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The Sun is currently turning hydrogen into helium. There are no other nuclear reactions taking place at any significant rate in the Sun. The Sun will not start to make heavier elements until it reaches the tip of the red giant branch in about 7 billion years time.
The elements heavier than helium that are present in the Sun were almost all made inside other ...
14
No. There is no consensus.
The discrepancy between the predicted big bang nucleosynthetic abundance of Lithium 7 and the measured value can be summarised as follows.
If we take what we know about the the baryonic mass density of the universe and the Hubble constant, we get a self-consistent picture between the cosmic microwave background, observations of ...
12
H$\delta$ absorption is formed when hydrogen in the level $n=2$ is excited to $n=6$.
To get strong H$\delta$ absorption lines you need large amounts of hydrogen in the first excited state $n=2$ and a radiation field that contains large numbers of photons with an energy equal to the difference between the $n=6$ and $n=2$ states.
These requirements are ...
12
The photosphere of the sun does produce an emission spectrum (a Planck spectrum according to its temperature of about 6000K). It is only that the atmosphere above the photosphere (the chromosphere) scatters light out of the line of sight at those frequencies where the scattering cross section is very high (i.e. at the atomic resonance frequencies of a given ...
11
The Ca triplet in the near infrared are extremely strong resonance absorption lines. They are by far the strongest features in the near infrared spectra of cool G,K,M type dwarfs and giants, which will be the majority of the stars observed by the Gaia RVS.
The Ca triplet lines are so strong that even in low metallicity halo stars, that have little Ca in ...
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Here is a link to a diffraction grating that can do what you want. It is mounted in a 1.25" filter ring that attaches to an eyepiece, or to most astro cameras. I believe they also sell adapters for other cameras and software to extract spectra from the photos.
Here is a link to a spectra of Vega that I took with a 120 mm refractor using the Star Analyser ...
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Close, but not quite right - the blue light is indeed emission from CO$^+$, but it's from the CO$^+$ ions themselves, with no need for recombination to CO; that (ionized) molecule has a strong set of energy transitions around 425 nm (4250 Angstroms), which is in the blue part of the visible spectrum:
Spectrum of Comet C/2016 R2 (Pan-STARRS), Figure 2 from ...
10
This is a rather broad question and this will not be a fully comprehensive answer.
There is no single temperature to the solar corona. The coronal temperature varies by an order of magnitude from place-to-place. It is hottest ($\sim 10^{7}$ K) in magneticloops undergoing flares, which tend to be anchored in low latitude regions. It is coolest (a bit less ...
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Looks like you can measure the cosmological redshift of quasars using that equipment and an 14" reflector:
http://www.rspec-astro.com/sample-projects/
(halfway down the page)
So the answer appears to be yes, you can do it, assuming that page isn't a total fabrication. Seems plausible to me: the redshift of that quasar is sizable at 0.15 (though that's ...
9
I suspect that the record holder (as of 14/2/2017) is HD 10180 which has at least 7 planets and possible evidence for as many as 9.
Lovis et al. (2011) announced the initial discovery based on 190 radial velocity measurements taken over 6 years. The precision of the measurements was 0.3-0.9 m/s.
Section 4 of that paper describes how they go about finding ...
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As your question is based on the plot you posted, I suggest you to look for a lower wavelength range of the atmospheric electromagnetic absorption. A quick search in google gave me this paper, which says:
The importance of molecular nitrogen as the most abundant species in the Earth's atmosphere is evident. The strong absorption bands in the range 80–100 nm ...
8
You are correct that the characteristic emission and absorption lines we see in stars' spectra are from electrons that are bound to atoms making transitions between different energy levels. That is possible because the elements in a star's photosphere are not fully ionized. Hydrogen - the easiest element to fully ionize because its nucleus only has a ...
8
Essentially what they did was assume that normally when observing with their telescope the spectral absorptions they see are due to the Earth's atmosphere. Which is a pretty good assumption. They then normalize the data to those absorption, so if there was any phosphine gas within the package of atmosphere they are looking through, it will be taken into ...
7
The ESA states it pretty clearly (although their figure of 855.2 nm is incorrect; it should be 866.2 nm):
The RVS wavelength range, 847-874 nm, has been selected to coincide with the energy-distribution peaks of G- and K-type stars which are the most abundant RVS targets. For these late-type stars, the RVS wavelength interval displays, besides numerous ...
