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Emission spectrum of hydrogen

The Balmer Rydberg equation explains the line spectrum of hydrogen. A line spectrum is a series of lines that represent the different energy levels of the an atom. In this video, we'll use the Balmer-Rydberg equation to solve for photon energy for n=3 to 2 transition. Created by Jay.

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  • aqualine ultimate style avatar for user BrownKev787
    In a hydrogen atom, why would an electron fall back to any energy level other than the n=1, since there are no other electrons stopping it from falling there?
    (29 votes)
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    • piceratops ultimate style avatar for user Just Keith
      The electron can only have specific states, nothing in between. By releasing a photon of a particular amount of energy, an electron can drop into one of the lower energy levels. If it happens to drop to an intermediate level, not n=1, the it is still in an excited state (albeit a lower excited state than it previously had). It will, if conditions allow, eventually drop back to n=1.

      So, it is not a matter of the electron not returning to n=1, it is just that it might do so in a number of steps instead of all at once.
      (34 votes)
  • aqualine ultimate style avatar for user Zachary
    So if an electron went from n=1 to n=2, no light would be emitted because it is absorbing light, not emitting light correct?
    (27 votes)
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  • aqualine seed style avatar for user ishita bakshi
    what is meant by the statement "energy is quantized"?
    (7 votes)
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  • leaf green style avatar for user Tom Pelletier
    Just as an observation, it seems that the bigger the energy level drop that the electron makes (nj to n=2), the higher the energy of the wave that is emitted by the electron.
    ex:
    n3 to n2 : red
    n4 to n2 : cyan
    etc

    But, the drop from n2 to n1 creates a wave with higher energy than n6 to n2 (122nm vs 410nm). Why is that ? is the energy difference between n2 and n1 really bigger than n6 to n2 ? or am i makinga big mistake ? haha

    Thanks a lot !
    (11 votes)
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    • male robot hal style avatar for user Aiman Khan
      As the number of energy levels increases, the difference of energy between two consecutive energy levels decreases. See this,
      Number of orbits-----Energy of orbits
      n=1 -13.6 eV
      n=2 -3.4 eV
      n=3 -1.51 eV
      n=4 -0.85 eV
      n=5 -0.544 eV
      n=6 -0.3778 eV
      See how the difference of energy decreases. That is why, more energy is released when there is a transition from n2 to n1 rather than from n6 to n2.
      Hope that helps.
      (15 votes)
  • aqualine ultimate style avatar for user Aditya Raj
    What is the relation between [(the difference between emission and absorption spectra) and (the difference between continuous and line/atomic spectra)]?
    (5 votes)
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    • blobby green style avatar for user Roger Taguchi
      Line spectra are produced when isolated atoms (e.g. in outer space or in high-vacuum tubes) emit or absorb only certain frequencies of energy (photons). Continuous spectra (absorption or emission) are produced when (1) energy levels are not quantized, but continuous, or (2) when zillions of energy levels are so close they are essentially continuous. The first occurs, for example, in plasmas like the Sun, where the temperatures are so high that the electrons are free to travel in straight lines until they encounter other electrons or positive ions. During these collisions, the electrons can gain or lose any amount of energy (within limits dictated by the temperature), so the spectrum is continuous (all frequencies or wavelengths of light are emitted or absorbed). The second case occurs in condensed states (solids and liquids), where the electrons are influenced by many, many electrons and nuclei in nearby atoms, and not just the closest ones. Because the electric force decreases as the square of the distance, it becomes weaker the farther apart the electric charged particles are, but there are many such particles, with the result that there are zillions of energy levels very close together, and transitions between all possible levels give rise to continuous spectra. Because solids and liquids have finite boiling points, the spectra of only a few (e.g. metals like tungsten, or oxides like cerium oxide in lantern mantles) include visible radiation. However, all solids and liquids at room temperature emit and absorb long-wavelength radiation in the infrared (IR) range of the electromagnetic spectrum, and the spectra are continuous, described accurately by the formula for the Planck black body spectrum. Infrared photons are invisible to the human eye, but can be felt as "heat rays" emitted from a hot solid surface like a cooling stove element (a red-hot stove or oven element gives off a small amount of visible light, red, but most of the energy emitted is in the infrared range). Hope this helps.
      (14 votes)
  • piceratops tree style avatar for user ANTHNO67
    My textbook says that there are 2 rydberg constant 2.18 x 10^-18 and 109,677
    my teacher says that rydberg found 2 constants 'probably'
    (3 votes)
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    • piceratops ultimate style avatar for user Just Keith
      They are related constants.
      The Rydberg constant is:
      R∞ = 109737 cm⁻¹ (though it is more often expressed in m⁻¹)

