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  4. How can we see ultraviolet light in Balmer series?
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How can we see ultraviolet light in Balmer series?

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Offline Bored chemist

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Re: How can we see ultraviolet light in Balmer series?
« Reply #40 on: 03/01/2022 10:02:47 »
Quote from: hamdani yusuf on 03/01/2022 08:24:28
Quote from: Bored chemist on 02/01/2022 22:53:31
Quote from: hamdani yusuf on 02/01/2022 21:49:08
I guess it's significantly lower than melting point of glass.
Very roughly 10,000 degrees.
Hot enough to boil glass (and anything else).

Hydrogen discharge tubes are available on line. They produce Balmer spectrum without causing the glass to boil.
The glass is much much colder than the gas.
Quote from: hamdani yusuf on 03/01/2022 08:15:32
So, this diagram is misleading then?
It depends on the gas and other details.
Chlorine is a yellow colour; you would see absorption lines in the blue and violet regions of the spectrum.
Trifluoronitrosomethane is blue- you would see absorption lines from about 500 to 700 nm.
And hydrogen is colourless, so you see no absorption at all.

Whether or not you see an absorption or emission spectrum in the case marked "absorption spectrum of hydrogen atom" depends on how bright the white light source is.
But you would also see the spectra of other things like H2, H2+ H3+ etc
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Offline hamdani yusuf (OP)

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Re: How can we see ultraviolet light in Balmer series?
« Reply #41 on: 03/01/2022 12:47:54 »
Quote from: Bored chemist on 03/01/2022 10:02:47
The glass is much much colder than the gas.
How did you determine the temperature of the gas?
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Online evan_au

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Re: How can we see ultraviolet light in Balmer series?
« Reply #42 on: 04/01/2022 02:23:07 »
The Sun is not contained in glass - but there are different layers at different temperatures. This led to the discovery of:
- Helium (named after the ancient Greek Sun god Helios): An element which makes up about 25% of the Sun, but which had not been previously discovered and purified on Earth
- Other "new" elements which turned out to be known elements, but at extreme temperatures and low pressures that had never been produced on Earth (eg highly ionised iron atoms in the Sun's corona, at a temperature of millions of degrees)
« Last Edit: 04/01/2022 10:40:19 by evan_au »
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Offline hamdani yusuf (OP)

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Re: How can we see ultraviolet light in Balmer series?
« Reply #43 on: 04/01/2022 04:17:05 »
Quote from: Bored chemist on 28/12/2021 12:11:49
The definition of "ultraviolet" is a bit flexible.
OK. We can read it in another Wikipedia article that the limits of visible spectrum may be different among individuals and conditions.
Quote
https://en.wikipedia.org/wiki/Visible_spectrum
A typical human eye will respond to wavelengths from about 380 to about 750 nanometers.[1] In terms of frequency, this corresponds to a band in the vicinity of 400–790 terahertz. These boundaries are not sharply defined and may vary per individual.[2] Under optimal conditions these limits of human perception can extend to 310 nm (ultraviolet) and 1100 nm (near infrared).[3][4] The optical spectrum is sometimes considered to be the same as the visible spectrum, but some authors define the term more broadly, to include the ultraviolet and infrared parts of the electromagnetic spectrum as well.[5]
I think the most probable reason to set the limit at 400 nm as quoted in the first post is due to rounding value.
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Offline Bored chemist

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Re: How can we see ultraviolet light in Balmer series?
« Reply #44 on: 04/01/2022 08:31:11 »
Quote from: hamdani yusuf on 03/01/2022 12:47:54
How did you determine the temperature of the gas?
Doppler broadening and also the spectrum. The hotter the material, the more shorter wavelength radiation it produces
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Offline hamdani yusuf (OP)

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Re: How can we see ultraviolet light in Balmer series?
« Reply #45 on: 04/01/2022 12:18:59 »
Quote from: Bored chemist on 04/01/2022 08:31:11
Quote from: hamdani yusuf on 03/01/2022 12:47:54
How did you determine the temperature of the gas?
Doppler broadening and also the spectrum. The hotter the material, the more shorter wavelength radiation it produces
How does the glass of the discharge lamp withstand the high temperature? Why doesn't it evaporate when touching the glowing hydrogen plasma?
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Offline Bored chemist