7
Light that is not light
That's meaningless. All light is electromagnetic radiation. A finite part of the infinitely large range of the electromagnetic spectrum is visible light. So you should talk about EM radiation and to discuss the visible spectrum just say visible spectrum.
Stars emit lots of energy at frequencies that are outside the visible range.
...
7
You need to compare it with the spectrum of a similar galaxy at a known redshift, that would probably enable you to identify features with known rest wavelengths.
If you can find such a template, then the best way of estimating a redshift for a galaxy spectrum like this, consisting of mostly weak and blended absorption features, is to cross-correlate your ...
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Eric Jensen has already provided a nice link to a description of the basic structure of the ${\rm CO}_{2}$ spectrum, so I'll focus on the question of why there's a "spike" at 15.0 microns in the Earth spectrum, but not in the Venus or Mars spectra.
If you look at the link in Eric's answer, the very first image shows a high-resolution version of the ...
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The hotter layers above the solar photosphere do have an emission spectrum. The emission spectrum is much fainter than the visible photosphere and so is not easily seen through broadband filters in the optical spectrum, though it can be observed through very narrow filters centred on the emission lines (e.g. H$\alpha$ from the chromosphere) in question.
The ...
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All this means is that you need to bin your spectra in equal intervals of log wavelength for each pixel to be a constant interval in velocity.
First consider the case whereeachpixel is worth a constant interval in linear wavelength. Here we have
$$ \frac{\Delta \lambda}{\lambda} = \frac{\Delta v}{c},$$
and the $\Delta v$ represented by each pixel depends on ...
6
According to Cropper and Katz 2011 part 2.2,
the RVS working group considered other bands,
but the ~850 nm band is relatively unaffected by absorption in the Earth's atmosphere,
facilitating ground-based preparation and follow-up.
In addition to the strong Ca II triplet,
this band is rich in lines enabling study of astrophysical quantities other than radial ...
6
No, the blue and red shift of stars is not possible to detect with the naked eye. There are a couple of reasons for this. First the effect is slight: even for an object moving at thousands of km/s (way faster than stars in the milky way move), it is only a tiny change in hue. Next, since the light of stars is a mixture of many different colours (a black body ...
6
This was a coronal mass ejection.
Those 1973 astronomers weren't looking at the picture correctly. They didn't have the tools at that time to look at the picture correctly. Coronal mass ejections (the term used now) were only discovered a couple of years prior to that picture taken from Skylab, via the Orbiting Solar Observatory 7 satellite. Those early ...
6
When we look at spectral lines in a star's spectrum, we're actually looking for absorption lines, not emission lines. A star's spectrum usually resembles that of a black body, continuous and smooth. However, there are elements in the star's atmosphere that absorb some of the emitted light; these create characteristic absorption lines in the spectrum we ...
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What you're missing is that the resolution of a prism isn't high enough to resolve the relatively narrow spectral lines. What's more, the light that gets generated by fusion reactions doesn't reach the surface of the sun for a very long time, and it get's scattered and split a large number of times along the way, removing any signature of fusion from their ...
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Under conditions of very high temperature and very low pressure forbidden conditions can occur, they shift the color of the excitation radiation from the ultraviolet into the visible spectrum giving nebula their beautiful colors. These are represented using brackets.
See Wikipedia's Selection Rules, section Angular Momentum:
"Semi-forbidden transitions (...
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In general, you can't. If obtaining spectra in regions where there is expected to be a spatially varying background then you either need to do long-slit spectroscopy so that you have a good measurement of the ISM contribution either side of your source, or you do integral field spectroscopy with the same idea.
The problem is that the line strengths for the ...
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You can probably get most if not all of your questions answered by perusing the main DESI web site, which I encourage you to check out.
There is, for example, a nice video describing the assembly of the main focal plane elements (the fibers and the associated robot positioners) here.
But in simple terms: the circular focal plane is divided into ten wedges (...
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It appears to be a conventional label that is applied to transitions between the ground state and another energy level (some definitions specify the first excited level) of an atom and is used in all the physical sciences, not just astrophysics.
e.g. He I (58.4 nm) is a transition from $^1$P to the $^1$S ground state.
In fact all atomic/ionic transitions can ...
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As @ELNJ answer pointed out, fully ionized the atoms at the star surface are not. It is not hot enough. Star cores are another case, but we usually don't see them. There, both pressure and temperature make impossible the existence of the usual atoms.
Atoms and molecules usually emit their characteristic wavelengths because of the electrons' energy levels...
...
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