My teacher explained today that unlike the spectrum obtained by analyzing the sunlight, the spectra of atoms are not continuous.

I have a question about this - the sunlight is emitted by the atoms of the elements composing the sun. So, spectrum obtained by sunlight is continuous despite its atomic spectra. In order for sunlight spectrum to be continuous and atomic spectra to be discontinuous, can we assume that sun consists of all those elements (sodium, helium, neon, mercury, etc) which emit the colors of frequency belonging to visible region?

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/46080/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Dec 16, 2013 at 12:02
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    $\begingroup$ Also relevant physics.stackexchange.com/q/71503 $\endgroup$
    – leonbloy
    Commented May 4, 2016 at 19:27
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    $\begingroup$ As Jim states below the spectrum from the sun is due to blackbody radiation. $\endgroup$
    – JQK
    Commented Sep 9, 2018 at 22:54
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    $\begingroup$ @JQK blackbody radiation is not a mechanism. To get a blackbody spectrum you need a continuum source of opacity. $\endgroup$
    – ProfRob
    Commented Aug 9, 2019 at 12:06

5 Answers 5


The sun's spectrum is very complex, and indeed there are a lot of "lines"—both light and dark (emission and absorption)—amidst a sea of what looks to be continuous frequencies.

Note that the atoms you study in a textbook are idealizations. In a hot object such as the sun, some photons come to us by way of atomic emissions, but the speeds of the atoms that emitted them are distributed continuously (something like the Maxwell–Boltzmann distribution), so there is a Doppler shift to each emitted photon. This "broadens" the spectral line, i.e., turns a discrete frequency into a continuum. This is called Doppler broadening or thermal broadening. However, this is not sufficient to produce a near-completely thermal black body spectrum (thanks to gigacyan for pointing out that this wasn't clear).

Other photons were emitted by nuclear processes and have been bouncing around inside the sun for many years (an astrophysicist could probably tell you how many), and each of these collisions has shifted its energy in a somewhat random fashion. Furthermore, this energy from fusion, along with gravitational effects, leaves most of the sun in the plasma state, where ions and electrons are separate from one another. Because this plasma is extremely hot, collisions and recombinations produce even more radiation which is the primary source of the light that reaches us. What we see is called the photosphere, which is the region of this plasma from which light can escape.

The overall effect is called thermalization, where the energy gets moved around in bits and pieces until everything is in thermal equilibrium. In the case of the sun it is only approximate, as different parts have different temperatures, so it is merely a tendency.

  • $\begingroup$ an astrophysicist could probably tell you how many - A few tens of thousands of years at least, maybe upwards of a hundred thousand depending on how you estimate things :) $\endgroup$
    – user10851
    Commented Dec 6, 2013 at 0:10
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    $\begingroup$ While this is true what you say, it has nothing to do with continuous spectrum of the Sun which is close to the black body spectrum. (tungsten lamp has continuous spectrum even thought tungsten atoms are hardly moving and photons do not bounce inside the filament for years). $\endgroup$
    – gigacyan
    Commented Dec 6, 2013 at 20:59
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    $\begingroup$ The point is that knowing what atoms do in isolation is not enough to predict the spectrum for a many-body system. A tungsten lamp's spectrum also looks nothing like the spectrum of a single tungsten atom in isolation. $\endgroup$ Commented Dec 7, 2013 at 0:17
  • $\begingroup$ @gigacyan, I've edited my answer to make this more clear. $\endgroup$ Commented Dec 7, 2013 at 0:27
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    $\begingroup$ This isn't a correct account of the continuous spectrum of the Sun, which is attributable to H minus ions in the visible photosphere. $\endgroup$
    – ProfRob
    Commented Aug 8, 2019 at 21:40

The continuous spectrum of the visible photosphere of the Sun is attributable to the radiative equilibrium of the $\mathrm{H}^{-}$ ion. This has been recognised for at least 80 years (Wildt 1939).

