# Tag Info

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What you're trying to do is called radiometric calibration. The problem with doing it with an ordinary incandescent light bulb is that the bulb itself would have to be radiometrically calibrated to get a precise spectral calibration of the detector. That's because the bulb's spectral emissivity deviates from an ideal blackbody source with an emissivity ...

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These relations are based on the fact that both the position and the momentum distributions are centred around zero, which is in turn due to the symmetry of the atom. Given that, the width of the position and momentum distributions ($\Delta x$ and $\Delta p$) is of the same order as a typical position or momentum within those distributions ($r$ and $p$).

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user26143 gives a correct argument about the effects of entanglement in two-atom processes, but specifically to your question, I have to reiterate the answer 1) You are not talking about entanglement 2) Yes, this effect gets effectively blocked out in the cases you consider 3) QFT does not really bring any new insight to this problem In theory, all ...

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Hydrogen bonding arises when a chemical bond is polarised to one end of it has a slight positive charge and the other has a slight negative charge. In the case of o-nitrophenol it's mainly the OH bond that is polarised - the H atom has a slight positive charge and the O atom has a slight negative charge. The charge separation means the OH bond has an ...

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The way this is justified is as follows: We start with the uncertainty principle, which can be roughly stated as $$\Delta x \Delta p \geq \hbar$$ For this rough estimate, we will ignore some factors of perhaps $2$ or $\pi$, but we're interested in some order of magnitude, not the exact result. Now, our second assumption will be that the ground state of the ...

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... then (condense) the hydrogen into a liquid and then again used electrolysis. What would happen ? Nothing. You already separated the hydrogen and gave it's electron back to it. Hydrogen will not accept another electron (it's not stable).

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Assuming you could get down to liquid hydrogen and figure out how to do electrolysis at 33 K....it seems like the conductivity of molecular hydrogen would be quite low, and nothing would happen. Similar to the case of very pure water.

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