Why do atoms (iron eg) glow with all frequencies of light when exposed to enough thermal radiation?












26












$begingroup$


Correct me if I'm wrong, but objects (made of constituent atoms) glow with a particular frequency of light which our eyes relate to as colour.



They glow when a particle in a higher energy quantum state gets converted into a lower one by the emission of a photon. And the energy difference between the two states will correlate to the frequency of the photon.



So when we look at an emission spectrum we look at many colours being seen from a sample, now why are there so many (more than one)? Is it because there are many energy levels and the difference between these energy levels vary? Is it because of electrons being promoted and demoted from n=2 to n=1, n=3 to n=2, n=4 to n=3? But aren't these energy states unstable also, won’t they all emit photons till they reach n=2?



Is that why iron's electromagnetic radiation is first within the infrared range and then progresses to the visible light range because the electrons are now in high enough energy levels that the frequency of the photons can be detected by our eyes?










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$endgroup$








  • 4




    $begingroup$
    Blackbody radiation is a phenomenon of bulk matter, not individual isolated atoms.
    $endgroup$
    – PM 2Ring
    Feb 14 at 15:23






  • 1




    $begingroup$
    Are you asking about a vapor or bulk phase?
    $endgroup$
    – Lewis Miller
    Feb 14 at 15:34










  • $begingroup$
    #PM 2ring It's still coming from vibrational modes of the atoms in the material being converted into radiation
    $endgroup$
    – Jan Bos
    Feb 14 at 16:06
















26












$begingroup$


Correct me if I'm wrong, but objects (made of constituent atoms) glow with a particular frequency of light which our eyes relate to as colour.



They glow when a particle in a higher energy quantum state gets converted into a lower one by the emission of a photon. And the energy difference between the two states will correlate to the frequency of the photon.



So when we look at an emission spectrum we look at many colours being seen from a sample, now why are there so many (more than one)? Is it because there are many energy levels and the difference between these energy levels vary? Is it because of electrons being promoted and demoted from n=2 to n=1, n=3 to n=2, n=4 to n=3? But aren't these energy states unstable also, won’t they all emit photons till they reach n=2?



Is that why iron's electromagnetic radiation is first within the infrared range and then progresses to the visible light range because the electrons are now in high enough energy levels that the frequency of the photons can be detected by our eyes?










share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Blackbody radiation is a phenomenon of bulk matter, not individual isolated atoms.
    $endgroup$
    – PM 2Ring
    Feb 14 at 15:23






  • 1




    $begingroup$
    Are you asking about a vapor or bulk phase?
    $endgroup$
    – Lewis Miller
    Feb 14 at 15:34










  • $begingroup$
    #PM 2ring It's still coming from vibrational modes of the atoms in the material being converted into radiation
    $endgroup$
    – Jan Bos
    Feb 14 at 16:06














26












26








26


5



$begingroup$


Correct me if I'm wrong, but objects (made of constituent atoms) glow with a particular frequency of light which our eyes relate to as colour.



They glow when a particle in a higher energy quantum state gets converted into a lower one by the emission of a photon. And the energy difference between the two states will correlate to the frequency of the photon.



So when we look at an emission spectrum we look at many colours being seen from a sample, now why are there so many (more than one)? Is it because there are many energy levels and the difference between these energy levels vary? Is it because of electrons being promoted and demoted from n=2 to n=1, n=3 to n=2, n=4 to n=3? But aren't these energy states unstable also, won’t they all emit photons till they reach n=2?



Is that why iron's electromagnetic radiation is first within the infrared range and then progresses to the visible light range because the electrons are now in high enough energy levels that the frequency of the photons can be detected by our eyes?










share|cite|improve this question











$endgroup$




Correct me if I'm wrong, but objects (made of constituent atoms) glow with a particular frequency of light which our eyes relate to as colour.



They glow when a particle in a higher energy quantum state gets converted into a lower one by the emission of a photon. And the energy difference between the two states will correlate to the frequency of the photon.



So when we look at an emission spectrum we look at many colours being seen from a sample, now why are there so many (more than one)? Is it because there are many energy levels and the difference between these energy levels vary? Is it because of electrons being promoted and demoted from n=2 to n=1, n=3 to n=2, n=4 to n=3? But aren't these energy states unstable also, won’t they all emit photons till they reach n=2?



