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Notice that the boundary of the high-fallout region in Idaho, Colorado, Oklahoma, and Illinois is blotchy and blobby and goes gradually from the red through green to blue. However, there is a boundary at the northern edges of Montana and North Dakota that looks like it falls precipitously from the maximum color to a background color exactly at the ...

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Electrons are identical particles, and an electron from any source will interact with the rest of the universe in exactly the same way. However, there is a transient difference between a beam of "beta rays" being emitted from a radioisotope and a beam of electrons with the same energy distribution produced via some other method. That difference ...

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That is because there is no finite $t_1$ such that $A(t_1)=0$. That is precisely what the math is telling you. $A(t)$ gets closer and closer to 0, but it never reaches it. It only approaches 0 asymptotically.

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The question probably could do with a little clarification as to what context and what you mean by 'due to radiation'? I have assumed you mean 'Do protons and electrons decay via radiation to lose energy?' In which case: -Electrons are fundamental, there is really nothing to decay to. They can lose energy if they have a mass-energy above their rest mass, but ...

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Binding energy of a proton in a nucleus is negative in the same way it is negative for an electron trapped in the potential well of a nucleus. A free particle with no kinetic energy has 0 total energy. A bound proton must have energy added to become free. "I would have added the oxygen binding energy and kinetic energy of the proton together and then ...

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First I start with the component formulation of the (monochromatic) radiative flux tensor $\mathbf{F_\nu}$, i.e. $\left( \mathbf{F_\nu} \right)^{i} \equiv F_\nu^{\;i}$, $$F_\nu^{\;i} := \oint_\Omega I_\nu(\mathbf{k}) \, k^{i} \, d\Omega \quad.$$ If we express the direction vector $\mathbf{k}$ in cartesian coordiantes, i.e. $\mathbf{k} = (k^i)_{i \in \{x,y,... 1 The answer is D, but not because of anyone's malice -- it's just the hand nature has dealt us. The neutrinoless double beta decay rate is proportional to$|m_{\beta\beta}|^2$, where$m_{\beta\beta}$is the "effective Majorana mass", which in turn depends on the elements of the PMNS matrix. Here's a plot of$|m_{\beta\beta}|\$ as a function of the ...

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