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Cause of continuous spectral lines [on hold]


Atomic propertiesPhotoelectric effect and electron lossAlternative to a protonWhat is generally meant by distinct line spectrum?Table of expanded electronic structures of atomsDetermination of initial excited state via luminescenceLennard-Jones diagram - what is on the x-axis?How were weights of elements and compounds determined?How isotopes atomic mass was defined?How much heat is created by absorbent-type sunscreens?













3












$begingroup$


We know different types of atoms have different spectral lines. But what exactly are the spectral lines representing? What causes white light to have a continuous spectra while other atoms show discontinuous spectral lines?










share|improve this question







New contributor



NightcoRohak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$



put on hold as too broad by Mithoron, Jon Custer, Poutnik, Buck Thorn, Karsten Theis 13 hours ago


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

















  • $begingroup$
    As a corollary to your question and answers : white light is not necessarily a continuous spectrum, the brain only needs stimuli of the three different receptor cells in a pattern to "see white". So white light is rarely a continuous spectrum, but it is broad enough to stimulate all the receptors.
    $endgroup$
    – Stian Yttervik
    Jun 2 at 17:26















3












$begingroup$


We know different types of atoms have different spectral lines. But what exactly are the spectral lines representing? What causes white light to have a continuous spectra while other atoms show discontinuous spectral lines?










share|improve this question







New contributor



NightcoRohak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$



put on hold as too broad by Mithoron, Jon Custer, Poutnik, Buck Thorn, Karsten Theis 13 hours ago


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

















  • $begingroup$
    As a corollary to your question and answers : white light is not necessarily a continuous spectrum, the brain only needs stimuli of the three different receptor cells in a pattern to "see white". So white light is rarely a continuous spectrum, but it is broad enough to stimulate all the receptors.
    $endgroup$
    – Stian Yttervik
    Jun 2 at 17:26













3












3








3


2



$begingroup$


We know different types of atoms have different spectral lines. But what exactly are the spectral lines representing? What causes white light to have a continuous spectra while other atoms show discontinuous spectral lines?










share|improve this question







New contributor



NightcoRohak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




We know different types of atoms have different spectral lines. But what exactly are the spectral lines representing? What causes white light to have a continuous spectra while other atoms show discontinuous spectral lines?







atoms light






share|improve this question







New contributor



NightcoRohak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










share|improve this question







New contributor



NightcoRohak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share|improve this question




share|improve this question






New contributor



NightcoRohak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








asked Jun 2 at 6:35









NightcoRohakNightcoRohak

565




565




New contributor



NightcoRohak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




New contributor




NightcoRohak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






put on hold as too broad by Mithoron, Jon Custer, Poutnik, Buck Thorn, Karsten Theis 13 hours ago


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.









put on hold as too broad by Mithoron, Jon Custer, Poutnik, Buck Thorn, Karsten Theis 13 hours ago


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    As a corollary to your question and answers : white light is not necessarily a continuous spectrum, the brain only needs stimuli of the three different receptor cells in a pattern to "see white". So white light is rarely a continuous spectrum, but it is broad enough to stimulate all the receptors.
    $endgroup$
    – Stian Yttervik
    Jun 2 at 17:26
















  • $begingroup$
    As a corollary to your question and answers : white light is not necessarily a continuous spectrum, the brain only needs stimuli of the three different receptor cells in a pattern to "see white". So white light is rarely a continuous spectrum, but it is broad enough to stimulate all the receptors.
    $endgroup$
    – Stian Yttervik
    Jun 2 at 17:26















$begingroup$
As a corollary to your question and answers : white light is not necessarily a continuous spectrum, the brain only needs stimuli of the three different receptor cells in a pattern to "see white". So white light is rarely a continuous spectrum, but it is broad enough to stimulate all the receptors.
$endgroup$
– Stian Yttervik
Jun 2 at 17:26




$begingroup$
As a corollary to your question and answers : white light is not necessarily a continuous spectrum, the brain only needs stimuli of the three different receptor cells in a pattern to "see white". So white light is rarely a continuous spectrum, but it is broad enough to stimulate all the receptors.
$endgroup$
– Stian Yttervik
Jun 2 at 17:26










2 Answers
2






active

oldest

votes


















6












$begingroup$

The spectral lines represent transitions between pairs of discrete energy levels in the atom; an electron is excited by absorbing the energy of a photon and a transition from one of these levels to another occurs as the photon is destroyed. Emission lines occur when the reverse process happens.



