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-1

Iron-57 is a stable isotope , so it does not emit any gamma ray . You measure a prompt gamma from a nuclear reaction ( n , gamma ) .


2

I don't know the origin of either convention, so I am reluctant to make any absolute statements, but as an experimenter the question I ask myself is "What material has I learned is present in the sample?" or perhaps "What material do I bring into the lab to observe this line?". The answer to these questions would be Berylium-7 in the former case and ...


1

The question operates under a false premise. It is not the case that fitting a blackbody spectrum to the Sun gives you its "surface temperature". The Sun does not have a black body spectrum, although sometimes that approximation is made. A star also does not have a single "surface temperature". What you can do is divide the luminosity of the Sun by $4\pi ...


3

Many spectral lines are very sensitive to the surface gravity of the star - which enables a distinction between dwarfs and giants, because a giant's surface gravity is factors of $\sim 100$ lower than that of a dwarf of the same temperature. The reason that surface gravity plays a role is via hydrostatic equilibrium; the densities and pressures in a gas ...


1

There are many mechanisms which can contribute to broadening spectral lines. Usually, one way or another, the atoms have a broad range of random velocities which cause doppler shifts of varying amounts, broadening the line. One the more fundamental cases is simple 'thermal broadening', where the velocity is from thermal motion. The hotter the gas is, the ...


1

The local oscillator field and the signal field "interact" at the detector. It may be easier to understand if the field intensity is written as follows: $$ I=|E_{LO}+E_s|^2=E_{LO}^2+E_s^2+E_{LO}^*E_se^{i\omega T}+E_{LO}E_s^*e^{-i\omega T} $$ What is happening here is that the LO and signal pulses are both being spectrally dispersed in the spectrometer and ...


2

Assuming you are trying to do vibrational Raman and IR then these two experiments are done in completely different frequency regions. This can be seen in your diagram where the arrows in the Raman experiment are much longer (higher frequency) than the arrows in the IR experiment. Because of this you would need two different light sources to get both spectra. ...


2

High pressure gases do not produce a true continuum, and they do not become closer to the blackbody case with increasing pressure. The individual emission lines only become broader with pressure and make the emission spectrum appear continuous because the lines start to overlap. The number of molecular transitions does not change with pressure. There is a ...


0

Sure it can be negative. It simply means that there is a dynamics (reaction, energy transfer, ...) between species, so that there is an exponential rise at some wavelengths


2

The motion does affect the observed continuous spectrum. The flux at each wavelength is doppler shifted to a new wavelength. The result is of course a new continuum, so a doppler shifted continuous spectrum is rather hard to distinguish from one that is not. In particular, unless the continuum has some feature or break in its slope, then it can be ...



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