# Question re: Deuterium Emission Spectrum

• Anti-Crackpot
In summary, the Deuterium Emission Spectrum is a graph that shows the wavelengths of light emitted by deuterium gas when it is heated to high temperatures. This spectrum is used to study the atomic structure of deuterium and can provide valuable information about its energy levels and transitions between them. It is an essential tool in the field of spectroscopy and can also be used to identify and measure the abundance of deuterium in different substances.
Anti-Crackpot
Deuterium Spectrum (nm) 410.07, 433.93, 486.01, 656.11

I came upon these figures on a commercial product site for Deuterium lamps. So, question: Are they correct? If so, then...

364.5068222*(3^2/(3^2 - 4)) = 656.11
364.5068222*(4^2/(4^2 - 4)) = 486.01
364.5068222*(5^2/(5^2 - 4)) = 433.93
364.5068222*(6^2/(6^2 - 4)) = 410.07

364.5068222... = 4 / .010973731568539 where 1.0973731568539*10^7 m^-1 is the Rydberg Constant (R). Yes, I know 4/R is supposed to equal B, the Balmer constant, but I also know there is a Balmer_H, for Hydrogen, which is different from 4/R exactly, so it seems to me this is a bit of an odd coincidence and I'm not quite sure what to make of it.

Compounding my confusion is that, empirically, here is how you achieve a match with light hydrogen Fraunhofer lines in the visible spectrum:

364.6006*(3^2/(3^2 - 4)) = 656.281 = H_alpha
364.6006*(4^2/(4^2 - 4)) = 486.134 = H_beta
364.6006*(5^2/(5^2 - 4)) = 434.048 ~ H_gamma (vs. 434.047)
364.6006*(6^2/(6^2 - 4)) = 410.176 ~ H_delta (vs. 410.175)

4 / 364.6006 = 0.010970909, but I can't seem to find 1.097090 * 10^7 m^-1 anywhere in the literature. The figure I have seen for R_H is ~ 1.09678 * 10^7 m^-1.

In trying to sort this out I'm thinking the first step is simply to know if I am actually dealing with the right figures for the deuterium emission spectrum as per the question in the first paragraph and/or how they are mathematically derived. So any help there would be greatly appreciated.

TIA,
AC

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Have a look here:
http://en.wikipedia.org/wiki/Rydberg_constant

Specifically you should use ##R_M## instead of ##R_\infty## for precise values.
However ##R_M## is different for H and D.
The formulas are given in the Wikipedia article, so you can check yourself.

In response to Dr. Du's feedback, and on the basis of CODATA 2010 values...

9.10938291*10^-31 kg: electron mass
1.672621777*10^-27 kg: proton mass
1.674927351*10^-27 kg: neutron mass
1.0973731568539 * 10^7 m^-1: Rydberg Constant

... here is what I calculate to be the first five lines in the Protium and Deuterium spectrums.

PROTIUM
4/((1.0973731568539*10^7)/(1 + 9.10938291*10^-31/1.672621777*10^-27)) = 3.64705336839 * 10^-7 = Balmer_H (B_H)

4/(3.64705336839 * 10^-7) = 1.096775834066 * 10^7 = Rydberg Constant for Protium (R_H)

Wavelengths (in nanometers)
364.705336839*(n^2/(n^2 - 4))
656.469, 486.273, 434.173, 410.293, 397.123
n = 3, 4, 5, 6, 7

DEUTERIUM
4/((1.0973731568539*10^7)/(1 + 9.10938291*10^-31/(1.672621777*10^-27 +1.674927351*10^-27) + 1.674927351*10^-27)) = 3.646060101671 * 10^-7 = Balmer_D (B_D)

4/(3.646060101671 * 10^-7) = 1.097074619852 * 10^7 = Rydberg Constant for Deuterium (R_D)

Wavelengths (in nanometers)
364.6060101671*(n^2/(n^2 - 4))
656.290, 486.141, 434.054, 410.181, 397.015
n = 3, 4, 5, 6, 7

- AC

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So, the next question, assuming I have calculated correctly above, is this:

How does one work into the equations the refractive index of air? Does one. for instance, simply divide wavelength by ~1.00029 to get observed values? e.g. 656.281 = H_alpha = 656.469 / 1.00028646 = 656.281.