      The number you quoted is the Rydberg constant for hydrogen, which is derived from the standard Rydberg constant and more than a few books get a little sloppy about and just call the "Rydberg constant". Its symbol is RH (the H is a subscript).
      RH = R∞ [ MH / (me+MH) = 109677.58 cm⁻¹
      Where,
      MH (H is a subscript) is the mass of a proton
      me (e is a subscript) is the mass of an electron

      If you multiply R∞ by hc, then you get the Rydberg unit of energy, Ry, which equals 2.17987×10⁻¹⁸ J

      Thus, Ry is derived from RH.
      So, one of your numbers was RH and the other was Ry.

      NOTE: I rounded off R∞, it is known to a lot of digits.
      (12 votes)
  • purple pi purple style avatar for user Arushi
    Do all elements have line spectrums or can elements also have continuous spectrums?
    (4 votes)
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    • blobby green style avatar for user Roger Taguchi
      Atoms in the gas phase (e.g. in outer space or in high vacuum) have line spectra. However, atoms in condensed phases (solids or liquids) can have essentially continuous spectra. For example, the tungsten filaments in incandescent light bulbs give off all colours of the visible spectrum (although most of the electrical energy ends up emitted as infrared (IR) photons, explaining why tungsten filament light bulbs are only 5-10% energy efficient). The explanation comes from the band theory of the solid state: in metallic solids, the electronic energy levels of all the valence electrons form bands of zillions of energy levels packed really closely together, with the electrons essentially free to move from one to any other. All the possible transitions involve all possible frequencies, so the spectrum emitted is continuous. The band theory also explains electronic properties of semiconductors used in all popular electronics nowadays, so it is not BS.
      (6 votes)
  • duskpin ultimate style avatar for user Aquila Mandelbrot
    At , what is a Balmer Rydberg equation?
    (2 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      The Balmer-Rydberg equation or, more simply, the Rydberg equation is the equation used in the video.
      1/λ = R(1/i² -1/j²)
      It is usually written as
      1/λ = R(1/n₁² -1/n₂²), where n₁ < n₂.
      For the Balmer series of lines ( the visible lines in the hydrogen spectrum), n₁ = 2.
      So the Rydberg formula for the Balmer series of lines is
      1/λ = R(1/4 - 1/n₂²).
      (8 votes)
  • duskpin ultimate style avatar for user Advaita Mallik
    At -, Jay calls it a continuous spectrum. But, we know that the sun's light is created due to the emission spectra of different elements, mainly H, some He and even Lesser Li, along with trace amounts of Ca, Mg and a few others. While the sun's spectrum (so to call it) is continuous, Jay says at that H's spectrum would not be continuous, it'd be a line spectrum (as he says at ). I've also heard it being called a discrete spectrum. Why is it that though the same element creates the spectrum, one is continuous while the other isn't? Is it because the spectrum is made in different conditions (heat in case of the sun, while electricity in case of a spectrometer)? I don't see why that should happen, since energy is energy in whichever manner it is supplied, and because there is no way an electron would respond differently though it belonged to the nucleus of the same element. Please help. Thanks a lot.
    (3 votes)
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    • male robot hal style avatar for user Andrew M
      The discrete spectrum emitted by a H atom is a result of the energy levels within the atom, which arise from the way the electron interacts with the proton. To view the spectrum we need hydrogen in its gaseous form, so that the individual atoms are floating around, not interacting too much with one another.

      The sun is not just a big ball of H atoms loosely floating around. For one thing, the temperature is very high, so the atoms are ionized. In other words, you have a lot of H nuclei (ie protons) and a lot of electrons, and they aren't bound together to make a nice simple H atom. Also, you have enormous pressure, which forces all of the atoms to interact with one another in a way that they don't when they are in low-pressure gas form. The interactions between the particles create a different, much larger set of energy levels that permit a continuous rather than discrete spectrum. The sun radiates like a so-called "black body" with temperature around 5700K.