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Re: How can we see ultraviolet light in Balmer series?
« Reply #46 on: 04/01/2022 15:52:24 »
Quote from: hamdani yusuf on 04/01/2022 12:18:59
Quote from: Bored chemist on 04/01/2022 08:31:11
Quote from: hamdani yusuf on 03/01/2022 12:47:54
How did you determine the temperature of the gas?
Doppler broadening and also the spectrum. The hotter the material, the more shorter wavelength radiation it produces
How does the glass of the discharge lamp withstand the high temperature? Why doesn't it evaporate when touching the glowing hydrogen plasma?
Because it is being cooled by the air outside, but only heated by a near vacuum inside.
Also, have a good look at a discharge tube. The light is emitted from the plasma at the centre of the tube, not the cool bit near the glass.
Gases are poor conductors of heat.
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Re: How can we see ultraviolet light in Balmer series?
« Reply #47 on: 18/01/2022 11:33:33 »
I learned that the gas could turn into plasma when exposed to microwave, or high AC voltage from Tesla coil. But they are relatively low frequency radiation compared to what causes photoelectric effect on metals.
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Re: How can we see ultraviolet light in Balmer series?
« Reply #48 on: 18/01/2022 13:52:57 »
Frequency isn't the criterion: energy transfer is what matters. Most common gases require about 15 - 30 eV to ionise because the electrons are tightly bound to the molecules, but electrons in a solid metal are considerably freer and can be ejected by visible or ultraviolet photons from 2 eV upwards. The initiating mechanisms of "field" (static or alternating voltage) versus "impact" (photon) ionisation are different and often confused because we use static-field ion cascade multiplication (geiger or proportional counters) to detect single photon events.
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Offline hamdani yusuf (OP)

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Re: How can we see ultraviolet light in Balmer series?
« Reply #49 on: 18/01/2022 21:33:44 »
Quote from: alancalverd on 18/01/2022 13:52:57
Frequency isn't the criterion: energy transfer is what matters. Most common gases require about 15 - 30 eV to ionise because the electrons are tightly bound to the molecules, but electrons in a solid metal are considerably freer and can be ejected by visible or ultraviolet photons from 2 eV upwards. The initiating mechanisms of "field" (static or alternating voltage) versus "impact" (photon) ionisation are different and often confused because we use static-field ion cascade multiplication (geiger or proportional counters) to detect single photon events.
What determines the energy transfered?
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Re: How can we see ultraviolet light in Balmer series?
« Reply #50 on: 20/01/2022 14:33:32 »
Thanks for that one Alan.