H-alpha
http://en.wikipedia.org/wiki/H-alpha
Fraunhofer lines
http://en.wikipedia.org/wiki/Fraunhofer_linesTIA,
AC

P.S. @ Dr. Du. Thank you for your response.

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Well, getting closer to getting a fix on this (maybe), but not sure if I am theorizing correctly that the difference between theoretical values and observed values is due to the refractive index of air. Any help would be appreciated. In the meantime a bit of data...

Using Balmer_H (B_H) calculated two posts previously based on R_M, and then dividing by the approximate refractive index of air we get:

PROTIUM OBSERVED VALUES
See: Fraunhofer lines http://en.wikipedia.org/wiki/Fraunhofer_lines
364.705336839*(n^2/(n^2 - 4))/1.00028809
~ 656.469 / 1.00028809 = 656.281 vs. 656.281 calculated in initial post --> H_alpha
~ 486.273 / 1.00028809 = 486.134 vs. 486.134 calculated in initial post --> H_beta
~ 434.173 / 1.00028809 = 434.048 vs. 434.048 calculated in initial post --> H_gamma
~ 410.293 / 1.00028809 = 410.175 vs. 410.176 calculated in initial post --> H_delta
~ 397.123 / 1.00028809 = 397.009
364.705336839/1.00028809 = 364.600299 ~ Corrected Balmer_H

DEUTERIUM OBSERVED VALUES
Using Balmer_D (B_D) calculated two posts previously based on R_M, and then dividing by the approximate refractive index of air we get:
364.6060101671*(n^2/(n^2 - 4))/1.00028809
~ 656.290 / 1.00028809 = 656.102 vs. 656.11 found value in initial post (product site)
~ 486.141 / 1.00028809 = 486.001 vs. 486.01 found value in initial post (product site)
~ 434.054 / 1.00028809 = 433.930 vs. 433.93 found value in initial post (product site)
~ 410.181 / 1.00028809 = 410.064 vs. 410.07 found value in initial post (product site)
~ 397.015 / 1.00028809 = 396.901
364.6060101671/1.00028809 = 364.501001 ~ Corrected Balmer_D

Rounded to 5 digits, these new values: 656.10, 486.00, 433.93, 410.06 compare very favorably to the figures found here...
Physics 439 – Lab in Modern Physics: Atomic Spectra Experiment (McGill University 2006, p. 11)

In fact, the accord is 100%. Only problem is that the refractive index of air, from what I can seem to tell, is 1.000293, not 1.000288. That's close, but still not quite the cigar. What am I missing?

- AC

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Correcting for the refractive index of air is perfectly sensible as you've done it - it's the same sort of initial correction one would consider when examining data taken in from a spectrometer operating at SATP versus data obtained under high-vacuum conditions.

Practically speaking, I could imagine that there is a small residual impurity in the deuterium found in the arc lamp that is causing the perturbation, e.g., it's 99.8% D2 and the remainder is HD, with perhaps a trace of H2. I am not extremely familiar with the manufacture of deuterium lamps or with isotopically enriched gases, but my recollection - from years of browsing catalogs from isotope vendors - is that there is always a small impurity in their gases (as in their dry chemicals, naturally).

If it is really that much of a concern, I would contact the company directly. I have irritated countless applications scientists and engineers at numerous companies over the years tracking down discrepancies and clarifying the unclear - it is part of their job, after all.

The only problem is that when you specify the wavelength of a radiation then it is the wavelength in vacuo.
I suppose the slight deviations are due to other effects, most possibly predominantly corrections for the finite size of the nucleus.

In your calculation of the Rydberg constant for Deuterium you should use the mass of the deuterium nucleus, not the sum of the mass of the neutron and the proton. The two quantities do not coincide, a deuterium nucleus is lighter.

@ Mike and Dr. Du. Thanks for the responses. I re-ran the numbers with the correct Deuteron mass and the change was negligible.

Here is a link to a set of data from the Kitt Peak National Observatory using the correction of 1.000288 to correct angstrom measurements in vacuo to angstron measurements in air.
http://www.noao.edu/kpno/manuals/flmn/OH_lines.txt

I also came across the same figure in the following paper:
An optical mechanism for aberration of starlight, Robert A. Woodruff (9/28/2011)
http://arxiv.org/pdf/1110.4788.pdf

So it seems, for instance, that when H_alpha is cited as 656.281 nm, then it is the corrected for air figure that is being cited, as opposed to the in vacuo figure which is more along the lines of 1.000288*656.281 nm = 656.47 nm. See: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydfin.html .

