What is the cause for the fine structure?What does the $y$-axis represent in the atomic spectra and what is its significance?Orbital magnetic moment versus Biot-Savart lawFine Structure CorrectionFWHM of different spectra and separation in fine structureClarification: non-relativistic fine structure of a one-electron atomClassification of the fine and hyperfine structure?Why is spin-orbit coupling neglected in Helium atom?Magnetic moment of hydrogen atom in weak magnetic field from hyper fine structurePredicting spectral lineswhat is the cause of the first atom moving in moving?
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What is the cause for the fine structure?
What does the $y$-axis represent in the atomic spectra and what is its significance?Orbital magnetic moment versus Biot-Savart lawFine Structure CorrectionFWHM of different spectra and separation in fine structureClarification: non-relativistic fine structure of a one-electron atomClassification of the fine and hyperfine structure?Why is spin-orbit coupling neglected in Helium atom?Magnetic moment of hydrogen atom in weak magnetic field from hyper fine structurePredicting spectral lineswhat is the cause of the first atom moving in moving?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
What is the cause for the fine structure? I know that it describes the splitting of the spectral lines of atoms but I don't understand its cause,
any help is appreciated
quantum-mechanics atomic-physics spectroscopy
edited Jul 3 at 10:24
Qmechanic♦
111k12 gold badges213 silver badges1306 bronze badges
asked Jul 3 at 8:22
Alessio PopovicAlessio Popovic
467 bronze badges
$endgroup$
|
show 1 more comment
$begingroup$
What is the cause for the fine structure? I know that it describes the splitting of the spectral lines of atoms but I don't understand its cause,
any help is appreciated
quantum-mechanics atomic-physics spectroscopy
edited Jul 3 at 10:24
Qmechanic♦
111k12 gold badges213 silver badges1306 bronze badges
asked Jul 3 at 8:22
Alessio PopovicAlessio Popovic
467 bronze badges
$endgroup$
$begingroup$
In quantum mechanics, the spectral lines are usually derived using the Schrodinger equation which is non-relativistic. The fine structure is obtained when taking into account relativistic corrections.
$endgroup$
– Christophe
Jul 3 at 8:43
1
$begingroup$
@Christophe Strange comment because the atomic structure with their spins and magnetic dipoles are responsible for the fine splitting of energy levels of spectra. The Schrödinger equation is among others the result of the observation of fine structures.
$endgroup$
– HolgerFiedler
Jul 3 at 9:49
$begingroup$
have a look at the simple hydrogen hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydfin.html
$endgroup$
– anna v
Jul 3 at 10:35
1
$begingroup$
@HolgerFiedler The (relativistic) Dirac equation is expected to give the exact spectral lines. The lowest-order expansion in $v/c$ gives the Schroedinger equation and then the usual spectral lines. The next order gives three additional terms in the Schroedinger equation: the first correction to the kinetic energy, the spin-orbit coupling (coupling of the spin with the orbital kinetic momentum) and the so-called Darwin term. The contribution of these three terms to the spectral lines are known as the fine structure.
$endgroup$
– Christophe
Jul 3 at 13:55
$begingroup$
@Christophe No offense. And your last comment could be a good answer. I only came across the fact that you called the formula the cause of the fine structure spectra.
$endgroup$
– HolgerFiedler
Jul 3 at 14:52
|
show 1 more comment
$begingroup$
What is the cause for the fine structure? I know that it describes the splitting of the spectral lines of atoms but I don't understand its cause,
any help is appreciated
quantum-mechanics atomic-physics spectroscopy
edited Jul 3 at 10:24
Qmechanic♦
111k12 gold badges213 silver badges1306 bronze badges
asked Jul 3 at 8:22
Alessio PopovicAlessio Popovic
467 bronze badges
$endgroup$
What is the cause for the fine structure? I know that it describes the splitting of the spectral lines of atoms but I don't understand its cause,
any help is appreciated
quantum-mechanics atomic-physics spectroscopy
quantum-mechanics atomic-physics spectroscopy
edited Jul 3 at 10:24
Qmechanic♦
111k12 gold badges213 silver badges1306 bronze badges
asked Jul 3 at 8:22
Alessio PopovicAlessio Popovic
467 bronze badges
edited Jul 3 at 10:24
Qmechanic♦
111k12 gold badges213 silver badges1306 bronze badges
asked Jul 3 at 8:22
Alessio PopovicAlessio Popovic
467 bronze badges
edited Jul 3 at 10:24
Qmechanic♦
111k12 gold badges213 silver badges1306 bronze badges
edited Jul 3 at 10:24
Qmechanic♦
111k12 gold badges213 silver badges1306 bronze badges
edited Jul 3 at 10:24
Qmechanic♦
111k12 gold badges213 silver badges1306 bronze badges
111k12 gold badges213 silver badges1306 bronze badges
asked Jul 3 at 8:22
Alessio PopovicAlessio Popovic
467 bronze badges
asked Jul 3 at 8:22
Alessio PopovicAlessio Popovic
467 bronze badges
asked Jul 3 at 8:22
Alessio PopovicAlessio Popovic
467 bronze badges
467 bronze badges
$begingroup$
In quantum mechanics, the spectral lines are usually derived using the Schrodinger equation which is non-relativistic. The fine structure is obtained when taking into account relativistic corrections.
