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See What Really Happens in the Franck-Hertz Experiment with Mercury? The article explains how the curve depends upon the design of the tube.


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Do the calculation at the computer/calculator precision then quote result with the appropriate number of significant figures. Yes that is correct procedure. Otherwise you could introduce rounding error.


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I'm having trouble understanding what you're trying to say, but assuming you're just looking for period resonances of the form $a:b$ where $1\leq a,b\leq 5$, the following 4 lines of Mathematica code using your example list of extrasolar periods should suffice: A = {0.44, 0.8, 0.9, 0.9, 1.2, 2.0, 3.0}; n = Length[A]; d = 0.05; ResonanceMatrix = ...


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Fast rotation keeps the paper-symmetry protons equivalent. High concentration makes for fast proton exchange in the alcohols (sharp line, no coupling to methylene protons). Integration identifies populations. The difference in chemical shifts tells you the temperature of the sample, ...


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You are right when you say that one peak is due to $\mathrm{OH}$ groups and the other one to $\mathrm{CH_2}$ groups. The protons in each groups are chemically equivalent and contribute to the same peak. You should compute the area below each peak. Since there are 4 $\mathrm{CH_2}$ protons and only 2 $\mathrm{OH}$ protons, my guess is that one peak's area ...


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2.6 million counts, with each count producing a dead time of 90 nsec, gives a total dead time of .234 seconds / second. So the total number of counts should be 3.4 million counts (2.6 / (1 - .234), assuming no dead time. Since each QD emits 770 million photons per second, the detector efficiency is 3.4 / 770 (which I get as .44%).



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