- AC

Interesting! Thank you.

For anyone who comes along this way with a similar question or questions as I had when I began this thread, here, for better or for worse, is a summary of my recent calculations. Corrections or improvements to the data most welcome.

Based on CODATA 2010 values...
9.10938291(40)*10^-31 kg = electron mass
1.672621777(74)*10^-27 kg = proton mass
3.34358348(15)*10^-27 kg = deuteron mass
1.0973731568539(55) * 10^7 m^-1 = Rydberg Constant
1.00028809 ~ Correction for refractive index of air (empirical best fit)

PROTIUM - in vacuo
4/((1.0973731568539*10^7 m^-1)/(1 + (9.10938291*10^-31 kg/1.672621777*10^-27 kg))) = 3.64705336839 * 10^-7 m= Balmer_H = B_H
4/(3.64705336839 * 10^-7 m) = 1.096775834066 * 10^7 m^-1 = Rydberg Constant for Protium = R_H

Wavelengths (in nanometers)
364.705336839*(n^2/(n^2 - 4)) =
656.469, 486.273, 434.173, 410.293, 397.123
n = 3, 4, 5, 6, 7

DEUTERIUM - in vacuo
4/((1.0973731568539*10^7 m^-1)/(1 + (9.10938291*10^-31 kg/3.34358348*10^-27 kg))) = 3.646061278111 * 10^-7 m = Balmer_D = B_D
4/(3.646061278111 * 10^-7 m) = 1.097074265870 * 10^7 m^-1 = Rydberg Constant for Deuterium = R_D

Wavelengths (in nanometers)
364.6060101671*(n^2/(n^2 - 4)) =
656.290, 486.141, 434.054, 410.181, 397.015
n = 3, 4, 5, 6, 7

PROTIUM - in air
3.64705336839*10^-7 m/1.00028809 = 3.646003*10^-7 m = Balmer_H(air) = B_H(air)
4/(3.64705336839*10^-7 m/1.00028809) = 1.0970918042162 * 10^7 m^-1 = Rydberg Constant for Protium in air = R_H(air)

Wavelengths (in nanometers)
364.705336839*(n^2/(n^2 - 4))/1.00028809 =
~ 656.469/1.00028809 = 656.281
~ 486.273/1.00028809 = 486.134
~ 434.173/1.00028809 = 434.048
~ 410.293/1.00028809 = 410.175
~ 397.123/1.00028809 = 397.009
n = 3, 4, 5, 6, 7

DEUTERIUM - in air
3.646060101671*10^-7 m/1.00028809 = 3.645010*10^-7 m = Balmer_D(air) = B_D(air)
4/(3.646060101671*10^-7 m/1.00028809) = 1.0973906760797 * 10^7 m^-1 = Rydberg Constant for Deuterium in air = R_D(air)

Wavelengths (in nanometers)
364.6060101671*(n^2/(n^2 - 4))/1.00028809 =
~ 656.290/1.00028809 = 656.102
~ 486.141/1.00028809 = 486.001
~ 434.054/1.00028809 = 433.930
~ 410.181/1.00028809 = 410.064
~ 397.015/1.00028809 = 396.901
n = 3, 4, 5, 6, 7

- AC

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DrDu said:
In your calculation of the Rydberg constant for Deuterium you should use the mass of the deuterium nucleus, not the sum of the mass of the neutron and the proton. The two quantities do not coincide, a deuterium nucleus is lighter.

Why is it that the deuteron mass is less than the sum of it's parts? Or am I framing the question incorrectly?

- AC

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Anti-Crackpot said:
Why is it that the deuteron mass less than the sum of it's parts? Or am I framing the question incorrectly?

- AC

There is a mass defect due to the nuclear binding energy (see, for example, these notes on this topic).

Otherwise, I just wanted to mention that the reason I suggested that there might need to be an air/vacuum correction is that deuterium lamps are often sold as a UV source for conventional non-vacuum-system spectrophotometers (most of which are accurate to anywhere between 0.1 nm to 0.5 nm, in my experience, depending on the application - which is well above the discrepancies that seem to be under discussion here).

Anti-Crackpot said:
Why is it that the deuteron mass is less than the sum of it's parts? Or am I framing the question incorrectly?

- AC

The binding energy released corresponds to a mass defect by ##E=\Delta mc^2##.
The enormous binding energy of nucleons is the reason for nuclear fusion generating so much energy e.g. in the sun.