Is that why iron's electromagnetic radiation is first within the infrared range and then progresses to the visible light range because the electrons are now in high enough energy levels that the frequency of the photons can be detected by our eyes?







photon-emission






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share|cite|improve this question













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edited Feb 15 at 5:13









CJ Dennis

438413




438413










asked Feb 14 at 15:03









user73837user73837

13125




13125








  • 4




    $begingroup$
    Blackbody radiation is a phenomenon of bulk matter, not individual isolated atoms.
    $endgroup$
    – PM 2Ring
    Feb 14 at 15:23






  • 1




    $begingroup$
    Are you asking about a vapor or bulk phase?
    $endgroup$
    – Lewis Miller
    Feb 14 at 15:34










  • $begingroup$
    #PM 2ring It's still coming from vibrational modes of the atoms in the material being converted into radiation
    $endgroup$
    – Jan Bos
    Feb 14 at 16:06














  • 4




    $begingroup$
    Blackbody radiation is a phenomenon of bulk matter, not individual isolated atoms.
    $endgroup$
    – PM 2Ring
    Feb 14 at 15:23






  • 1




    $begingroup$
    Are you asking about a vapor or bulk phase?
    $endgroup$
    – Lewis Miller
    Feb 14 at 15:34










  • $begingroup$
    #PM 2ring It's still coming from vibrational modes of the atoms in the material being converted into radiation
    $endgroup$
    – Jan Bos
    Feb 14 at 16:06








4




4




$begingroup$
Blackbody radiation is a phenomenon of bulk matter, not individual isolated atoms.
$endgroup$
– PM 2Ring
Feb 14 at 15:23




$begingroup$
Blackbody radiation is a phenomenon of bulk matter, not individual isolated atoms.
$endgroup$
– PM 2Ring
Feb 14 at 15:23




1




1




$begingroup$
Are you asking about a vapor or bulk phase?
$endgroup$
– Lewis Miller
Feb 14 at 15:34




$begingroup$
Are you asking about a vapor or bulk phase?
$endgroup$
– Lewis Miller
Feb 14 at 15:34












$begingroup$
#PM 2ring It's still coming from vibrational modes of the atoms in the material being converted into radiation
$endgroup$
– Jan Bos
Feb 14 at 16:06




$begingroup$
#PM 2ring It's still coming from vibrational modes of the atoms in the material being converted into radiation
$endgroup$
– Jan Bos
Feb 14 at 16:06










4 Answers
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active

oldest

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$begingroup$


They glow when a particle in a higher energy quantum state gets converted into a lower one by the emission of a photon.




That is one method of emission. Because individual atoms (and small molecules) have a smallish number of stable configurations, the types of emissions possible from the decay of a single particle is limited.



But in dense, high-temperature systems, the emission from an isolated particle is no longer dominant. Instead, the collisions and interactions between the particles cause charges (electrons) to be accelerated. Accelerating charges emit radiation, and this radiation is not associated with change in the atomic/molecular configuration.



Because there is no discrete configuration involved, just various rates of acceleration, the discrete lines of an emission spectrum are not present.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much, this pointed out to a few caveats in my thinking. So the method of emission I mentioned is not feasible (negligible impact) in dense systems, rather the accelerating charges play a more pivotal role. And why do accelerated charges emit radiation? Forgive my ignorance, I’m still at the Igcse level.
    $endgroup$
    – user73837
    Feb 15 at 16:23








  • 1




    $begingroup$
    "why do accelerated charges emit radiation?" physics.stackexchange.com/questions/65339/…
    $endgroup$
    – BowlOfRed
    Feb 15 at 17:01










  • $begingroup$
    Could you explain with more detail the nature of how these interactions work? Does the collision actually knock the electron out of the atom?
    $endgroup$
    – John Hon
    Feb 16 at 6:11



















18












$begingroup$

Matter comes in phases: solid, liquid, gas, plasma



Individual atoms/molecules join into lattices when solid, are in collective states in liquid, free in gas, and ionized mostly in plasma.



Transitions in the atomic energy levels you envisage are detectable only in gases and plasma, there the changes in n,l,m and the resulting absorption and emission of spectral photons can be detected, although there is also continuum photons from interactions in the spill over electric and magnetic fields.



In solids, there are a large number of energy levels that are lattice related, this means that there will be transitions in rotational and vibrational states that have nothing to do with atomic transitions. These transitions are the black body radiation, and are energy dependent. They also exist in the gas due to the kinetic energy . As you were told in comments this is the black body radiation, which characterizes temperature of a body. The higher the temperature the more photons in the visible.



bbody



Note the high temperature it is the high temperatures that gives us the observed light of the sun.



So yes, there are many energy levels , but it is the kinetic energy that dominates at high temperatures and gives a continuum of frequencies according to black body , or approximately ( atomic spectral lines can be filtered in a plasma, but it is the black body type of radiation that dominates).



Now for iron and metals in general the colors will barely touch the visible, as the temperatures are between 770K and 1480K






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$endgroup$





















    0












    $begingroup$

    I suppose by this question what you're asking is "what is the mechanism of emission of thermal radiation by a hot object?"



