaviors. - **Physics Behind the Interaction**: This interaction arises because an electron moving in an atomic orbit sees an effective magnetic field due to its motion in the electric field of the nucleus.
- **Effect on Atomic Spectra**: It causes splitting of spectral lines and affects the energy levels of atoms. In more complex atoms, this creates fine structure in the atomic spectra.
- **Significance in Chemistry and Physics**: It explains phenomena like the splitting of spectral lines in magnetic fields and plays a role in determining atomic and molecular electronic structures. However, for the hydrogen ground state, this interaction is generally minimal or absent due to the spherical symmetry and lack of orbital motion.
Magnetic Field Calculation
Calculating magnetic fields in atomic physics involves applying the principles of electromagnetism to quantum systems. In the case of the 21 cm line, this calculation provides insight into hydrogen's unique behaviors.- **Basic Formula**: The magnetic field at the center of a current loop is given by the formula \[ B = \mu_0 \frac{I}{2R} \] where \( \mu_0 \) is the permeability of free space, \( I \) is the current, and \( R \) is the loop radius.
- **Application to the Electron-Proton System**: Seeing the electron's spin as a current loop, you can calculate the effective magnetic field impacting the proton.
- **Energy Implications**: This field dictates how the proton's magnetic moment aligns, affecting the interaction energy. Lower energy settings occur when the proton’s spin aligns with the electron, influencing overall atomic behavior and observable spectral lines.
- **Significance**: Understanding these calculations aids in comprehending how atomic structures result in macroscopically observable properties, such as the hyperfine splitting seen in the 21 cm line.

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Most popular questions from this chapter

In the Stern-Gerlach experiment, how much would a hydrogen atom emanating from a \(500 \mathrm{~K}\) oven \(\left(K E=\frac{3}{2} k_{B} T\right)\) be deflected in traveling 1 m through a magnetic field whose rate of change is \(10 \mathrm{~T} / \mathrm{m} ?\)

The well-known sodium doublet is two yellow spectral lines of very close wavelength. \(589.0 \mathrm{nm}\) and \(589.6 \mathrm{nm} .\) lt is caused by splitting of the \(3 p\) energy level. due to the spin-orbit interaction. In its ground state, sodium's single valence electron is in the \(3 s\) level. It may be excited to the next higher level, the \(3 p\), then emit a photon as it drops back to the \(3 s\). However. the \(3 \rho\) is actually two levels. in which \(L\) and \(S\) are aligned and antialigned. IIn the notation of Section 8.7 these are. respectively. the \(3 p_{3 / 2}\) and the \(3 p_{1 n}\) ) Because the transitions stan from slightly different initial energies yet have identical final energies(the \(3 s\) having no orbital angular momentum to lead to spin- orbit interaction), there are two differenl wavelengths possible for the emitted photon. Calculate the difference in energy between the two photons. From this, obtain a rough value of the average strength of the internal magnetic field experienced by sodium's valence electron.

A dipole withour angular momenturn can simply rotate to align with the field (though i would oscillate uniess it could shed energy). One with angular momentum cannot. Why?

Using a beam of electrons accelerated in an X-ray tube, we wish to knock an electron out of the \(K\) shell of a given element in a target. Section 7.8 gives the energies in a hydrogenlike atom as \(Z^{2}\left(-13.6 \mathrm{eV} / \mathrm{n}^{2}\right)\). Assume that for \(f\) airly high \(Z\), a \(K\) -shell electron can be treated as orbiting the nucleus alone, (a) A typical accelerating potential in an X-ray tube is \(50 \mathrm{kV}\). In roughly how high a \(Z\) could a hole in the \(K\) -shell be produced? (b) Could a hole be produced in elements of higher \(Z\) ?

Slater Determinant: A convenient and compact way of expressing multiparticle states of antisymmetric character for many fermions is the Slater determinant. lt is based on the fact that for \(N\) fermions there must be \(N\) different individual-particle states, or sets of quantum numbers. The ith state has sparial quantum numbers (which might be \(n_{i}, \ell_{i},\) and \(m_{c i}\) ) represented simply by \(n_{t}\) and spin quanturn number \(m_{s i^{i}}\). Were it occupied by the ith particle, the state would be \(\psi_{n}\left(x_{j}\right) m_{s i}\). A column corresponds to a given state and a row to a given particle. For instance, the first column corresponds to individualparticle state \(\psi_{n}(x,) m_{3},\) where \(j\) progresses (through the rows) from particle 1 to particle \(N\). The first row corresponds to particle I. which successively occupies all individual-particle states (progressing through the columns). (a) What property of determinants ensures that the multiparticle state is 0 if any two individualparticle states are identical? (b) What property of deterninants ensures that switching the labels on any two particles switches the sign of the multiparticle state?

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