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6

Your question is identical to a question posed to Feynman by his father. A two-minute monologue on that is here: https://www.youtube.com/watch?v=eebWoZkN3FQ The photons are genuinely created at some moments – and they may be similarly destroyed, too. They're not coming from anywhere and they are going nowhere. The number of photons $N_\gamma$ is simply ...

5

For many—but not all—shielding processes the parameters of interest are proportional to the areal density, $$\text{Range} \propto \frac{\text{areal density}}{\text{mass density}} = \frac{\sigma}{\rho} \,,$$ so a first expectation would be that the same areal density of material (of roughly the same composition) will have the same effect. I don't have the ...

4

Higher energy gamma and longer wavelength radio? Keep in mind that the different 'kinds' are merely human labeling conventions for a spectrum that is continuous in the mathematical sense. There is no feature of "radio" that distinguishes it objectively from microwaves. We just pick a boundary on the basis of some technological limitations that apply when ...

2

I had similar stupid doubt. It's coming from binding energy. The equation you give: Unstable atomic nuclei with an excess of protons may undergo β+ decay $$p → n + e^{+} + \nu_{e}$$ proton decays into neutron, positron, and electron neutrino. How do you think a proton can be converted to neutron which has greater mass? + you get positron and ...

2

The microscopic mechanism of emitting photon in a solid is the transformation of kinetic energy of atoms into EM energy. If an atom is in an excited state due to collisions among other atoms, then it will emit photon when it jumps into the ground state, and the energy of the photon is $$E=\varepsilon(\text{excited state})-\varepsilon(\text{ground ... 1 In addition to the answer by dmckee and to answer the question how high in energy you could get a photon it might be worth thinking about 'Gamma Ray Astronomy' where the highest energy photons are detected. The record highest photon energy observed is apparently currently 80 TeV, which corresponsds to a wavelength of 1.5 \times10^{-20}m wavelength (if I ... 1 Background The specific intensity or brightness, I_{\nu}, is defined as:$$ I_{\nu} = \frac{ dE }{ dA \ dt \ d\Omega \ d\nu } \tag{1}  where $\nu$ is the frequency, $dE$ the differential energy, $dA$ the differential area, $dt$ the differential time, $d\Omega$ the differential solid angle, and $d\nu$ the differential frequency. We can define a net flux ...

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