    The answer to this is that there is, effectively, no one mechanism, because thermal motion is, in a sense, the "most random" motion that a physical system can possibly have. From a classical mechanical point of view, charged particles jostling around randomly will emit a random field of radiation as they produce small fluctuations in the field nearby to them, which then must, by Maxwell's equations, propagate outward as waves. At the surface, these waves are able to escape the object. If you want to think about this, think about dipping your fingers in water and then fluttering them around. There is no one type of motion that produces this radiation - rather, it is due to all of their non-linear, accelerated motions together which are disturbing the electromagnetic field.



    Of course, the prediction one derives if one takes this seriously is that the power actually ends up being effectively infinite - the famous "ultraviolet catastrophe" - and that's obviously garbage, so we need to talk about quantum mechanics. In the case of quantum mechanics, particles are more structured with regard to how their energies can change and this is what obviates the difficulty, but nonetheless again all possible transitions, which can result in the emission of a photon, are fair game and all of them will contribute radiation, just as all movements which in classical mechanics would result in emission are fair game in the classical scenario. This can include shell transitions, but also can include, and especially in metals, transitions in band levels as well.



    In the most general sense, thermal emissions result from excitation of every single possible way that energy is capable of escaping from the system at a microscopic level - no, in fact, every possible way, period: in theory, they could even excite macroscopic escapes such as ejecting a portion of the material in a spontaneous self-fracturing, but the probabilities of these are incalculably small and thus, effectively, "suppressed". (The microscopic version of this - ejection of single atoms - does occur with relative frequency, and this results in evaporation.) This also even is not only how radiation is possible (thermal excitation of radiative escape paths) but also conduction, as well: such can be understood as thermal excitation of modes of escape where energy escapes by being kinetically transferred to neighboring matter.






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    $endgroup$





















      -1












      $begingroup$

      Like you can see all around you, at the low temperatures we find ourselves in, all solid bodies have a specific color. The Sun shines on them with the continuous spectrum of a black body at around 6000 (K), which means that there are a lot of photons with different frequencies coming in. In this case, all solid objects have a specific color because the electron transitions (excitation and relaxation) in atoms (or the constituent molecules of the object) you describe are different for all these objects, and made possible by the many different frequencies of the photons coming from the Sun (the black body radiation of these objects at the low temperature on Earth can be ignored; none of this radiation can be seen).



      Now if we would be much closer to the Sun, the temperature on Earth will rise and the temperature of all objects accordingly. A new dynamical thermal equilibrium with the Sun (dynamical, because of day and night on Earth, but that's another story; it is important though because if the Earth would always face the Sun, the surface temperature on the bright side of the Earth would become too hot for most objects to exist, while on the dark side there would be nothing to see) will be established. Most of the objects that have a color on the Earth we live in here and now would be burned (or evaporated) before turning into an object emitting black body radiation (if we're close enough to the Sun) of which we can see the visible part. But some objects will "survive" so we could see them glow (like many metals, or, say, glass) and overpower the emission-absorption spectrum.






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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

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        32












        $begingroup$


        They glow when a particle in a higher energy quantum state gets converted into a lower one by the emission of a photon.




        That is one method of emission. Because individual atoms (and small molecules) have a smallish number of stable configurations, the types of emissions possible from the decay of a single particle is limited.



        But in dense, high-temperature systems, the emission from an isolated particle is no longer dominant. Instead, the collisions and interactions between the particles cause charges (electrons) to be accelerated. Accelerating charges emit radiation, and this radiation is not associated with change in the atomic/molecular configuration.



        Because there is no discrete configuration involved, just various rates of acceleration, the discrete lines of an emission spectrum are not present.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Thank you so much, this pointed out to a few caveats in my thinking. So the method of emission I mentioned is not feasible (negligible impact) in dense systems, rather the accelerating charges play a more pivotal role. And why do accelerated charges emit radiation? Forgive my ignorance, I’m still at the Igcse level.
          $endgroup$
          – user73837
          Feb 15 at 16:23








        • 1




          $begingroup$
          "why do accelerated charges emit radiation?" physics.stackexchange.com/questions/65339/…
          $endgroup$
          – BowlOfRed
          Feb 15 at 17:01










        • $begingroup$
          Could you explain with more detail the nature of how these interactions work? Does the collision actually knock the electron out of the atom?
          $endgroup$
          – John Hon
          Feb 16 at 6:11
















        32












        $begingroup$


        They glow when a particle in a higher energy quantum state gets converted into a lower one by the emission of a photon.




        That is one method of emission. Because individual atoms (and small molecules) have a smallish number of stable configurations, the types of emissions possible from the decay of a single particle is limited.