Thank you again for the feedback Mike and Dr. Du. Interesting stuff and a clear example of how little things can add up to a lot. I can't help but wonder if, even without Einstein, one could have eventually deduced e=mc^2 from measurements of mass had, of course, those measurements been precise enough. In any case, a great jumping off point for me to learn more.

On another note, I'd like to share with you a mathematical coincidence I came across in the process of running these numbers. It's related to one you can find here involving the speed of light and the Rydberg constant:

Mathematical coincidence
http://en.wikipedia.org/wiki/Mathematical_coincidence

$\sqrt{\frac{\pi^2 * 10^{15} Hz}{3cR_{H(air)}}} = \frac{Spin_{2/2}}{Spin_{4/2}}\sqrt{\frac{\pi^2 * 10^{15} Hz}{cR_{H(air)}}} ≈ n_{(hydrogen)}$

$\sqrt{\frac{\pi^2 * 10^{15} Hz}{3*299792458m/s*1.0970918042162 * 10^7 m^{-1}}} = 1.0001321967531...$

... which rounds to 1.000132, the refractive index of hydrogen, or roughly the square root of the dielectric constant (relative permittivity) for hydrogen, 1.000264.

List of Refractive Indices
http://en.wikipedia.org/wiki/List_of_refractive_indices
Properties of Hydrogen (gives figure of 1.000264, Boltzmann and Clemencic)
http://www.bnl.gov/magnets/staff/gupta/cryogenic-data-handbook/Section3.pdf

Do either of you see any way this could be more than coincidence? For instance, is there any correction one could surmise in relation to the hydrogen emission spectrum related to what I'll call, for lack of a better term, "self-diffraction"? I realize it's a stretch, and I've not a clue how to justify it, but I found the coincidence rather odd in context.

- AC

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To anti-crackpot:

Your figure for R-infinity at 1.09 73731 6 * 10^7 m^-1 is correct. To get the figure for real hydrogen you correct for the reduced mass ratio of a proton/proton+electron, so you have to multiply by 1836.1/1837.1 which gives you the 1.09 6775 or 1.09 68 figure you give. Why this isn't the same as the reported Fraunhaufer H-alpha 4/364.006 I don't know. It should be.

The figures you have for deuterium lines (all a little shorter wavelength than H) are correct. To derive them from the corresponding hydrogen lines you multiply by the reduced mass for H to get to R-infinity, then use the reduced mass for D. This is a factor of [1837.1/1836.1] * [3670/3671], which all turns out to be 1.000272, the factor by which D frequencies and energies are larger. For the wavelengths, of course, you DIVIDE by that, so wavelengths are all a bit shorter. If deuterium had a Rydberg constant it would be R-infinity(3670/3671) = 1.09 70742 * 10^7 m^-1.

## 1. What is the Deuterium Emission Spectrum?

The Deuterium Emission Spectrum is a series of spectral lines that are emitted by the element deuterium when it is excited by energy. These lines are unique to deuterium and can be used to identify its presence in a sample.

## 2. How is the Deuterium Emission Spectrum produced?

The Deuterium Emission Spectrum is produced when deuterium atoms are excited by energy, either through heating or passing an electrical current through a gas containing deuterium. This causes the electrons in the deuterium atoms to jump to higher energy levels, and when they return to their ground state, they emit light at specific wavelengths.

## 3. What is the significance of the Deuterium Emission Spectrum?

The Deuterium Emission Spectrum is significant because it allows scientists to identify the presence and concentration of deuterium in a sample. This is important in various fields, such as chemistry, astrophysics, and nuclear physics, where deuterium is used as a tracer or a fuel source.

## 4. How is the Deuterium Emission Spectrum different from the Hydrogen Emission Spectrum?

The Deuterium Emission Spectrum is different from the Hydrogen Emission Spectrum because deuterium has an extra neutron in its nucleus compared to hydrogen. This results in a slightly different energy level structure and therefore, different wavelengths of light are emitted in the two spectra.

## 5. Can the Deuterium Emission Spectrum be used to study other elements?

Yes, the Deuterium Emission Spectrum can be used as a reference spectrum for studying other elements. Since the energy levels of deuterium are well-known and consistent, they can be used as a standard for comparing and identifying the spectral lines of other elements. This is known as spectroscopy and is a common technique used in many scientific fields.

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