This ion forms by the attachment of a free electron (with a continuous spectrum of energies) to a hydrogen atom, emitting a continuous spectrum of photons in the process. The reverse process, photo-detachment, occurs at the same rate and is the principal source of continuum opacity in the solar photosphere.

Absorption and emission by H$^{-}$ ions demands that the temperature is low enough not to dissociate the extra electron ($<10^{4}$K), but high enough that there is a supply of free electrons donated by the ionisation of alkali metals ($\gtrsim 2500$K) such as sodium and potassium.

Superimposed on this continuum are dark, discrete absorption features caused by transitions within atomic species (mainly metals, but also hydrogen). These are dark because the photons at these frequencies arrive here from higher up in the atmosphere at cooler temperatures.

The original question asks only about "sunlight", and the H$^{-}$ opacity mechanism is only effective between about 2,500-10,000K. Stars with photospheres with temperatures above 10,000K or below about 3,000K are dominated by different opacity and hence emission mechanisms.

In hotter stars the primary sources of continuous opacity at visible wavelengths, at temperatures exceeding 10,000K, are scattering by free electrons and the Paschen continuum arising from transitions between the $n=3$ state in hydrogen atoms and ionised hydrogen. There are also smaller contributions from free-free transitions of electrons in the electric fields of ions (bremsstrahlung).

There can of course be a cooler, overlying layer in hot stars where H$^{-}$ can form, but it has a lower density and a small optical depth (i.e. photons travel through it) and it therefore does not contribute significantly to the photospheric continuum.

In very cool photospheres, there are few free electrons and the atoms begin to form molecules, like carbon monoxide, water, molecular hydrogen, titanium oxide, vanadium oxide etc. There is actually very little true continuum absorption/emission in the visible part of the spectrum of these stars. Instead there is an overlapping mess of rotational and vibrational molecular transitions that form a pseudo-continuum when observed by instruments with finite spectral resolution. The dominant opacity in the visible spectrum at just below 3000K is due to TiO molecules.

At even lower temperatures (below 2500K, and approaching the substellar regime), absorption and emission by dust becomes important, although there is negligible flux at visible wavelengths.

  • $\begingroup$ Something's fishy: are you saying this mechanism can only work in the rather narrow range 3000K < T < 10 000K? The temperatures of stars would in general vary over much larger ranges. $\endgroup$ Commented Aug 9, 2019 at 13:53
  • $\begingroup$ @flippiefanus The question is asking about "sunlight". The principle opacity sources are different for photospheres that are hotter than 10,000K or cooler than 3000K. It is clear in my answer that I am referring to "the Sun". I may broaden my answer. $\endgroup$
    – ProfRob
    Commented Aug 9, 2019 at 14:12
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    $\begingroup$ So are you saying the sun is unique in this way and that other stars operate with different mechanisms? That does not sound reasonable. $\endgroup$ Commented Aug 10, 2019 at 5:12
  • $\begingroup$ @flippifanus No that is not a correct summary of what I've said. I have added to my answer to deal with hotter and cooler stars. The Sun isn't unique. The vast majority of stars have photospheres between 3000 and 10,000 K. $\endgroup$
    – ProfRob
    Commented Aug 10, 2019 at 8:30
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    $\begingroup$ @EmilioPisanty that is clearly a different question and I am not going to write a treatise on stellar atmospheres. I have absolutely answered the question that was asked. The answer to your Q is that the density-temperature profile in more massive stars results in an optical depth of 1 at higher temperatures. There is of course a cooler layer higher up, but it has a very small optical depth. $\endgroup$
    – ProfRob
    Commented Aug 10, 2019 at 8:55

This is quite a natural confusion. You are correct, were the Solar spectrum purely due to the spectral output of the atoms composing it, we would not be able to get a continuous spectrum. However, the light emitted by the Sun is due to its temperature. All objects that are above $-273.15^{\circ}C$ (so, all objects) emit radiation at a continuous spectrum that relates to their temperature, we call the temperature dependent spectrum a "blackbody" spectrum. The Solar spectrum is practically a perfect fit for this relation for a blackbody temperature of about $5250^{\circ}C$.