        But in dense, high-temperature systems, the emission from an isolated particle is no longer dominant. Instead, the collisions and interactions between the particles cause charges (electrons) to be accelerated. Accelerating charges emit radiation, and this radiation is not associated with change in the atomic/molecular configuration.



        Because there is no discrete configuration involved, just various rates of acceleration, the discrete lines of an emission spectrum are not present.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Thank you so much, this pointed out to a few caveats in my thinking. So the method of emission I mentioned is not feasible (negligible impact) in dense systems, rather the accelerating charges play a more pivotal role. And why do accelerated charges emit radiation? Forgive my ignorance, I’m still at the Igcse level.
          $endgroup$
          – user73837
          Feb 15 at 16:23








        • 1




          $begingroup$
          "why do accelerated charges emit radiation?" physics.stackexchange.com/questions/65339/…
          $endgroup$
          – BowlOfRed
          Feb 15 at 17:01










        • $begingroup$
          Could you explain with more detail the nature of how these interactions work? Does the collision actually knock the electron out of the atom?
          $endgroup$
          – John Hon
          Feb 16 at 6:11














        32












        32








        32





        $begingroup$


        They glow when a particle in a higher energy quantum state gets converted into a lower one by the emission of a photon.




        That is one method of emission. Because individual atoms (and small molecules) have a smallish number of stable configurations, the types of emissions possible from the decay of a single particle is limited.



        But in dense, high-temperature systems, the emission from an isolated particle is no longer dominant. Instead, the collisions and interactions between the particles cause charges (electrons) to be accelerated. Accelerating charges emit radiation, and this radiation is not associated with change in the atomic/molecular configuration.



        Because there is no discrete configuration involved, just various rates of acceleration, the discrete lines of an emission spectrum are not present.






        share|cite|improve this answer









        $endgroup$




        They glow when a particle in a higher energy quantum state gets converted into a lower one by the emission of a photon.




        That is one method of emission. Because individual atoms (and small molecules) have a smallish number of stable configurations, the types of emissions possible from the decay of a single particle is limited.



        But in dense, high-temperature systems, the emission from an isolated particle is no longer dominant. Instead, the collisions and interactions between the particles cause charges (electrons) to be accelerated. Accelerating charges emit radiation, and this radiation is not associated with change in the atomic/molecular configuration.



        Because there is no discrete configuration involved, just various rates of acceleration, the discrete lines of an emission spectrum are not present.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 14 at 17:24









        BowlOfRedBowlOfRed

        16.8k22642




        16.8k22642












        • $begingroup$
          Thank you so much, this pointed out to a few caveats in my thinking. So the method of emission I mentioned is not feasible (negligible impact) in dense systems, rather the accelerating charges play a more pivotal role. And why do accelerated charges emit radiation? Forgive my ignorance, I’m still at the Igcse level.
          $endgroup$
          – user73837
          Feb 15 at 16:23








        • 1




          $begingroup$
          "why do accelerated charges emit radiation?" physics.stackexchange.com/questions/65339/…
          $endgroup$
          – BowlOfRed
          Feb 15 at 17:01










        • $begingroup$
          Could you explain with more detail the nature of how these interactions work? Does the collision actually knock the electron out of the atom?
          $endgroup$
          – John Hon
          Feb 16 at 6:11


















        • $begingroup$
          Thank you so much, this pointed out to a few caveats in my thinking. So the method of emission I mentioned is not feasible (negligible impact) in dense systems, rather the accelerating charges play a more pivotal role. And why do accelerated charges emit radiation? Forgive my ignorance, I’m still at the Igcse level.
          $endgroup$
          – user73837
          Feb 15 at 16:23








        • 1




          $begingroup$
          "why do accelerated charges emit radiation?" physics.stackexchange.com/questions/65339/…
          $endgroup$
          – BowlOfRed
          Feb 15 at 17:01










        • $begingroup$
          Could you explain with more detail the nature of how these interactions work? Does the collision actually knock the electron out of the atom?
          $endgroup$
          – John Hon
          Feb 16 at 6:11
















        $begingroup$
        Thank you so much, this pointed out to a few caveats in my thinking. So the method of emission I mentioned is not feasible (negligible impact) in dense systems, rather the accelerating charges play a more pivotal role. And why do accelerated charges emit radiation? Forgive my ignorance, I’m still at the Igcse level.
        $endgroup$
        – user73837
        Feb 15 at 16:23






        $begingroup$
        Thank you so much, this pointed out to a few caveats in my thinking. So the method of emission I mentioned is not feasible (negligible impact) in dense systems, rather the accelerating charges play a more pivotal role. And why do accelerated charges emit radiation? Forgive my ignorance, I’m still at the Igcse level.
        $endgroup$
        – user73837
        Feb 15 at 16:23






