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Presumably the emitted light is spread out uniformly. In that case at a distance $\ell$ the light is spread out over a sphere with a surface area of $4\pi\ell^2$. If your detector has an area $A$ then the fraction of the light that hits your detector is just $A/(4\pi\ell^2)$.

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It is an approximation that is ok so long as $l \gg r$. The idea is that your detector will intercept some small fraction on the surface of a sphere of radius $l$, which has a surface area of $4\pi l^2$. If the luminosity $L$ is spread out isotropically over this sphere, then the flux at the edge of the sphere is $L/4\pi l^2$ per unit area. Strictly ...

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We take into account the inelastic interactions that take place between the respective type of particle and the material, i.e. interaction that can consume part of the energy of the radiation particle, and calculate the mean-free path of the respective type of particle in the material. In this calculus we also consider the density and the structure of the ...

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Free neutrons are unstable, with a half life of about 10 minutes. They almost always decay via $\beta$-decay: $$\text{n}^0 \rightarrow \text{p}^+ + \text{e}^-+\bar{\nu}_\text{e}$$ This is the same $\beta$-decay that occurs in unstable nuclei, and is possible outside the nucleus because free neutrons are more massive than free protons. The situation in a ...

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You're right that1+, 2+, 3+, 4+, and 5+ are all possible values for the final state following an E3 transition. However, if the final state were 1+,2+, or 3+, then an E1 transition would be possible. If the final state were 4+, then an M2 transition would be possible. So if the lowest multipole is E3, that leaves just one option.

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