12

[...] the protein luciferin is oxidized by the enzyme luciferase Luciferin is not a protein, but a benzothiazole with a thiazole attached to the carbon atom between nitrogen an sulfur. Upon oxidation at the thiazole ring, an oxetanone is formed. This four-membered ring breaks, releases carbon dioxide and turns the thiazole into a thiazolone. The resulting ...


9

There are two potential questions here: one is why is dimming a slow drawn-out process (incoherent, unlike a "switch"). The second question is why different stars may appear to lose their intensity at a different rate. The light-emission process behind your stars involves phosphorescence. The process involves relaxation of an electronically excited system ...


9

LPG is propane, butane or a mixture of both. Paraffin candle wax is $(CH_2)_n$ typically $C_{31}H_{64}$ (other, nonparaffin, waxes are also used in candles, like stearin, beeswax, etc.) However, LPG is a gas from the start and can mix with air (i.e., oxygen) before it begins to burn, while candle wax has to be heated by the flame before it can begin to burn,...


7

Klaus gave a nice summary on luciferin. The second part of your question was: Also, are there any chemical reactions which produce high energy electromagnetic waves, such as X-Rays? Any energy released with an emitted photon must previously have been added into the molecule (or atom) in question. For example, if singlet oxygen is created chemically, it ...


7

Fluorescent lamps do primarily work by fluorescence. According to this Wikipedia article: "The inner surface of the lamp is coated with a fluorescent (and often slightly phosphorescent) coating..." First off, an electric current excites the outer electrons of gas atoms (typically mercury) within the lamp to a higher energy state. When the atoms relax ...


7

The human eye is most sensitive in green wavelengths. Therefore, if one were to put the same amount of light energy into different wavelengths, the green portion of the spectrum would appear brightest, though it contained no more energy than the rest. If a phosphor or glow-stick manufacturer is seeking the "greatest bang for the buck", i.e. the most light ...


7

Yes it is very common particularly in the more rigid type of molecule. The best example is chlorophyll and this overlap of absorption and emission leads to energy transfer in photosynthesis. The reason for the overlap is that the excited state potential energy profile is very similar in shape but slightly shifted (as displacement) to that of the ground ...


7

I have read some answers on web which says that it is due to the insufficient supply of oxygen, so I tried burning candle beside LPG ( so there is no difference in supply of oxygen for both candle and LPG ) , but still candle gives a yellow flame. Why? You cannot conduct this experiment by yourself without using a proper apparatus. Otherwise comparing the ...


6

Both expressions are identical. It's just juggling of the signs, since $$E^{\mathrm{red}}(D^{\cdot+}/D) = - E^{\mathrm{ox}}(D/D^{\cdot+})$$ Relevant primary references would be: Rehm, D.; Weller, A. H. Isr. J. Chem. 1970, 8, 259-271. Rehm, D.; Weller, A. H. Ber. Bunsen-Ges. Phys. Chem. 1969, 73, 834-839. Weller, A. H. Z. Phys. Chem. NF 1982, 133, 93-98. ...


6

You are looking for the rate constant $1/\tau^0=k_2$. The rate equation for the $\ce{NO2^*}$ is (using $N^*$ for $\ce{[NO2^*]}$ and N for ground state NO2), $\displaystyle \frac{dN^*}{dt}=-k_2N^*$ and when quenched $\displaystyle \frac{dN^*}{dt}=-k_2N^* -N^*((k_3+k_5)N+k_4[Xe])$ which can both be integrated to give exponentials, i.e $N^*=N^*_0e^{-k_2t}$ and ...


6

In a very very broad reasoning, XRF is much more sensitive to the "mid-range" elements. Elements below Na are generally not detectable. The x-ray tubes must have a beryllium window to seal the tubs which absorbs the low energy x-rays. The lower the atomic number the less the fluorescent yield. Low energy x-rays don't penetrate from the sample ...


5

As mentioned in this comment, which links to this paper, and this comment, which links to this Wired article, there are at least two molecules: $\beta$-carboline: 4-methyl-7-hydroxycoumarin: $\tiny\text{Yes I poached those comments, they could have been answers.}$


5

By convention both donor (D) and acceptor (A) potentials are listed as reduction potentials. For a nice description, see Turro's explanation of the Rehm-Weller equation in his photochemical treatise "Modern Molecular Photochemistry of Organic Molecules" To clarify, the equation that is being referenced is not actually the Rehm-Weller equation but is the "...