        1




        1




        $begingroup$
        "why do accelerated charges emit radiation?" physics.stackexchange.com/questions/65339/…
        $endgroup$
        – BowlOfRed
        Feb 15 at 17:01




        $begingroup$
        "why do accelerated charges emit radiation?" physics.stackexchange.com/questions/65339/…
        $endgroup$
        – BowlOfRed
        Feb 15 at 17:01












        $begingroup$
        Could you explain with more detail the nature of how these interactions work? Does the collision actually knock the electron out of the atom?
        $endgroup$
        – John Hon
        Feb 16 at 6:11




        $begingroup$
        Could you explain with more detail the nature of how these interactions work? Does the collision actually knock the electron out of the atom?
        $endgroup$
        – John Hon
        Feb 16 at 6:11











        18












        $begingroup$

        Matter comes in phases: solid, liquid, gas, plasma



        Individual atoms/molecules join into lattices when solid, are in collective states in liquid, free in gas, and ionized mostly in plasma.



        Transitions in the atomic energy levels you envisage are detectable only in gases and plasma, there the changes in n,l,m and the resulting absorption and emission of spectral photons can be detected, although there is also continuum photons from interactions in the spill over electric and magnetic fields.



        In solids, there are a large number of energy levels that are lattice related, this means that there will be transitions in rotational and vibrational states that have nothing to do with atomic transitions. These transitions are the black body radiation, and are energy dependent. They also exist in the gas due to the kinetic energy . As you were told in comments this is the black body radiation, which characterizes temperature of a body. The higher the temperature the more photons in the visible.



        bbody



        Note the high temperature it is the high temperatures that gives us the observed light of the sun.



        So yes, there are many energy levels , but it is the kinetic energy that dominates at high temperatures and gives a continuum of frequencies according to black body , or approximately ( atomic spectral lines can be filtered in a plasma, but it is the black body type of radiation that dominates).



        Now for iron and metals in general the colors will barely touch the visible, as the temperatures are between 770K and 1480K






        share|cite|improve this answer











        $endgroup$


















          18












          $begingroup$

          Matter comes in phases: solid, liquid, gas, plasma



          Individual atoms/molecules join into lattices when solid, are in collective states in liquid, free in gas, and ionized mostly in plasma.



          Transitions in the atomic energy levels you envisage are detectable only in gases and plasma, there the changes in n,l,m and the resulting absorption and emission of spectral photons can be detected, although there is also continuum photons from interactions in the spill over electric and magnetic fields.



          In solids, there are a large number of energy levels that are lattice related, this means that there will be transitions in rotational and vibrational states that have nothing to do with atomic transitions. These transitions are the black body radiation, and are energy dependent. They also exist in the gas due to the kinetic energy . As you were told in comments this is the black body radiation, which characterizes temperature of a body. The higher the temperature the more photons in the visible.



          bbody



          Note the high temperature it is the high temperatures that gives us the observed light of the sun.



          So yes, there are many energy levels , but it is the kinetic energy that dominates at high temperatures and gives a continuum of frequencies according to black body , or approximately ( atomic spectral lines can be filtered in a plasma, but it is the black body type of radiation that dominates).



          Now for iron and metals in general the colors will barely touch the visible, as the temperatures are between 770K and 1480K






          share|cite|improve this answer











          $endgroup$
















            18












            18








            18





            $begingroup$

            Matter comes in phases: solid, liquid, gas, plasma



            Individual atoms/molecules join into lattices when solid, are in collective states in liquid, free in gas, and ionized mostly in plasma.



            Transitions in the atomic energy levels you envisage are detectable only in gases and plasma, there the changes in n,l,m and the resulting absorption and emission of spectral photons can be detected, although there is also continuum photons from interactions in the spill over electric and magnetic fields.



            In solids, there are a large number of energy levels that are lattice related, this means that there will be transitions in rotational and vibrational states that have nothing to do with atomic transitions. These transitions are the black body radiation, and are energy dependent. They also exist in the gas due to the kinetic energy . As you were told in comments this is the black body radiation, which characterizes temperature of a body. The higher the temperature the more photons in the visible.



            bbody



            Note the high temperature it is the high temperatures that gives us the observed light of the sun.



            So yes, there are many energy levels , but it is the kinetic energy that dominates at high temperatures and gives a continuum of frequencies according to black body , or approximately ( atomic spectral lines can be filtered in a plasma, but it is the black body type of radiation that dominates).



            Now for iron and metals in general the colors will barely touch the visible, as the temperatures are between 770K and 1480K






            share|cite|improve this answer











            $endgroup$



            Matter comes in phases: solid, liquid, gas, plasma



            Individual atoms/molecules join into lattices when solid, are in collective states in liquid, free in gas, and ionized mostly in plasma.