$endgroup$
– Christophe
Jul 3 at 8:43
1
$begingroup$
@Christophe Strange comment because the atomic structure with their spins and magnetic dipoles are responsible for the fine splitting of energy levels of spectra. The Schrödinger equation is among others the result of the observation of fine structures.
$endgroup$
– HolgerFiedler
Jul 3 at 9:49
$begingroup$
have a look at the simple hydrogen hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydfin.html
$endgroup$
– anna v
Jul 3 at 10:35
1
$begingroup$
@HolgerFiedler The (relativistic) Dirac equation is expected to give the exact spectral lines. The lowest-order expansion in $v/c$ gives the Schroedinger equation and then the usual spectral lines. The next order gives three additional terms in the Schroedinger equation: the first correction to the kinetic energy, the spin-orbit coupling (coupling of the spin with the orbital kinetic momentum) and the so-called Darwin term. The contribution of these three terms to the spectral lines are known as the fine structure.
$endgroup$
– Christophe
Jul 3 at 13:55
$begingroup$
@Christophe No offense. And your last comment could be a good answer. I only came across the fact that you called the formula the cause of the fine structure spectra.
$endgroup$
– HolgerFiedler
Jul 3 at 14:52
|
show 1 more comment
$begingroup$
In quantum mechanics, the spectral lines are usually derived using the Schrodinger equation which is non-relativistic. The fine structure is obtained when taking into account relativistic corrections.
$endgroup$
– Christophe
Jul 3 at 8:43
1
$begingroup$
@Christophe Strange comment because the atomic structure with their spins and magnetic dipoles are responsible for the fine splitting of energy levels of spectra. The Schrödinger equation is among others the result of the observation of fine structures.
$endgroup$
– HolgerFiedler
Jul 3 at 9:49
$begingroup$
have a look at the simple hydrogen hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydfin.html
$endgroup$
– anna v
Jul 3 at 10:35
1
$begingroup$
@HolgerFiedler The (relativistic) Dirac equation is expected to give the exact spectral lines. The lowest-order expansion in $v/c$ gives the Schroedinger equation and then the usual spectral lines. The next order gives three additional terms in the Schroedinger equation: the first correction to the kinetic energy, the spin-orbit coupling (coupling of the spin with the orbital kinetic momentum) and the so-called Darwin term. The contribution of these three terms to the spectral lines are known as the fine structure.
$endgroup$
– Christophe
Jul 3 at 13:55
$begingroup$
@Christophe No offense. And your last comment could be a good answer. I only came across the fact that you called the formula the cause of the fine structure spectra.
$endgroup$
– HolgerFiedler
Jul 3 at 14:52
$begingroup$
In quantum mechanics, the spectral lines are usually derived using the Schrodinger equation which is non-relativistic. The fine structure is obtained when taking into account relativistic corrections.
$endgroup$
– Christophe
Jul 3 at 8:43
$begingroup$
In quantum mechanics, the spectral lines are usually derived using the Schrodinger equation which is non-relativistic. The fine structure is obtained when taking into account relativistic corrections.
$endgroup$
– Christophe
Jul 3 at 8:43
1
1
$begingroup$
@Christophe Strange comment because the atomic structure with their spins and magnetic dipoles are responsible for the fine splitting of energy levels of spectra. The Schrödinger equation is among others the result of the observation of fine structures.
$endgroup$
– HolgerFiedler
Jul 3 at 9:49
$begingroup$
@Christophe Strange comment because the atomic structure with their spins and magnetic dipoles are responsible for the fine splitting of energy levels of spectra. The Schrödinger equation is among others the result of the observation of fine structures.
$endgroup$
– HolgerFiedler
Jul 3 at 9:49
$begingroup$
have a look at the simple hydrogen hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydfin.html
$endgroup$
– anna v
Jul 3 at 10:35
$begingroup$
have a look at the simple hydrogen hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydfin.html
$endgroup$
– anna v
Jul 3 at 10:35
1
1
$begingroup$
@HolgerFiedler The (relativistic) Dirac equation is expected to give the exact spectral lines. The lowest-order expansion in $v/c$ gives the Schroedinger equation and then the usual spectral lines. The next order gives three additional terms in the Schroedinger equation: the first correction to the kinetic energy, the spin-orbit coupling (coupling of the spin with the orbital kinetic momentum) and the so-called Darwin term. The contribution of these three terms to the spectral lines are known as the fine structure.
