# Tag Info

8

Why not send in white light (with all the wavelengths at once)? This is certainly and routinely done today. The main thing is how much price are we willing to pay? In a rigorous sense, colorimetry refers to the fact that we use optical filters to isolate wavelengths rather than monochromators. When the latter are used, you call them spectrophotometers. ...

7

As has already been mentioned in comments, those letters and numbers for each state correspond to their symmetry, specifically the symmetry of the (purely electronic) wavefunction. For more information regarding symmetry notation, see here, and see here for more discussion about how symmetry can be applied to molecular orbitals. It might not be clear from ...

6

Surely you are aware that the calculation of this sort is but a very rough estimate, yet still I feel the urge to stress that once more. Having said that, let's look at the double bond in the first cycle: isn't it exocyclic with respect to the other cycle?

6

There are two main reasons for what you're describing. The first is the fundamental properties of the molecules involved, as mentioned in the comments. Even for isolated molecules, the transition is not always between the same vibrational states of the excited and ground state. Since the vibrational states are quantized, we would expect a series of very ...

6

The previous answers by @Buck Thorn and @M. Farooq are very good, but not quite complete. The main advantages of FT UV-Vis spectroscopy would be 1) accuracy of wavelength (or wavenumber) determination and 2) better spectral resolution of bands, relative to conventional UV Vis spectroscopy. But it is very hard to do well, and very expensive. The disadvantage ...

6

Metal chlorides are usually much easier volatilized than oxides or sulfates. They melt or sublime at temperatures much lower than 800°C. So they are volatilized in any flame like Bunsen burners. Platinum is not attacked by hot chlorides or acids, and is not oxidized in hot air. So it can be used for a great number of consecutive tests in the flame. It will ...

5

To quote the key points of an easy to read publication by Salgado and Vargas-Hernández (doi 10.4236/ajac.2014.517135, open access): All starts with the dissociation equilibrium between the acid $\ce{HA}$ and its anion $\ce{A-}$ for which you write $$\ce{HA + H2O <=> H3O+ + A-}$$ By consequence of this, the recorded total absorbance $A_\mathrm{t}$ ...

5

Woodward's rules are an empirical method for predicting the maximum absorption of a molecule based on the functional groups present (that is, conjugated functional groups which possess a distinctive maximum absorbance such as a diene or an unsaturated carbonyl). Should the marked double bond considered as an exocyclic double bond? Exocyclic is defined ...

5

It seems that here we only need to consider the lowest energy transitions, i.e. HOMO–LUMO transitions. The difference between benzene and the linear polyenes is that both the HOMO and LUMO of benzene are doubly degenerate, whereas the linear polyenes do not have degeneracy. If we think about this HOMO–LUMO excitation in the case of benzene, we can ...

4

Most light doesn't result from chemical reactions While there are many chemical reactions that can emit light and many related reactions where the emission is related to chemistry (for example the light from some flames is caused by electronic transitions), most light you see isn't. An incandescent light bulb emit light because of black-body radiation ...

4

An explanation is provided in this abstract (1): Fourier transform spectrometry in the UV-Vis region (FT/UV-Vis), because it is source shot-noise limited, has a signal-to-noise ratio (S/N) disadvantage in comparison to dispersive spectrometry, especially with dense spectra. At the expense of poorer S/N, FT/UV-Vis can be satisfactory for high-...

4

After radiation is absorbed and the electron is at an excited state in the molecule there are several pathways for de-excitation to occur (see fig.). The pathway of choice depends on its rate, ie how fast it can happen. It turns out that the fastest de-excitation pathways are radiationless (wavy arrows in fig) such as internal conversion (IC) that happens ...

3

There are several ways to answer the question, so here are my three. The first answer involves using a simple integer wavelengths approximation, i.e., noticing that Figs. 1 and 2 have 1 nm point spacing, and then using a spreadsheet to perform a summation. The answer explained in this part also serves to define the terms used in the second answer, which ...

3

The classical concept of oscillator strength $f$ is useful here, but it should be used only in a qualitative way. This is defined as $f=a\int\epsilon_\nu d\nu$ where $a$ is set of units with value $4.3 10^{-9}$ if the extinction coefficient is in units dm$^3$/mol/cm and $\nu$ is in wavenumbers. The maximum value of $f$ is unity and is close to this for an ...

3

Iodine is highly polarisable and will form a (Mulliken) charge-transfer complex with many aromatics. Very many pairs of donor - acceptor pairs do this and have been studied for approx $100$ years, see for example tetracyanoethylene–pyrene (J. Chem. Phys. 105,1996 p2287) for an ultrafast time resolved study. In absorbing a photon the ground state complex DA ...

