We’re rewarding the question askers & reputations are being recalculated! Read more.
74

Based on your description, I may have found the article you originally saw, or at least one very similar. Researchers from Dartmouth College published a paper$\mathrm{^1}$ in which they report, among other things, the results of viewing sunlit white paper through two 3 meter lengths of plexiglass; one filled with $\ce{H2O}$ and one with $\ce{D2O}$. Sure ...


65

This does seem to be the case. I don't have images of the different types of water, but I did find this overlaid IR-visible spectrum of water and heavy water: As you stated, the presence of deuterium shifts the absorbance spectrum of heavy water further into the IR region, rendering it colorless. The website I found this on (http://www.webexhibits.org/...


50

Substances which absorb almost all the light falling on them appear black. Therefore you are looking for the blackest known compound. The record is currently held by Vantablack[1], a substance composed of vertically aligned carbon nanotubes, which absorbs up to 99.965% of visible light incident upon it. As you can see it is really black. Previously, the ...


50

In theory, yes, you can heat objects to a high enough temperature to emit x-rays or gamma rays. You cannot do this to food, and you certainly cannot do this in your kitchen (or probably any kitchen). Let's take the lowest energy x-ray out there and see what it would take. X-rays range in frequency from $30 \times 10^{ 16}$ to $30\times 10^{10}$ hertz. The ...


29

It has nothing to do with what you were going for, but there is a small, but non-trivial amount of x- and gamma-ray output for most food and so the answer is trivially "yes". In particular any food containing potassium will have the usual admixture of K-40 (with it's 1.3 and 1.5 MeV gamma lines). When I worked in a lab with a low-background, high-...


22

All metals are capable of absorbing photons of any wavelength below hard ultraviolet, as ideally there are allowed electronic transitions of arbitrarily small energy between states in the unfilled valence band. This means metals are theoretically the materials with the widest wavelength range for photon absorption (except for plasmas). However, this comes ...


21

Yes, you can use a common stove to test for cations. But a stove is designed to minimize the risk of incomplete combustion (which could lead to production of carbon monoxide), hence its flame always appears as an intense blue flame. Such color contamination could be problematic when testing for cations. In contrast, the combustion (and the color of the flame)...


18

For vibrational spectra the primary transition under investigation is the $v = 1 \leftarrow 0$ excitation (because $\hbar\omega >> k_\mathrm{B}T$, so excited states have negligible thermal population). The intensity of the transition depends on the transition dipole moment: $$R_{10} = \langle 1 | \hat{\mu} | 0 \rangle$$ in that the intensity $I \...


16

The terms arose back in the early days of quantum physics when spectral lines that were expected to be singlets were actually observed to be more complex (doublets, triplets, etc.). An electron can have a spin quantum number of $+\frac12$ or $-\frac12$. For a system that exists as a singlet, all spins are paired and the total spin for the system is $S=0$....


16

TMS was first proposed as a reliable internal chemical shift reference in 1958 by Tiers. Back in them good ol' days, 1H NMR was called proton nuclear spin resonance, or nsr, and the tau scale was used for reporting chemical shifts (10ppm in the delta scale was set to 0 and positive values were read to the right. So TMS came at t10.0), and CCl4 was the one ...


16

As Tyberius noted, the projection formula does not work for infinite order groups (this is because the Hermitian form on characters is defined to be G-invariant by averaging over all elements in a group. That is, it works because it is possible to hit all the elements of the group in some order). How can one work in the infinite dimensional groups then? One ...


15

As the electrons fall from higher levels to lower levels, they release photons. Different "falls" create different colors of light. A larger transition releases higher energy (short wavelength) light, while smaller transitions release lower energies (longer wavelength). The visible wavelengths are caused a by single electron making the different ...


15

TMS has 12 protons which are all equivalent and four carbons, which are also all equivalent. This means that it gives a single, strong signal in the spectrum, which turns out to be outside the range of most other signals, especially from organic compounds. Although the chemical shift scales are still zeroed at the TMS peak, most spectra are now calibrated ...


15

This is not in general true Consider molecules a point group not containing inversion symmetry, e.g. $C_2$ hydrogen peroxide The $C_2$ group has only two elements, $E$ and $C_2$, and the $C_2$ rotation operation maps between two identical arrangements of atoms. Both the symmetry number and order of the group are 2. Matrix representations All rotations ...


14

I'm not familiar with the computational packages -- I'm an experimentalist, not a theoretician. As an example for why both might be present, we can turn to gaseous infrared spectroscopy: one will quickly find that the line-widths in vibration-rotation spectra depend on pressure. At lower pressures, say less than a torr, Doppler-broadening is the main ...