      When astronomers study distant stars they actually look at ABSORPTION spectra rather than emission spectra. The absorption spectrum is what's left after the white light passes through the outer layers of the star, where the pressure is lower and the protons can join with the electrons to make complete H atoms. The absorption spectrum is like the negative of the emission spectrum, it has dark lines where the emission spectrum has light. We identify the H in the star by seeing its characteristic absorption spectrum.
      (5 votes)
  • blobby green style avatar for user Rosalie Briggs
    What happens when the energy higher than the energy needed for an electron to jump to the next energy level is supplied to the atom? Does it not change its position at all, or does it jump to the higher energy level, but is very unstable?
    (1 vote)
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    • male robot hal style avatar for user Charles LaCour
      Nothing happens. For an electron to jump from one energy level to another it needs the exact amount of energy. The only exception to this is when the electron is freed from the atom and then the excess energy goes into the momentum of the electron.
      (5 votes)

Video transcript

- [Voiceover] I'm sure that most of you know the famous story of Isaac Newton where he took a narrow beam of light and he put that narrow beam of light through a prism and the prism separated the white light into all the different colors of the rainbow. And so if you did this experiment, you might see something like this rectangle up here so all of these different colors of the rainbow and I'm gonna call this a continuous spectrum. It's continuous because you see all these colors right next to each other. So they kind of blend together. So that's a continuous spectrum If you did this similar thing with hydrogen, you don't see a continuous spectrum. So, if you passed a current through a tube containing hydrogen gas, the electrons in the hydrogen atoms are going to absorb energy and jump up to a higher energy level. When those electrons fall down to a lower energy level they emit light and so we talked about this in the last video. This is the concept of emission. If you use something like a prism or diffraction grating to separate out the light, for hydrogen, you don't get a continuous spectrum. You'd see these four lines of color. So, since you see lines, we call this a line spectrum. So this is the line spectrum for hydrogen. So you see one red line and it turns out that that red line has a wave length. That red light has a wave length of 656 nanometers. You'll also see a blue green line and so this has a wave length of 486 nanometers. A blue line, 434 nanometers, and a violet line at 410 nanometers. And so this emission spectrum is unique to hydrogen and so this is one way to identify elements. And so this is a pretty important thing. And since line spectrum are unique, this is pretty important to explain where those wavelengths come from. And we can do that by using the equation we derived in the previous video. So I call this equation the Balmer Rydberg equation. And you can see that one over lamda, lamda is the wavelength of light that's emitted, is equal to R, which is the Rydberg constant, times one over I squared, where I is talking about the lower energy level, minus one over J squared, where J is referring to the higher energy level. For example, let's say we were considering an excited electron that's falling from a higher energy level n is equal to three. So let me write this here. So we have an electron that's falling from n is equal to three down to a lower energy level, n is equal to two. All right, so it's going to emit light when it undergoes that transition. So let's look at a visual representation of this. Now let's see if we can calculate the wavelength of light that's emitted. All right, so if an electron is falling from n is equal to three to n is equal to two, I'm gonna go ahead and draw an electron here. So an electron is falling from n is equal to three energy level down to n is equal to two, and the difference in those two energy levels are that difference in energy is equal to the energy of the photon. And so that's how we calculated the Balmer Rydberg equation in the previous video. All right, let's go ahead and calculate the wavelength of light that's emitted when the electron falls from the third energy level to the second. So, we have one over lamda is equal to the Rydberg constant, as we saw in the previous video, is one point zero nine seven times ten to the seventh. The units would be one over meter, all right? One over I squared. So, I refers to the lower energy level, all right? So the lower energy level is when n is equal to two. So we plug in one over two squared. And then, from that, we're going to subtract one over the higher energy level. That's n is equal to three, right? So this would be one over three squared. So one over two squared minus one over three squared. Let's go ahead and get out the calculator and let's do that math. So one over two squared, that's one fourth, so that's point two five, minus one over three squared, so that's one over nine. So, one fourth minus one ninth gives us point one three eight repeating. And if we multiply that number by the Rydberg constant, right, that's one point zero nine seven times ten to the seventh, we get one five two three six one one. So let me go ahead and write that down. So now we have one over lamda is equal to one five two three six one one. So to solve for lamda, all we need to do is take one over that number. So one over that number gives us six point five six times ten to the negative seven and that would now be in meters. So we have lamda is equal to six point five six times ten to the negative seventh meters. So let's convert that into, let's go like this, let's go 656, that's the same thing as 656 times ten to the negative ninth meters. And so that's 656 nanometers. 