            Transitions in the atomic energy levels you envisage are detectable only in gases and plasma, there the changes in n,l,m and the resulting absorption and emission of spectral photons can be detected, although there is also continuum photons from interactions in the spill over electric and magnetic fields.



            In solids, there are a large number of energy levels that are lattice related, this means that there will be transitions in rotational and vibrational states that have nothing to do with atomic transitions. These transitions are the black body radiation, and are energy dependent. They also exist in the gas due to the kinetic energy . As you were told in comments this is the black body radiation, which characterizes temperature of a body. The higher the temperature the more photons in the visible.



            bbody



            Note the high temperature it is the high temperatures that gives us the observed light of the sun.



            So yes, there are many energy levels , but it is the kinetic energy that dominates at high temperatures and gives a continuum of frequencies according to black body , or approximately ( atomic spectral lines can be filtered in a plasma, but it is the black body type of radiation that dominates).



            Now for iron and metals in general the colors will barely touch the visible, as the temperatures are between 770K and 1480K







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Feb 14 at 19:02

























            answered Feb 14 at 15:58









            anna vanna v

            159k8152450




            159k8152450























                0












                $begingroup$

                I suppose by this question what you're asking is "what is the mechanism of emission of thermal radiation by a hot object?"



                The answer to this is that there is, effectively, no one mechanism, because thermal motion is, in a sense, the "most random" motion that a physical system can possibly have. From a classical mechanical point of view, charged particles jostling around randomly will emit a random field of radiation as they produce small fluctuations in the field nearby to them, which then must, by Maxwell's equations, propagate outward as waves. At the surface, these waves are able to escape the object. If you want to think about this, think about dipping your fingers in water and then fluttering them around. There is no one type of motion that produces this radiation - rather, it is due to all of their non-linear, accelerated motions together which are disturbing the electromagnetic field.



                Of course, the prediction one derives if one takes this seriously is that the power actually ends up being effectively infinite - the famous "ultraviolet catastrophe" - and that's obviously garbage, so we need to talk about quantum mechanics. In the case of quantum mechanics, particles are more structured with regard to how their energies can change and this is what obviates the difficulty, but nonetheless again all possible transitions, which can result in the emission of a photon, are fair game and all of them will contribute radiation, just as all movements which in classical mechanics would result in emission are fair game in the classical scenario. This can include shell transitions, but also can include, and especially in metals, transitions in band levels as well.



                In the most general sense, thermal emissions result from excitation of every single possible way that energy is capable of escaping from the system at a microscopic level - no, in fact, every possible way, period: in theory, they could even excite macroscopic escapes such as ejecting a portion of the material in a spontaneous self-fracturing, but the probabilities of these are incalculably small and thus, effectively, "suppressed". (The microscopic version of this - ejection of single atoms - does occur with relative frequency, and this results in evaporation.) This also even is not only how radiation is possible (thermal excitation of radiative escape paths) but also conduction, as well: such can be understood as thermal excitation of modes of escape where energy escapes by being kinetically transferred to neighboring matter.






                share|cite|improve this answer











                $endgroup$


















                  0












                  $begingroup$

                  I suppose by this question what you're asking is "what is the mechanism of emission of thermal radiation by a hot object?"



                  The answer to this is that there is, effectively, no one mechanism, because thermal motion is, in a sense, the "most random" motion that a physical system can possibly have. From a classical mechanical point of view, charged particles jostling around randomly will emit a random field of radiation as they produce small fluctuations in the field nearby to them, which then must, by Maxwell's equations, propagate outward as waves. At the surface, these waves are able to escape the object. If you want to think about this, think about dipping your fingers in water and then fluttering them around. There is no one type of motion that produces this radiation - rather, it is due to all of their non-linear, accelerated motions together which are disturbing the electromagnetic field.



                  Of course, the prediction one derives if one takes this seriously is that the power actually ends up being effectively infinite - the famous "ultraviolet catastrophe" - and that's obviously garbage, so we need to talk about quantum mechanics. In the case of quantum mechanics, particles are more structured with regard to how their energies can change and this is what obviates the difficulty, but nonetheless again all possible transitions, which can result in the emission of a photon, are fair game and all of them will contribute radiation, just as all movements which in classical mechanics would result in emission are fair game in the classical scenario. This can include shell transitions, but also can include, and especially in metals, transitions in band levels as well.



