# Tag Info

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Answering the Question as Asked The raw data is available from a unreasonable and unmanageable slew of papers published in endless venues over the last upteen decades. Ouch. The processed data is available in the form of ENDFs (evaluated nuclear data files) maintained by several agencies (i.e. these is a Japanese adgency that maintains the JENDFs), but ...

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This is a really good question. I have also thought about this quite a bit. At present, it does not seem like it is possible with a standard experimental probe. Inelastic x-ray scattering and electron energy loss spectroscopy only measure the longitudinal response function at finite frequency and momentum. As you said though, if you are only seeking small ...

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There is no reason why you can't measure the rate frequently. However, in order to estimate the half life, you need to see a change in the rate of decay. How long you need to measure for, and how far apart you need to change your measurements, depends on the number of decays per second that you observe as well as the required accuracy. For example, if you ...

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you should resolve the differential equation of N. If you do that you'll get that $$A = A_0 e^{\lambda (t-t_0)},$$ where $A_0$ is the activity at time $t_0$. From there you can obtain the value of $\lambda$ from two measurements of the activity whatever the time interval.

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Firstly, the activity formula is in fact: $$-\frac{dN}{dt}=A=λN,$$ because $\frac{dN}{dt}<0$. [...] is there any particular reason why our time interval for measuring the number of remaining Radionuclides should be close to the half-life of the substance? No and that's not how it's done in practice. $\lambda$ and the half-life are determined by ...

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Not an expert in this but I can try to answer some parts of your question. As you pointed out, one can find the optical conductivity to measure the current-current correlation. For that if you use linear response then you can conclude the following (for an isotropic material): \sigma^{ij} = \sigma_L \frac{q^i q^j}{q^2} + \sigma_T \left( \delta^{ij} - ...

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Are you using the Apex-Gamma software? If so, a fitted singlet is a line the software has identified as being a single distinct line rather than a peak in a region where there are multiple overlapping lines. As I recall, Apex-Gamma identifies lines that are overlapping by $M$ or $m$ (presumably for multiplet).

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That means that the peaks could be fit with single Gaussian, the simplest fit. They did not have to fit as a doublet (two lines close together). Ortec has a technical paper on line at Deconvolution of Gamma-Ray Peak Doublets as a Function of Peak Separation and Relative Amplitude

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You may be able to modify a yeti cup, they are vacuum insulated. Just have to make a proper lid. Just brain storming, it might be worth a try.

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