656 nanometers, and that should sound familiar to you. All right, so let's go back up here and see where we've seen 656 nanometers before. 656 nanometers is the wavelength of this red line right here. So, that red line represents the light that's emitted when an electron falls from the third energy level down to the second energy level. So let's go back down to here and let's go ahead and show that. So we can say that a photon, right, a photon of red light is given off as the electron falls from the third energy level to the second energy level. So that explains the red line in the line spectrum of hydrogen. So how can we explain these other lines that we see, right? So we have these other lines over here, right? We have this blue green one, this blue one, and this violet one. So if you do the math, you can use the Balmer Rydberg equation or you can do this and you can plug in some more numbers and you can calculate those values. So those are electrons falling from higher energy levels down to the second energy level. So let's go ahead and draw them on our diagram, here. So, let's say an electron fell from the fourth energy level down to the second. All right, so that energy difference, if you do the calculation, that turns out to be the blue green line in your line spectrum. So, I'll represent the light emitted like that. And if an electron fell from the fifth energy level down to the second energy level, that corresponds to the blue line that you see on the line spectrum. And then, finally, the violet line must be the transition from the sixth energy level down to the second, so let's go ahead and draw that in. And so now we have a way of explaining this line spectrum of hydrogen that we can observe. And since we calculated this Balmer Rydberg equation using the Bohr equation, using the Bohr model, I should say, the Bohr model is what allowed us to do this. So the Bohr model explains these different energy levels that we see. So when you look at the line spectrum of hydrogen, it's kind of like you're seeing energy levels. At least that's how I like to think about it 'cause you're, it's the only real way you can see the difference of energy. All right, so energy is quantized. We call this the Balmer series. So this is called the Balmer series for hydrogen. But there are different transitions that you could do. For example, let's think about an electron going from the second energy level to the first. All right, so let's get some more room here If I drew a line here, again, not drawn to scale. Think about an electron going from the second energy level down to the first. So from n is equal to two to n is equal to one. Let's use our equation and let's calculate that wavelength next. So this would be one over lamda is equal to the Rydberg constant, one point zero nine seven times ten to the seventh, that's one over meters, and then we're going from the second energy level to the first, so this would be one over the lower energy level squared so n is equal to one squared minus one over two squared. All right, so let's get some more room, get out the calculator here. So, one over one squared is just one, minus one fourth, so that's point seven five and so if we take point seven five of the Rydberg constant, let's go ahead and do that. So one point zero nine seven times ten to the seventh is our Rydberg constant. Then multiply that by point seven five, right? So three fourths, then we should get that number there. So that's eight two two seven five zero zero. So let's write that down. One over the wavelength is equal to eight two two seven five zero. So to solve for that wavelength, just take one divided by that number and that gives you one point two one times ten to the negative seven and that'd be in meters. So the wavelength here is equal to one point, let me see what that was again. One point two one five. One point two one five times ten to the negative seventh meters. And so if you move this over two, right, that's 122 nanometers. So this is 122 nanometers, but this is not a wavelength that we can see. So 122 nanometers, right, that falls into the UV region, the ultraviolet region, so we can't see that. We can see the ones in the visible spectrum only. And so this will represent a line in a different series and you can use the Balmer Rydberg equation to calculate all the other possible transitions for hydrogen and that's beyond the scope of this video. So, here, I just wanted to show you that the emission spectrum of hydrogen can be explained using the Balmer Rydberg equation which we derived using the Bohr model of the hydrogen atom. So even thought the Bohr model of the hydrogen atom is not reality, it does allow us to figure some things out and to realize that energy is quantized.