2

Certainly not The problem is not, strictly, one of physics but one of perception. The eye doesn't detect individual wavelengths of light but the relative amount of red, green and blue light (but the sensors in the eye cover a relatively wide and overlapping range of wavelengths centred on what we might measure as "pure" blue, green and red. If a "pure" ...

2

Let's suppose a hypothetical irreversible reaction: $$\ce{B + H -> D + F}$$ where H is a reactant with constant concentration during the reaction and that don't absorb light in the working frequencies. The absorbance of a species $i$ related with the concentration by the Lambert-Beer Law: $$A_{i}=c_{i}\varepsilon_{i}l$$ where $c_i$ is the concentration of ...

2

Found the following value (see the very comprehensive reference below): molar absorption coefficient $\pu{\epsilon = 24 000 M-1 cm-1}$ ($\pu{510 nm}$) in water CARMOISINE (E122), CAS Number 3567-69-9, (3~{E})-4-oxo-3-[(4-sulfonaphthalen-1-yl)hydrazinylidene]naphthalene-1-sulfonic acid, SMILES(canonical): C1=CC=C2C(=C1)C(=CC=C2S(=O)(=O)O)NN=C3C=C(C4=CC=CC=...

2

I'm collecting ideas from the excellent comments and adding some of my own: The typical UV/Vis spectrum has a narrow range On the low energy side, it ends where the visible spectrum ends, at about 800 nm. On the high energy side, it ends where water and quartz start absorbing too much, at about 220 nm. Compared to an IR spectrum, it has a narrow range. ...

2

I would imagine that you could just use one of the spectrophotometers you have available to determine the molar absorptivity constant experimentally. After all, the molar absortivity constant is simply the constant which equates the terms (those being Absorbance and Concentration) which are already proportional. So, using something like the beer-lambert law.....

2

Absorbance spectrum of colored solutions may be boring for a "sight seeing" student, as you call yourself. Emission spectrum is far more interesting for visual learners. Anything luminous, which you can see with your eyes, should have a spectrum in the visible range. This includes the sky, moon, a red hot stove, fluorescent lamps, computer screen, flames, ...

1

The selection rules are related to the transition dipole moment that determines the intensity of a given transition. If a given selection rule holds perfectly, all that it says is that a forbidden transition has zero intensity and an allowed transition has non-zero intensity. If the selection rule does not hold exactly, just like the in the case of the spin ...

1

I Think you are in right track using UV-vis spectroscopy as a tool to identify your alloys. Other than UV-vis spectroscopy, researchers around the world have used many other analytical techniques to evaluate the synthesized nanomaterials, including X-ray diffractometry (XRD), Fourier transform infrared spectroscopy (FTIR), X-ray photoelectron spectroscopy (...

1

The negative values can arise due to two major reasons: You did not mention that whether you are using a single beam instrument or a double beam instrument. In a double beam instrument, absorbance of the sample and the blank is continuously measured. In a single beam spectrophotometer, A(blank) is recorded first and later subtracted. In each case, the output ...

1

By definition (in Woodward-Fieser rule), an exocyclic double bond is a double bond where one of the $sp^2$ carbon atoms of the bond is a part of a ring, while the other $sp^2$ carbon of the bond is not a part of same ring (it could be a part of another ring). If both the $sp^2$ carbon atoms of the double bond is part of the same ring, then it is called ...

1

The law is not as you state although it often gets reported as this. The '$A$' you quote is the optical density. The Beer-Lambert law is $I_{tr}=I_0e^{-\epsilon_\lambda [C]\ell}$ where $I_{tr}$ is the intensity of transmitted light for a molecule at concentration $[C]$ at wavelength $\lambda$ and cell path length $\ell$, and $\epsilon_\lambda$ is the ...

1

The colour we perceive for any compound, such as vermillion is not associated with a single wavelength, but rather with all of the wavelengths reflected by the compound that are in the visible portion of the spectrum. The reflectance spectra of vermillion under different conditions is show below. Source We can see that most visible light with wavelengths ...

1

In contrast to a typical sample provided by a synthetic lab your sample consists of multiple components; milk -- already consisting of many chemicals for its own, plus urea added, plus hydrogen peroxide added (as aqueous solution). Using ATR-FT-IR is indeed a good choice for characterizing such samples, both for qualitative, as well as quantitative analysis ...

1

Simply put, the reason the emitted color is different from the absorbed one is because there is a loss of vibrational energy that takes place in addition to the photon emitted by "falling back to ground state". This site has a good detailed explanation: https://micro.magnet.fsu.edu/primer/lightandcolor/fluoroexcitation.html Essentially, the "π MOs to π∗ ...

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