                  In the most general sense, thermal emissions result from excitation of every single possible way that energy is capable of escaping from the system at a microscopic level - no, in fact, every possible way, period: in theory, they could even excite macroscopic escapes such as ejecting a portion of the material in a spontaneous self-fracturing, but the probabilities of these are incalculably small and thus, effectively, "suppressed". (The microscopic version of this - ejection of single atoms - does occur with relative frequency, and this results in evaporation.) This also even is not only how radiation is possible (thermal excitation of radiative escape paths) but also conduction, as well: such can be understood as thermal excitation of modes of escape where energy escapes by being kinetically transferred to neighboring matter.






                  share|cite|improve this answer











                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    I suppose by this question what you're asking is "what is the mechanism of emission of thermal radiation by a hot object?"



                    The answer to this is that there is, effectively, no one mechanism, because thermal motion is, in a sense, the "most random" motion that a physical system can possibly have. From a classical mechanical point of view, charged particles jostling around randomly will emit a random field of radiation as they produce small fluctuations in the field nearby to them, which then must, by Maxwell's equations, propagate outward as waves. At the surface, these waves are able to escape the object. If you want to think about this, think about dipping your fingers in water and then fluttering them around. There is no one type of motion that produces this radiation - rather, it is due to all of their non-linear, accelerated motions together which are disturbing the electromagnetic field.



                    Of course, the prediction one derives if one takes this seriously is that the power actually ends up being effectively infinite - the famous "ultraviolet catastrophe" - and that's obviously garbage, so we need to talk about quantum mechanics. In the case of quantum mechanics, particles are more structured with regard to how their energies can change and this is what obviates the difficulty, but nonetheless again all possible transitions, which can result in the emission of a photon, are fair game and all of them will contribute radiation, just as all movements which in classical mechanics would result in emission are fair game in the classical scenario. This can include shell transitions, but also can include, and especially in metals, transitions in band levels as well.



                    In the most general sense, thermal emissions result from excitation of every single possible way that energy is capable of escaping from the system at a microscopic level - no, in fact, every possible way, period: in theory, they could even excite macroscopic escapes such as ejecting a portion of the material in a spontaneous self-fracturing, but the probabilities of these are incalculably small and thus, effectively, "suppressed". (The microscopic version of this - ejection of single atoms - does occur with relative frequency, and this results in evaporation.) This also even is not only how radiation is possible (thermal excitation of radiative escape paths) but also conduction, as well: such can be understood as thermal excitation of modes of escape where energy escapes by being kinetically transferred to neighboring matter.






                    share|cite|improve this answer











                    $endgroup$



                    I suppose by this question what you're asking is "what is the mechanism of emission of thermal radiation by a hot object?"



                    The answer to this is that there is, effectively, no one mechanism, because thermal motion is, in a sense, the "most random" motion that a physical system can possibly have. From a classical mechanical point of view, charged particles jostling around randomly will emit a random field of radiation as they produce small fluctuations in the field nearby to them, which then must, by Maxwell's equations, propagate outward as waves. At the surface, these waves are able to escape the object. If you want to think about this, think about dipping your fingers in water and then fluttering them around. There is no one type of motion that produces this radiation - rather, it is due to all of their non-linear, accelerated motions together which are disturbing the electromagnetic field.



                    Of course, the prediction one derives if one takes this seriously is that the power actually ends up being effectively infinite - the famous "ultraviolet catastrophe" - and that's obviously garbage, so we need to talk about quantum mechanics. In the case of quantum mechanics, particles are more structured with regard to how their energies can change and this is what obviates the difficulty, but nonetheless again all possible transitions, which can result in the emission of a photon, are fair game and all of them will contribute radiation, just as all movements which in classical mechanics would result in emission are fair game in the classical scenario. This can include shell transitions, but also can include, and especially in metals, transitions in band levels as well.



                    In the most general sense, thermal emissions result from excitation of every single possible way that energy is capable of escaping from the system at a microscopic level - no, in fact, every possible way, period: in theory, they could even excite macroscopic escapes such as ejecting a portion of the material in a spontaneous self-fracturing, but the probabilities of these are incalculably small and thus, effectively, "suppressed". (The microscopic version of this - ejection of single atoms - does occur with relative frequency, and this results in evaporation.) This also even is not only how radiation is possible (thermal excitation of radiative escape paths) but also conduction, as well: such can be understood as thermal excitation of modes of escape where energy escapes by being kinetically transferred to neighboring matter.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Feb 15 at 6:54

























                    answered Feb 15 at 6:49









                    The_SympathizerThe_Sympathizer

                    3,909923




                    3,909923























                        -1












                        $begingroup$

                        Like you can see all around you, at the low temperatures we find ourselves in, all solid bodies have a specific color. The Sun shines on them with the continuous spectrum of a black body at around 6000 (K), which means that there are a lot of photons with different frequencies coming in. In this case, all solid objects have a specific color because the electron transitions (excitation and relaxation) in atoms (or the constituent molecules of the object) you describe are different for all these objects, and made possible by the many different frequencies of the photons coming from the Sun (the black body radiation of these objects at the low temperature on Earth can be ignored; none of this radiation can be seen).