14

I am told that carbon dioxide is IR inactive. You're right, that's not true. Since carbon dioxide is linear it has $3n-5 = 4$ vibrations and they are pictured below. The symmetric stretch does not result in a change (of the initially zero dipole moment), so it is ir-inactive. The asymmetric stretch does result in a change in dipole moment so it is ir-...


13

Any black compound absorbs photons in all the visible spectrum; that is why essentially it appears black to our eyes. So, for instance, iron(II,III) oxide, $\ce{Fe3O4}$ will do so; it is even used as a black pigment. It is very unlikely though that it will absorb each and every photon of any wavelength in the visible spectrum, but the search for such a ...


13

Unfortunately, although the answer given by bon provides a very simplistic answer to a fairly common NMR-101 problem, it is not quite correct. It is fine for the propane case, but falls short for butane. While a very simple molecule, butane has a complicated spectrum because, though the two methylene groups are chemically equivalent, they are in fact ...


13

A deviation within the low picometres is nothing to worry about, there are many reasons for this. Primary literature, like peer-reviewed journals, will always publish an analysis of the obtained values along with any error bars, i.e. how accurate the measurement is. When you have disagreeing values it is worthwhile to have a look at these, too. Secondary ...


13

I will provide a full quantum mechanical explanation here.[1] Warning: rather MathJax heavy. Hopefully, this lends some insight into how the diagrams that long and porphyrin posted come about. Finding the states between which transitions occur The Hamiltonian for two coupled spins is $$\hat{H} = \omega_1\hat{I}_{\!1z} + \omega_2\hat{I}_{\!2z} + \frac{2\pi ...


13

In short, there are two obvious problems with the setup OP uses for TD-DFT calculations: B3LYP functional is not a good choice for TD-DFT. 6-31G(d) basis is usually too small. At M06-2X/Def2-TZVP level I get a maximum at ~160 nm, which, taking the accuracy of the TD-DFT approximations into account, is close enough to the experimental value.


12

Yes. For some chiral compounds, each enantiomer in a racemic mixture absorbs certain frequencies of light differently depending on the circular polarization type of the light. That means that if you can pump in sufficiently intense and energetic (e.g. ultraviolet) circular polarized light pulses into the racemic mix of such compounds, you may in some cases ...


12

The $\ce{C=C}$ stretch is responsible for this ir peak. For an ir absorbtion to occur, the absorption must result in a change in dipole moment. If we examine the $\ce{C=C}$ stretch in cis- and trans-2-butene we find that image source cis-2-butene has a dipole moment (0.33 D) and upon stretching the double bond the dipole moment will change; therefore we ...


12

This question goes along the line of what does it mean when it is said that an sp3 orbital has 25% s-character. It also intrigued me so I have tried to find answer, which would not break my hybridised orbital view. What you clearly see from the spectrum is that there are two bands, corresponding to different energy levels in the molecule and their intensity ...


12

Very technically? Yes. Realistically? The probability is small enough that even if it does happen, the peaks for the multiple transitions are going to be small enough that we cannot really observe them on the spectrum. The lifetime of a given excited state is so small compared to the analogous time in the ground state that it can basically be considered ...


12

The two observed C=O frequencies are due to the symmetric and asymmetric stretching modes of the anhydride. Source: Introduction to Spectroscopy, Pavia and Lampman You can see that the lower frequency symmetric stretch occurs where both C=O bonds are lengthening and shortening in tandem, whilst the higher frequency asymmetric stretch occurs when one C=...


12

1. Weak-field and strong-field limits I will adopt the description used in Figgis and Hitchman's Ligand Field Theory and Its Applications (p 5), because I cannot really phrase it better: It is useful to recognise the relative importance of the terms in the Hamiltonian for different systems: First- and second-transition series: $\mathbf{H}_\mathrm{LF}...


11

Sorry for the late answer. I just discovered this question. Preludium Firstly, I think you might have an error or some non-standard notation in your formula for $Q_{\mathrm{rot}}$. The rotational energy levels are given by \begin{align} E_{\mathrm{rot}} = \frac{\hbar^2}{2 I} J ( J + 1 ) \ , \end{align} where $I$ is the moment of inertia. Then the ...


11

In order for an electronic transition to be allowed (occur with strong intensity), certain "selection rules" must be obeyed. You may already be familiar with the rule that the electron spin quantum number cannot change during a transition. It is this rule that forbids singlet-to-triplet absorption or emission (emission follows the same rules as absorption) ...


11

Time-dependent DFT can be used to predict excitation energies through a linear-response formulation. In this Gaussian result, beyond the first line, you are looking at the largest coefficients in the configuration-interaction (CI) style expansion. (It's not strictly CI, but the implementation of time-dependent HF or RPA is essentially the same for TDDFT or ...


Only top voted, non community-wiki answers of a minimum length are eligible