If enough energy is absorbed the atom can be ionised and an electron with un-quantised energy can be liberated and moves to a position of complete separation where attractive forces between electron and nucleus is insignificant. As the liberated electron's energy is un-quantised the absorption giving rise to ionisation is continuous. The absorption is truly continuous, no matter how high a spectral resolution is used discrete absorption lines will not be observed.



Black body radiation is continuous emission from a hot body. The analysis of this emission lead Plank in 1901 to propose, for the first time, the idea that the photons were quantised, thus avoiding the 'ultra-violet catastrophe'. He did this by allowing ultra-violet photons to have more energy than visible ones which in turn had more energy that infra-red ones. Nowadays we know that he energy is directly proportional to frequency via the Planck constant, $E=hv$.



In the black body, all the numerous energy levels in the body are highly excited and transitions between them continuously occur as the body is being continuously heated. These transition energies overlap one another in energy so that the spectrum is continuous.



(In molecules there are also discrete energy levels, in the ground state these are due to the vibrational and rotational motion of the molecule. A molecule may also have electronically excited states and these can also have discrete vibrational and rotational energy levels. Furthermore there are dissociative excited states that have continuous absorption spectra.)






share|improve this answer











$endgroup$












  • $begingroup$
    just as a side comment (it might be of interest for the OP), the continuum observed in electrons' energy when ionization occurs is the photoelectric effect; all energy excess increases electron's speed.
    $endgroup$
    – santimirandarp
    Jun 2 at 13:47







  • 1




    $begingroup$
    yes; a good point
    $endgroup$
    – porphyrin
    yesterday


















4












$begingroup$

Natural white light had continuous spectrum, as the source - black radiator - has continuous emission in all wavelengths.



Atoms emit or absorb only at the wavelengths, that are corresponding to energy differences between discreet energy levels of electrons in atoms.






share|improve this answer









$endgroup$



















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    The spectral lines represent transitions between pairs of discrete energy levels in the atom; an electron is excited by absorbing the energy of a photon and a transition from one of these levels to another occurs as the photon is destroyed. Emission lines occur when the reverse process happens.



    If enough energy is absorbed the atom can be ionised and an electron with un-quantised energy can be liberated and moves to a position of complete separation where attractive forces between electron and nucleus is insignificant. As the liberated electron's energy is un-quantised the absorption giving rise to ionisation is continuous. The absorption is truly continuous, no matter how high a spectral resolution is used discrete absorption lines will not be observed.



    Black body radiation is continuous emission from a hot body. The analysis of this emission lead Plank in 1901 to propose, for the first time, the idea that the photons were quantised, thus avoiding the 'ultra-violet catastrophe'. He did this by allowing ultra-violet photons to have more energy than visible ones which in turn had more energy that infra-red ones. Nowadays we know that he energy is directly proportional to frequency via the Planck constant, $E=hv$.



    In the black body, all the numerous energy levels in the body are highly excited and transitions between them continuously occur as the body is being continuously heated. These transition energies overlap one another in energy so that the spectrum is continuous.