If your interested in a bit more information including how we are able to see a continuous spectrum, check out a previous post of mine here.

  • $\begingroup$ " All objects that are above −273.15∘C (so, all objects) emit radiation at a continuous spectrum that relates to their temperature, we call the temperature dependent spectrum a "blackbody" spectrum." This is not true. $\endgroup$
    – ProfRob
    Commented Aug 9, 2019 at 8:07
  • $\begingroup$ @RobJeffries Why is this not true? Please explain. $\endgroup$ Commented Aug 10, 2019 at 9:06
  • $\begingroup$ @AlexTrounev it just isn't. Take a spectrum of any object or light source. It won't be a blackbody spectrum. A blackbody spectrum is an ideal that is only emitted by an object that absorbs everything incident upon it and which is in thermal equilibrium. $\endgroup$
    – ProfRob
    Commented Aug 10, 2019 at 9:11
  • $\begingroup$ @RobJeffries But we use this radiation model to solve various problems. And there is an experimental database for different materials. In the case of the Sun, it is obvious that radiation with a continuous spectrum predominates and the maximum corresponds to temperature of 5,777K, see en.wikipedia.org/wiki/Sun#/media/… $\endgroup$ Commented Aug 10, 2019 at 9:26
  • $\begingroup$ @AlexTrounev Even the sun does not exhibit a blackbody spectrum. The "maximum" does not correspond to 5777K. This is in fact the effective temperature, which is defined as $(L/4\pi R^2 \sigma)^{0.25}$. The solar spectrum arises from material at a range of temperatures. Most other objects depart from a BB spectrum even more. $\endgroup$
    – ProfRob
    Commented Aug 10, 2019 at 11:16

We have two type of atoms comprising most of the sun's structure: hydrogen and helium. And these have a noncontinuous spectrum.

But the sun has a continuous spectrum like a black body radiation because of free to bond or free to free radiation. There are three types of transition:

  1. bond to bond (which is discrete)

  2. free to bond (a free electron bond to an atom and radiate photons with continuous frequency)

  3. free to free (a free electron change its speed due to interaction with a potential and radiate photons with continuous frequency)

The last one also called Bremsstrahlung radiation.

So free electrons in the plasma medium in the sun's corona radiate Bremsstrahlung radiation due to statistical collisions and the output result is a continuous spectrum like back body radiation.

The tungsten in bulb light also behaves in this manner. Electrons from electrical current collide with heavy atoms and free-free transitions occurred (Bremsstrahlung radiation) and we have a continuous light due to this interaction.

This is also true for continuous radiation of hot objects and metals.

It is also important to note that Doppler broadening is not the answer because it has a very small bandwidth ($\ll1$ nm) for gas speed distribution.


Certain wavelengths of electromagnetic spectrum is emitted when the electrons in an atom to move from a higher level to a lower. The wavelength that is emitted depends upon the number of shells the electrons move down. When an electron/electrons move into certain number shells, white light is emitted i.e. when two hydrogen atoms fuse into an Helium atom, visible light is produced. As we know, visible light contains of all the wavelengths of colours from red to Blue. When a spectrometer is used, we are able to see a range of frequencies and therefore, calling it "Continuous Spectrum".

However, spectrometer is used only to analyze the light that is visible to us. You can't analyze the frequency that aren't visible to us. As you said, sodium, neon and mercury do exist in sun but, they don't emit light that is visible to us. Since only we see white light, you just see that light split apart into a continuous range of colours

  • $\begingroup$ 'You can't analyze the frequesncy that aren't visible to us' This sounds very wrong to me, and not just grammatically. $\endgroup$
    – Danu
    Commented Dec 5, 2013 at 23:33
  • $\begingroup$ When hydrogen atoms fuse, gamma rays are produced. The visible light is really just along for the ride... $\endgroup$ Commented Dec 6, 2013 at 0:45
  • $\begingroup$ Sorry for any grammar mistakes. I was typing the answer in a hurry. $\endgroup$
    – The DON
    Commented Dec 6, 2013 at 1:17

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