$endgroup$
– Christophe
Jul 3 at 13:55
$begingroup$
@HolgerFiedler The (relativistic) Dirac equation is expected to give the exact spectral lines. The lowest-order expansion in $v/c$ gives the Schroedinger equation and then the usual spectral lines. The next order gives three additional terms in the Schroedinger equation: the first correction to the kinetic energy, the spin-orbit coupling (coupling of the spin with the orbital kinetic momentum) and the so-called Darwin term. The contribution of these three terms to the spectral lines are known as the fine structure.
$endgroup$
– Christophe
Jul 3 at 13:55
$begingroup$
@Christophe No offense. And your last comment could be a good answer. I only came across the fact that you called the formula the cause of the fine structure spectra.
$endgroup$
– HolgerFiedler
Jul 3 at 14:52
$begingroup$
@Christophe No offense. And your last comment could be a good answer. I only came across the fact that you called the formula the cause of the fine structure spectra.
$endgroup$
– HolgerFiedler
Jul 3 at 14:52
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
The splitting of spectral lines into close pairs is due to the electron's angular momentum and the fact that an electron can have either of two spin states.
answered Jul 3 at 10:12
Michael WalsbyMichael Walsby
1,0151 silver badge6 bronze badges
$endgroup$
$begingroup$
And the coupling between them. Spin-orbit coupling can be seen as the magnetic field due to the current of the nuclear charge around the electron magnetic dipole moment.
$endgroup$
– Pieter
Jul 3 at 10:27
add a comment |
$begingroup$
When we observed the emission spectrum of hydrogen, we were able to see a fine splitting in the spectrum. This was something where Bohr's model of atom failed to explain. This was arising due to the interaction of the intrinsic spin of the electron with the magnetic field created by the orbital motion of the electron around the atom. The angular momentum of electrons the units of reduced plank's constant is 1/2. this has only 2 possible Stern-Gerlach experiment showed (space quantization) direction either parallel or antiparallel to the magnetic field of the atom. Which increases or decreases the energy of the electron. Hope this solved your doubt.
answered Jul 3 at 11:45
Sharun ShajiSharun Shaji
387 bronze badges
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
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$begingroup$
The splitting of spectral lines into close pairs is due to the electron's angular momentum and the fact that an electron can have either of two spin states.
answered Jul 3 at 10:12
Michael WalsbyMichael Walsby
1,0151 silver badge6 bronze badges
$endgroup$
$begingroup$
And the coupling between them. Spin-orbit coupling can be seen as the magnetic field due to the current of the nuclear charge around the electron magnetic dipole moment.
$endgroup$
– Pieter
Jul 3 at 10:27
add a comment |
$begingroup$
The splitting of spectral lines into close pairs is due to the electron's angular momentum and the fact that an electron can have either of two spin states.
answered Jul 3 at 10:12
Michael WalsbyMichael Walsby
1,0151 silver badge6 bronze badges
$endgroup$
$begingroup$
And the coupling between them. Spin-orbit coupling can be seen as the magnetic field due to the current of the nuclear charge around the electron magnetic dipole moment.
$endgroup$
– Pieter
Jul 3 at 10:27
add a comment |
$begingroup$
The splitting of spectral lines into close pairs is due to the electron's angular momentum and the fact that an electron can have either of two spin states.
answered Jul 3 at 10:12
Michael WalsbyMichael Walsby
1,0151 silver badge6 bronze badges
$endgroup$
The splitting of spectral lines into close pairs is due to the electron's angular momentum and the fact that an electron can have either of two spin states.
answered Jul 3 at 10:12
Michael WalsbyMichael Walsby
1,0151 silver badge6 bronze badges
answered Jul 3 at 10:12
Michael WalsbyMichael Walsby
1,0151 silver badge6 bronze badges
answered Jul 3 at 10:12
Michael WalsbyMichael Walsby
1,0151 silver badge6 bronze badges
answered Jul 3 at 10:12
Michael WalsbyMichael Walsby
1,0151 silver badge6 bronze badges
1,0151 silver badge6 bronze badges
$begingroup$
And the coupling between them. Spin-orbit coupling can be seen as the magnetic field due to the current of the nuclear charge around the electron magnetic dipole moment.
$endgroup$
– Pieter
Jul 3 at 10:27
add a comment |
$begingroup$
And the coupling between them. Spin-orbit coupling can be seen as the magnetic field due to the current of the nuclear charge around the electron magnetic dipole moment.