5

Concentrating on fluorescence intensity for the moment the equation becomes $I_O = I_D $ where $I_O$ is that for the free dye and $I_D$ for that bound to the DNA. If $\alpha$ dissociates the equilibrium is then $$\begin{align} &I_O \quad \rightleftharpoons &I_D \\ &1-\alpha &\alpha \end{align}$$ and the equilibrium constant $K=I_D/I_O=\...


4

Source of light I imagine that the cause of the stars becoming dimmer is that some electrons are relaxing down before others. This statement could be interpreted in two different ways, one incorrect and one correct. It could mean that electrons have an "on" state where they continuously emit light. That is not the case. Instead, they emit light once ...


4

As you mentioned, Gaussian can do it but is proprietary. I believe ezSpectrum (https://github.com/iopenshell/ezSpectrum) might be able to do what you are looking at but I have not used it myself so far. There used to be a program available called FCClasses which would run the Franck-Condon computations for you based on (I think) Gaussian output, but it ...


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 ...


4

A propane molecule $\ce{C3H8}$ in the gas phase must find $7$ molecules $\ce{O2}$ to burn totally. A wax molecule $\ce{C_{31}H_{64}}$ must find $63$ molecules $\ce{O2}$ to burn completely. During its trip through the flame (having maybe $10 - 20$ cm length), the propane molecule will have enough time to meet $7$ molecules $\ce{O2}$. The wax molecule may not ...


3

Based on your description, it sounds like you are measuring the production of hydrogen peroxide by an oxidase enzyme. In that case, it is not necessary for the Amplex Red to be at the same concentration as the substrate. You only need enough reagent to determine the initial rate of reaction at each substrate concentration, ie when only a small portion of the ...


3

If you want to convert visible light into UV, you have two options. Two-photon excited fluorescence. The process is inherently wasteful, but the field is studied due to applications in microscopy. See example paper on fluorophores https://pubs.acs.org/doi/abs/10.1021/ol017150e . Also, there are commercially available dyes. Optical frequency multiplier. The ...


3

It is not possible in a single molecule as energy conservation will forbid it. There is not enough thermal energy, $k_BT \approx 210 $ wavenumbers, visible to uv thousands of wavenumbers. It is sort of possible using triplet-triplet annihilation ($\ce{T + T \to S^* + S}$ ) but this needs significant initial excitation and long lived triplets, which means de-...


3

The process is called fluorescence. Some substances, such as calcium sulfide, $\ce{CaS}$, absorb light and release it slowly in a process called phosphorescence, glowing for some time after the source of light is removed.


3

In part of the naphthalene molecule you draw you indicate a transition moment. This is the direction for absorption into the lowest excited state $S_1$. As naphthalene is highly symmetric the fluorescence is also emitted from a transition moment aligned in the same direction. The second excited state $S_2$ has a transition dipole set perpendicularly to the ...


3

A fluorescence quencher is any species that causes your fluorophore to stop fluorescing. There are several ways this can happen, but there are two broad categories: chemical reactions and energy transfer. I will give simple examples of each category below. However, which method is operating is going to depend on the fluorophore. Bromide and fluoide will not ...


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.


3

The after-glow is called phosphorescence. Fluorescence stops immediately the moment light is turned off because the process is ultrafast! Phosphorescence on the other hand requires a change in the spin of the electron in a molecule, such transitions are forbidden, this is why it is slow. You can check the Jablonski diagram to understand this better, see here ...


2

Ca cannot be detected in General Metal Mode. You have to change to "Mining Mode" In this mode, you can measure Ca properly.


2

A low quantum yield would indicate that there are efficient intra-molecular non-radiative decay pathways, likely internal conversion or spin-orbit coupling. These decay pathways may be faster that FRET, rendering FRET essentially a non-competetive process. A small extinction coefficient would indicate a small transition dipole moment. If the transition ...


2

This phenomenon is thermoluminescence. The incoming photons excite electrons of the material to a higher level, a metastable state, which fall back to a lower state and emit photons. The metastable state may last for only nano- or picoseconds, in which case it is termed fluorescence. If the electrons need a bit of a "nudge" and hang around for a few seconds ...


2

A nonfluorescent derivative of dihydroresorufin, 10-acetyl-3,7-dihydroxyphenoxazine has been marketed under the name Amplex® Red (AR). AR is regarded as the best fluorogenic substrate for peroxidase because it is highly specific and stable. It has been used to detect Reactive Oxygen Species (ROS) such as $\ce{H2O2}$ in various experimental enzymatic systems (...


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