                        Now if we would be much closer to the Sun, the temperature on Earth will rise and the temperature of all objects accordingly. A new dynamical thermal equilibrium with the Sun (dynamical, because of day and night on Earth, but that's another story; it is important though because if the Earth would always face the Sun, the surface temperature on the bright side of the Earth would become too hot for most objects to exist, while on the dark side there would be nothing to see) will be established. Most of the objects that have a color on the Earth we live in here and now would be burned (or evaporated) before turning into an object emitting black body radiation (if we're close enough to the Sun) of which we can see the visible part. But some objects will "survive" so we could see them glow (like many metals, or, say, glass) and overpower the emission-absorption spectrum.






                        share|cite|improve this answer









                        $endgroup$


















                          -1












                          $begingroup$

                          Like you can see all around you, at the low temperatures we find ourselves in, all solid bodies have a specific color. The Sun shines on them with the continuous spectrum of a black body at around 6000 (K), which means that there are a lot of photons with different frequencies coming in. In this case, all solid objects have a specific color because the electron transitions (excitation and relaxation) in atoms (or the constituent molecules of the object) you describe are different for all these objects, and made possible by the many different frequencies of the photons coming from the Sun (the black body radiation of these objects at the low temperature on Earth can be ignored; none of this radiation can be seen).



                          Now if we would be much closer to the Sun, the temperature on Earth will rise and the temperature of all objects accordingly. A new dynamical thermal equilibrium with the Sun (dynamical, because of day and night on Earth, but that's another story; it is important though because if the Earth would always face the Sun, the surface temperature on the bright side of the Earth would become too hot for most objects to exist, while on the dark side there would be nothing to see) will be established. Most of the objects that have a color on the Earth we live in here and now would be burned (or evaporated) before turning into an object emitting black body radiation (if we're close enough to the Sun) of which we can see the visible part. But some objects will "survive" so we could see them glow (like many metals, or, say, glass) and overpower the emission-absorption spectrum.






                          share|cite|improve this answer









                          $endgroup$
















                            -1












                            -1








                            -1





                            $begingroup$

                            Like you can see all around you, at the low temperatures we find ourselves in, all solid bodies have a specific color. The Sun shines on them with the continuous spectrum of a black body at around 6000 (K), which means that there are a lot of photons with different frequencies coming in. In this case, all solid objects have a specific color because the electron transitions (excitation and relaxation) in atoms (or the constituent molecules of the object) you describe are different for all these objects, and made possible by the many different frequencies of the photons coming from the Sun (the black body radiation of these objects at the low temperature on Earth can be ignored; none of this radiation can be seen).



                            Now if we would be much closer to the Sun, the temperature on Earth will rise and the temperature of all objects accordingly. A new dynamical thermal equilibrium with the Sun (dynamical, because of day and night on Earth, but that's another story; it is important though because if the Earth would always face the Sun, the surface temperature on the bright side of the Earth would become too hot for most objects to exist, while on the dark side there would be nothing to see) will be established. Most of the objects that have a color on the Earth we live in here and now would be burned (or evaporated) before turning into an object emitting black body radiation (if we're close enough to the Sun) of which we can see the visible part. But some objects will "survive" so we could see them glow (like many metals, or, say, glass) and overpower the emission-absorption spectrum.






                            share|cite|improve this answer









                            $endgroup$



                            Like you can see all around you, at the low temperatures we find ourselves in, all solid bodies have a specific color. The Sun shines on them with the continuous spectrum of a black body at around 6000 (K), which means that there are a lot of photons with different frequencies coming in. In this case, all solid objects have a specific color because the electron transitions (excitation and relaxation) in atoms (or the constituent molecules of the object) you describe are different for all these objects, and made possible by the many different frequencies of the photons coming from the Sun (the black body radiation of these objects at the low temperature on Earth can be ignored; none of this radiation can be seen).



                            Now if we would be much closer to the Sun, the temperature on Earth will rise and the temperature of all objects accordingly. A new dynamical thermal equilibrium with the Sun (dynamical, because of day and night on Earth, but that's another story; it is important though because if the Earth would always face the Sun, the surface temperature on the bright side of the Earth would become too hot for most objects to exist, while on the dark side there would be nothing to see) will be established. Most of the objects that have a color on the Earth we live in here and now would be burned (or evaporated) before turning into an object emitting black body radiation (if we're close enough to the Sun) of which we can see the visible part. But some objects will "survive" so we could see them glow (like many metals, or, say, glass) and overpower the emission-absorption spectrum.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 15 at 6:51









                            descheleschilderdescheleschilder

                            4,10021241




                            4,10021241






























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