# Tag Info

## Hot answers tagged uv-vis-spectroscopy

9

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. ...

8

Disclaimer: I'm not affiliated with any company However, Ocean Optics cuvette spectrophotometers are a good choice This model is used in some undergraduate labs and is good because the cuvette holder is built in (i.e. it doesn't require a fiber optic cable). As long as the optics don’t get jumbled, the prism detector mechanism work well and reliably ...

8

Good spectrometers in "new, unopened" condition are out of your budget. Their prices start at about 2000 USD. This also includes the spectrometers specifically made for education—like the ones from PASCO (example; they are usually based on Ocean Optics components), and from Ocean Optics themselves (example). One option for your budget is to buy a used ...

7

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 ...

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

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 ...

6

We can measure the UV-Vis spectra for π → π* or n → π* electronic transitions (other transitions are forbidden according to selection rule) [1, pp. 577–578]. But in diethyl ether, being a saturated compound, it has no π-bonds. This successfully dismisses the idea of having π* orbital. And, as there is no π* orbital, we can not have a transition to the π* ...

6

I think this was basically an educational experiment. Your analytical method assumes that the hydrolysis was complete (this is the key requirement of a successful calibration curve). Your experiment did not include a test to assess the completion of hydrolysis. Acetylsalicylic acid begins to hydrolyze in alkaline solution to salicylic acid and basically it ...

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

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-...

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 ...

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 ...

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 ...

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

I just did the experiment suggested by @porphyin. I used a 405 nm laser diode built into a toy "Space Gun" and the orange long pass filter shown in the figure below: The results are in the next dark figure: The spots on the paper are not fluorescence of the paper (see the violet fluorescence of the paper around the dark shadow of the filter) and are not ...

4

Mirror image spectra are only observed in solution and then only if the ground and excited state potential energies have almost exactly the same shape. Thus in rigid molecules such as anthracene a good mirror image is observed but not in more flexible molecules such as stilbene ($\phi-C=C-\phi$). The spectra of each vibrational level is wide in solution due ...

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

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 ...

3

It is because the water blank absorbs so much light that the usable scale is compressed, thus, causing a poor accuracy. For water, the absorption coefficients $\alpha\ [\mathrm{m^{-1}}]$ at $\lambda$ = 680, 580, 535, 480 and 380 nm are 0.47, 0.09, 0.045, 0.013 and 0.011. (Source: R. M. Pope, E. S. Fry, Applied Optics, 1997, 36, 8710-8723) That is not a ...

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

At the risk of providing a somewhat circular answer, in the harmonic approximation of an interatomic potential, the force constant is a measure of the force of attraction between the atoms (in a classical description they would obey Hooke's Law $|\vec{F}|=-k(x-x_\circ$)). The answer to the question is therefore that the attractive interatomic force in the ...

3

From the Jablonski diagram, The mirror image is only true if you are talking about transitions from $S_0$ to $S_1$ (absorption) and $S_1$ to $S_0$. Also read about Kasha's rule. Quinine is the most famous fluorescent molecule, its $complete$ absorption spectrum is not a mirror image of its emission spectrum.

2

The double bond B is exocyclic to ring 2 as it is attached to an atom which is shared between ring 1 and ring 2 while the double bond A is not connected to any ring 2 atoms and is within just one ring, hence making it endocyclic for further references visit https://pharmaxchange.info/2012/08/ultraviolet-visible-uv-vis-spectroscopy-%E2%80%93-woodward-fieser-...

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.....

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