    (In molecules there are also discrete energy levels, in the ground state these are due to the vibrational and rotational motion of the molecule. A molecule may also have electronically excited states and these can also have discrete vibrational and rotational energy levels. Furthermore there are dissociative excited states that have continuous absorption spectra.)






    share|improve this answer











    $endgroup$












    • $begingroup$
      just as a side comment (it might be of interest for the OP), the continuum observed in electrons' energy when ionization occurs is the photoelectric effect; all energy excess increases electron's speed.
      $endgroup$
      – santimirandarp
      Jun 2 at 13:47







    • 1




      $begingroup$
      yes; a good point
      $endgroup$
      – porphyrin
      yesterday















    6












    $begingroup$

    The spectral lines represent transitions between pairs of discrete energy levels in the atom; an electron is excited by absorbing the energy of a photon and a transition from one of these levels to another occurs as the photon is destroyed. Emission lines occur when the reverse process happens.



    If enough energy is absorbed the atom can be ionised and an electron with un-quantised energy can be liberated and moves to a position of complete separation where attractive forces between electron and nucleus is insignificant. As the liberated electron's energy is un-quantised the absorption giving rise to ionisation is continuous. The absorption is truly continuous, no matter how high a spectral resolution is used discrete absorption lines will not be observed.



    Black body radiation is continuous emission from a hot body. The analysis of this emission lead Plank in 1901 to propose, for the first time, the idea that the photons were quantised, thus avoiding the 'ultra-violet catastrophe'. He did this by allowing ultra-violet photons to have more energy than visible ones which in turn had more energy that infra-red ones. Nowadays we know that he energy is directly proportional to frequency via the Planck constant, $E=hv$.



    In the black body, all the numerous energy levels in the body are highly excited and transitions between them continuously occur as the body is being continuously heated. These transition energies overlap one another in energy so that the spectrum is continuous.



    (In molecules there are also discrete energy levels, in the ground state these are due to the vibrational and rotational motion of the molecule. A molecule may also have electronically excited states and these can also have discrete vibrational and rotational energy levels. Furthermore there are dissociative excited states that have continuous absorption spectra.)






    share|improve this answer











    $endgroup$












    • $begingroup$
      just as a side comment (it might be of interest for the OP), the continuum observed in electrons' energy when ionization occurs is the photoelectric effect; all energy excess increases electron's speed.
      $endgroup$
      – santimirandarp
      Jun 2 at 13:47







    • 1




      $begingroup$
      yes; a good point
      $endgroup$
      – porphyrin
      yesterday













    6












    6








    6





    $begingroup$

    The spectral lines represent transitions between pairs of discrete energy levels in the atom; an electron is excited by absorbing the energy of a photon and a transition from one of these levels to another occurs as the photon is destroyed. Emission lines occur when the reverse process happens.



    If enough energy is absorbed the atom can be ionised and an electron with un-quantised energy can be liberated and moves to a position of complete separation where attractive forces between electron and nucleus is insignificant. As the liberated electron's energy is un-quantised the absorption giving rise to ionisation is continuous. The absorption is truly continuous, no matter how high a spectral resolution is used discrete absorption lines will not be observed.



    Black body radiation is continuous emission from a hot body. The analysis of this emission lead Plank in 1901 to propose, for the first time, the idea that the photons were quantised, thus avoiding the 'ultra-violet catastrophe'. He did this by allowing ultra-violet photons to have more energy than visible ones which in turn had more energy that infra-red ones. Nowadays we know that he energy is directly proportional to frequency via the Planck constant, $E=hv$.



    In the black body, all the numerous energy levels in the body are highly excited and transitions between them continuously occur as the body is being continuously heated. These transition energies overlap one another in energy so that the spectrum is continuous.



    (In molecules there are also discrete energy levels, in the ground state these are due to the vibrational and rotational motion of the molecule. A molecule may also have electronically excited states and these can also have discrete vibrational and rotational energy levels. Furthermore there are dissociative excited states that have continuous absorption spectra.)






    share|improve this answer











    $endgroup$



    The spectral lines represent transitions between pairs of discrete energy levels in the atom; an electron is excited by absorbing the energy of a photon and a transition from one of these levels to another occurs as the photon is destroyed. Emission lines occur when the reverse process happens.