" ! Not sure about extended violet sensitivity, but with the cataracts gone, I'm amazed how much more air traffic there is nowadays. "
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Re: How can we see ultraviolet light in Balmer series?
« Reply #51 on: 09/02/2022 13:51:15 »
In atomic physics, the Balmer series (or Balmer lines) is one of six named series that describe the spectral line emissions of the hydrogen atom. The Balmer series is determined using the Balmer formula, which was established by Johann Balmer in 1885 and is an empirical equation.
Four wavelengths in the visible spectrum of hydrogen light, 410 nm, 434 nm, 486 nm, and 656 nm, corresponding to photon emissions by excited electrons transitioning to the quantum level defined by the primary quantum number n = 2. With wavelengths less than 400 nm, there are numerous noteworthy UV Balmer lines. As the UV spectrum approaches a limit of 364.5 nm, the number of these lines becomes an endless continuum. The model also clarifies the Balmer formula for hydrogen spectral lines. In the Bohr formula, light energy is the difference in energies between the two orbits. The Balmer series in a hydrogen atom connects the wavelength of the emission seen by scientists to the probable electron transitions down to the n = 2 positions. When electrons travel between different energy levels surrounding the atom (defined by the primary quantum number, n ), they either emit or absorb a photon, according to quantum physics. The Balmer series defines the wavelengths of emitted photons as well as transitions from higher energy levels to the second energy level. The Rydberg formula can be used to compute this.
The Balmer series' "visible" hydrogen emission spectrum lines. The red line on the right is H-alpha. The viewable range consists of four lines (counting from the right). Lines 5 and 6 are visible to the human eye, although they are classified as ultraviolet since their wavelengths are shorter than 400 nm.
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Re: How can we see ultraviolet light in Balmer series?
« Reply #52 on: 09/02/2022 18:52:04 »
Quote from: Iannguyen on 09/02/2022 13:51:15
In atomic physics, the Balmer series (or Balmer lines) is one of six named series that describe the spectral line emissions of the hydrogen atom. The Balmer series is determined using the Balmer formula, which was established by Johann Balmer in 1885 and is an empirical equation.
Four wavelengths in the visible spectrum of hydrogen light, 410 nm, 434 nm, 486 nm, and 656 nm, corresponding to photon emissions by excited electrons transitioning to the quantum level defined by the primary quantum number n = 2. With wavelengths less than 400 nm, there are numerous noteworthy UV Balmer lines. As the UV spectrum approaches a limit of 364.5 nm, the number of these lines becomes an endless continuum. The model also clarifies the Balmer formula for hydrogen spectral lines. In the Bohr formula, light energy is the difference in energies between the two orbits. The Balmer series in a hydrogen atom connects the wavelength of the emission seen by scientists to the probable electron transitions down to the n = 2 positions. When electrons travel between different energy levels surrounding the atom (defined by the primary quantum number, n ), they either emit or absorb a photon, according to quantum physics. The Balmer series defines the wavelengths of emitted photons as well as transitions from higher energy levels to the second energy level. The Rydberg formula can be used to compute this.
The Balmer series' "visible" hydrogen emission spectrum lines. The red line on the right is H-alpha. The viewable range consists of four lines (counting from the right). Lines 5 and 6 are visible to the human eye, although they are classified as ultraviolet since their wavelengths are shorter than 400 nm.

Why have you paraphrased the wiki page?
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Offline hamdani yusuf (OP)

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Re: How can we see ultraviolet light in Balmer series?
« Reply #53 on: 10/02/2022 02:39:20 »
Does anyone know how to derive Balmer series from Schrodinger's wave equation?
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Offline Bored chemist

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Re: How can we see ultraviolet light in Balmer series?
« Reply #54 on: 10/02/2022 08:40:20 »
Quote from: hamdani yusuf on 10/02/2022 02:39:20
Does anyone know how to derive Balmer series from Schrodinger's wave equation?
The internet knows.
https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/4%3A_Quantum_Theory/4.10%3A_The_Schr%C3%B6dinger_Wave_Equation_for_the_Hydrogen_Atom
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Offline hamdani yusuf (OP)

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Re: How can we see ultraviolet light in Balmer series?
« Reply #55 on: 15/02/2022 10:29:51 »
Quote
https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/4%3A_Quantum_Theory/4.10%3A_The_Schr%C3%B6dinger_Wave_Equation_for_the_Hydrogen_Atom
It is interesting to compare the results obtained by solving the Schrödinger equation with Bohr’s model of the hydrogen atom. There are several ways in which the Schrödinger model and Bohr model differ.

  • First, and perhaps most strikingly, the Schrödinger model does not produce well-defined orbits for the electron. The wavefunctions only give us the probability for the electron to be at various directions and distances from the proton.
  • Second, the quantization of angular momentum is different from that proposed by Bohr. Bohr proposed that the angular momentum is quantized in integer units of ℏ , while the Schrödinger model leads to an angular momentum of d68efabc562f44c7a8ccef7a40905306.gif.
  • Third, the quantum numbers appear naturally during solution of the Schrödinger equation while Bohr had to postulate the existence of quantized energy states. Although more complex, the Schrödinger model leads to a better correspondence between theory and experiment over a range of applications that was not possible for the Bohr model.

I wonder how different values of angular momenta in different models can lead to the same result in hydrogen spectrum.
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Offline hamdani yusuf (OP)

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Re: How can we see ultraviolet light in Balmer series?
« Reply #56 on: 15/02/2022 10:35:53 »
I think this animation can help us understand the problem instead of just memorizing formulas.

« Last Edit: 15/02/2022 10:46:34 by hamdani yusuf »
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