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3

Because the electron's initial state is a superposition of states with different energies, the electron does not have definite energy. Measuring its energy, you would find either $E_{4s}$ (with probability $\frac{x^2}{x^2+(1-x)^2}$) or $E_{3p}$ (with probability one minus that). Since it's energy isn't a definite number, we have to resort to talking about ...


4

A "sharp" tip typically has a finite curvature; there will be a very small part of the "tip" that is therefore angled at such a way that light will be reflected off it. The sharper the tip, the smaller the radius of curvature, and the smaller the "twinkle" or glint. The second effect is diffraction: Light that passes an object will be diffracted. For ...


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According to the contemporary but standard mathematics of AIAS.US, a Physics site dedicated to the revision and particularly, the rebuilding of false axioms of physics across the board; and according to my own research [largely consisting of the physical building of 3D models that mimic accordingly, the rotation and dispersion of stellar nebulae about the ...


0

The Wigner Threshold law describes the yield or cross-section of an ionization or detachment process where the incoming particles are just above the energy required for the reaction to occur (the threshold). It seems that Wigner described the case where there are only two outgoing particles (such as the ones you mention), and others extended his theory to ...


4

There are, in general, no closed form solutions (aka formulas) for the spectra of multi-electron atoms. There are reasonably precise formulas for special cases, like approximate values of x-ray transitions from inner shell electrons, though. Unlike in case of hydrogen and Rydberg atoms, which can be treated as a non-relativistic one-body problems (i.e. for ...


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You are basically asking about nebular astrophysics! You are correct that the positions of the emission lines are characteristic of the chemical elements present in the gas (a discharge tube in this case I guess) and in what ionisation state(s) the element is in. It is also of course possible to get emission lines that are characteristic of molecules - ...


1

It is what it claims to be: intensity of light at that wavelength. However there are a lot of different ways to measure intensity. The "proper" way might be to measure the emitted power in watts into a particular wavelength bin, but that takes a lot of careful algebra and calibration to get right. If your light meter is a digitized CCD readout, however, it ...


1

Here is the spectrum taken as a photo: Note the difference in the visible intensity of the lines registered. In your spectrum instead of a film there is a counter which measures the number of hits at that wavelength from the excited helium. The location of the excitations on the wavelength axis identifies the atom uniquely, like a fingerprint a person ...



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