$endgroup$
– Pieter
Jul 3 at 10:27
$begingroup$
And the coupling between them. Spin-orbit coupling can be seen as the magnetic field due to the current of the nuclear charge around the electron magnetic dipole moment.
$endgroup$
– Pieter
Jul 3 at 10:27
$begingroup$
And the coupling between them. Spin-orbit coupling can be seen as the magnetic field due to the current of the nuclear charge around the electron magnetic dipole moment.
$endgroup$
– Pieter
Jul 3 at 10:27
add a comment |
$begingroup$
When we observed the emission spectrum of hydrogen, we were able to see a fine splitting in the spectrum. This was something where Bohr's model of atom failed to explain. This was arising due to the interaction of the intrinsic spin of the electron with the magnetic field created by the orbital motion of the electron around the atom. The angular momentum of electrons the units of reduced plank's constant is 1/2. this has only 2 possible Stern-Gerlach experiment showed (space quantization) direction either parallel or antiparallel to the magnetic field of the atom. Which increases or decreases the energy of the electron. Hope this solved your doubt.
answered Jul 3 at 11:45
Sharun ShajiSharun Shaji
387 bronze badges
$endgroup$
add a comment |
$begingroup$
When we observed the emission spectrum of hydrogen, we were able to see a fine splitting in the spectrum. This was something where Bohr's model of atom failed to explain. This was arising due to the interaction of the intrinsic spin of the electron with the magnetic field created by the orbital motion of the electron around the atom. The angular momentum of electrons the units of reduced plank's constant is 1/2. this has only 2 possible Stern-Gerlach experiment showed (space quantization) direction either parallel or antiparallel to the magnetic field of the atom. Which increases or decreases the energy of the electron. Hope this solved your doubt.
answered Jul 3 at 11:45
Sharun ShajiSharun Shaji
387 bronze badges
$endgroup$
add a comment |
$begingroup$
When we observed the emission spectrum of hydrogen, we were able to see a fine splitting in the spectrum. This was something where Bohr's model of atom failed to explain. This was arising due to the interaction of the intrinsic spin of the electron with the magnetic field created by the orbital motion of the electron around the atom. The angular momentum of electrons the units of reduced plank's constant is 1/2. this has only 2 possible Stern-Gerlach experiment showed (space quantization) direction either parallel or antiparallel to the magnetic field of the atom. Which increases or decreases the energy of the electron. Hope this solved your doubt.
answered Jul 3 at 11:45
Sharun ShajiSharun Shaji
387 bronze badges
$endgroup$
When we observed the emission spectrum of hydrogen, we were able to see a fine splitting in the spectrum. This was something where Bohr's model of atom failed to explain. This was arising due to the interaction of the intrinsic spin of the electron with the magnetic field created by the orbital motion of the electron around the atom. The angular momentum of electrons the units of reduced plank's constant is 1/2. this has only 2 possible Stern-Gerlach experiment showed (space quantization) direction either parallel or antiparallel to the magnetic field of the atom. Which increases or decreases the energy of the electron. Hope this solved your doubt.
answered Jul 3 at 11:45
Sharun ShajiSharun Shaji
387 bronze badges
answered Jul 3 at 11:45
Sharun ShajiSharun Shaji
387 bronze badges
answered Jul 3 at 11:45
Sharun ShajiSharun Shaji
387 bronze badges
answered Jul 3 at 11:45
Sharun ShajiSharun Shaji
387 bronze badges
387 bronze badges
add a comment |
add a comment |
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In quantum mechanics, the spectral lines are usually derived using the Schrodinger equation which is non-relativistic. The fine structure is obtained when taking into account relativistic corrections.
$endgroup$
– Christophe
Jul 3 at 8:43
1
$begingroup$
@Christophe Strange comment because the atomic structure with their spins and magnetic dipoles are responsible for the fine splitting of energy levels of spectra. The Schrödinger equation is among others the result of the observation of fine structures.
$endgroup$
– HolgerFiedler
Jul 3 at 9:49
$begingroup$
have a look at the simple hydrogen hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydfin.html
$endgroup$
– anna v
Jul 3 at 10:35
1
$begingroup$
@HolgerFiedler The (relativistic) Dirac equation is expected to give the exact spectral lines. The lowest-order expansion in $v/c$ gives the Schroedinger equation and then the usual spectral lines. The next order gives three additional terms in the Schroedinger equation: the first correction to the kinetic energy, the spin-orbit coupling (coupling of the spin with the orbital kinetic momentum) and the so-called Darwin term. The contribution of these three terms to the spectral lines are known as the fine structure.
$endgroup$
– Christophe
Jul 3 at 13:55
$begingroup$
@Christophe No offense. And your last comment could be a good answer. I only came across the fact that you called the formula the cause of the fine structure spectra.
$endgroup$
– HolgerFiedler
Jul 3 at 14:52