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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

This answer was suggested by Ed V. It resolves the apparent problem nicely and readers with some fluency in the subject would probably not find the transition from $\lambda$ (medium) to $\lambda$ (vacuum) too troublesome. We begin with $c = \nu \lambda_o$ in a vacuum and $\frac{c}{n} =\nu\lambda $ in a medium, in which $n$ is th refractive index. In ...


0

My general understanding may help. First, the UV photolysis of methanol: $\ce{CH3OH + hv -> .CH3 + .OH}$ The attack of a starting organic (say RH) with the created hydroxyl radical usually results in the extraction of a hydrogen atom forming water: $\ce{RH + .OH -> .R + H2O}$ This process continues with a succession of intermediate radicals and ...


4

Ed V's answer above--which I am accepting--is correct and also allowed me to reach the conclusion on my own. I am just posting this because my confusion was not too hard to resolve once I saw the quantities involved and someone else may have the same confusion. The short of it is: student polarimeters measure circular birefringence (CB), usually at a ...


10

It gets very complicated, there are many papers and books on the topic and this little table, from Jensen et al. 1, gives a glimpse of the anisotropic effects: An extremely short answer is this: when linearly polarized light is incident on an optically active medium, it can be thought of as consisting of equal components of right circularly polarized (RCP) ...


-2

Beer's law limitation can be more easily understood if the coefficient $\epsilon$ is considered as a cross-section effect. Let's start with a broad beam or a bundle of rays having a section $A$ (in $m^2$). Its measured intensity is $I_o$. Now suppose there exist plenty of tiny obstacles suspended in the bundle of rays, and that each such particle behave like ...


4

If you use R for data analysis, take a look at the R package hyperSpec. Example: install.packages('hyperSpec') library('hyperSpec') spectrum <- read.spc('<your spc file>') ... plot(spectrum) You can also use Sys.glob() function for wildcard search of your files, then read all of them with read.spc(), and call collapse() to build a single hyperSpec ...


2

Do all noble gases or alkaline earth metals have similar spectral lines considering the above points? The question is interesting after you modified it. The basic set of reasoning you provided is the main story. Each element has a different nuclear charge and the outermost electron(s) is responsible for the atomic emission spectrum. Since the nuclear ...


1

Half the fwhm has to be added/subtracted to the central wavelength and then this change in wavelength converted to an energy. Thus the limits on the wavelength are $\lambda \pm \Delta \lambda /2) $ and the energy change $\Delta E=hc/(\lambda - \Delta \lambda /2)-hc/(\lambda + \Delta \lambda /2)$ Joules with $c$ in m/s and $h$ in J s. Plugging in the numbers ...


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