    If enough energy is absorbed the atom can be ionised and an electron with un-quantised energy can be liberated and moves to a position of complete separation where attractive forces between electron and nucleus is insignificant. As the liberated electron's energy is un-quantised the absorption giving rise to ionisation is continuous. The absorption is truly continuous, no matter how high a spectral resolution is used discrete absorption lines will not be observed.



    Black body radiation is continuous emission from a hot body. The analysis of this emission lead Plank in 1901 to propose, for the first time, the idea that the photons were quantised, thus avoiding the 'ultra-violet catastrophe'. He did this by allowing ultra-violet photons to have more energy than visible ones which in turn had more energy that infra-red ones. Nowadays we know that he energy is directly proportional to frequency via the Planck constant, $E=hv$.



    In the black body, all the numerous energy levels in the body are highly excited and transitions between them continuously occur as the body is being continuously heated. These transition energies overlap one another in energy so that the spectrum is continuous.



    (In molecules there are also discrete energy levels, in the ground state these are due to the vibrational and rotational motion of the molecule. A molecule may also have electronically excited states and these can also have discrete vibrational and rotational energy levels. Furthermore there are dissociative excited states that have continuous absorption spectra.)







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Jun 2 at 11:54

























    answered Jun 2 at 8:28









    porphyrinporphyrin

    18.9k3358




    18.9k3358











    • $begingroup$
      just as a side comment (it might be of interest for the OP), the continuum observed in electrons' energy when ionization occurs is the photoelectric effect; all energy excess increases electron's speed.
      $endgroup$
      – santimirandarp
      Jun 2 at 13:47







    • 1




      $begingroup$
      yes; a good point
      $endgroup$
      – porphyrin
      yesterday
















    • $begingroup$
      just as a side comment (it might be of interest for the OP), the continuum observed in electrons' energy when ionization occurs is the photoelectric effect; all energy excess increases electron's speed.
      $endgroup$
      – santimirandarp
      Jun 2 at 13:47







    • 1




      $begingroup$
      yes; a good point
      $endgroup$
      – porphyrin
      yesterday















    $begingroup$
    just as a side comment (it might be of interest for the OP), the continuum observed in electrons' energy when ionization occurs is the photoelectric effect; all energy excess increases electron's speed.
    $endgroup$
    – santimirandarp
    Jun 2 at 13:47





    $begingroup$
    just as a side comment (it might be of interest for the OP), the continuum observed in electrons' energy when ionization occurs is the photoelectric effect; all energy excess increases electron's speed.
    $endgroup$
    – santimirandarp
    Jun 2 at 13:47





    1




    1




    $begingroup$
    yes; a good point
    $endgroup$
    – porphyrin
    yesterday




    $begingroup$
    yes; a good point
    $endgroup$
    – porphyrin
    yesterday











    4












    $begingroup$

    Natural white light had continuous spectrum, as the source - black radiator - has continuous emission in all wavelengths.



    Atoms emit or absorb only at the wavelengths, that are corresponding to energy differences between discreet energy levels of electrons in atoms.






    share|improve this answer









    $endgroup$

















      4












      $begingroup$

      Natural white light had continuous spectrum, as the source - black radiator - has continuous emission in all wavelengths.



      Atoms emit or absorb only at the wavelengths, that are corresponding to energy differences between discreet energy levels of electrons in atoms.






      share|improve this answer









      $endgroup$















        4












        4








        4





        $begingroup$

        Natural white light had continuous spectrum, as the source - black radiator - has continuous emission in all wavelengths.



        Atoms emit or absorb only at the wavelengths, that are corresponding to energy differences between discreet energy levels of electrons in atoms.






        share|improve this answer









        $endgroup$



        Natural white light had continuous spectrum, as the source - black radiator - has continuous emission in all wavelengths.



        Atoms emit or absorb only at the wavelengths, that are corresponding to energy differences between discreet energy levels of electrons in atoms.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jun 2 at 6:52









        PoutnikPoutnik

        3,281620




        3,281620













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