# Tag Info

## Hot answers tagged particle-physics

19

Hadronic jets deposit a significant fraction of their energy in the electromagnetic calorimeter, for example because they can contain neutral pions that decay as $\pi^0\to\gamma\gamma$, bottom/charm mesons with semi-leptonic decays... Therefore the jet reconstruction algorithm uses energy deposits from both electromagnetic and hadronic calorimeters, so that ...

18

Those of us who've worked at JLAB (and those who worked at SLAC) know that energetic electrons create a lot of hadronic junk when incident on significant amounts of matter. Think about Deep Inelastic Scattering. Once you have an electron with energy in the few GeV range or higher there is a significant chance of creating pions or other light mesons in the ...

7

Edit: I am leaving this in because some effort has been made to present how decisions are made on complicated channels.The simple answer by @atlas-insider clarifies the general point that the OP is asking. From the sample ATLAS paper given in the comments Search for supersymmetry at $\sqrt{s}=13\ \rm TeV$ in final states with jets and two same-sign ...

4

I want to offer a different perspective from the already existing answers, which all seem to somehow refer to the Standard Model or other specific physical theories to say that mass is not an integral multiple of some fundamental mass unit, hence not discretized. The reason why mass is not like that - and can indeed conceivably have continuous values in a ...

2

Electrons and holes occupy their states according to the Fermi-Dirac distribution, which has a single parameter $E_f$, the Fermi level (assume a fixed temperature). Provided $E_f$ is in within the band gap and far from the band edges, the (energy integral of) Fermi-Dirac takes an exponential form $\propto e^{E_f}$ for electrons and $\propto e^{-E_f}$ for ...

2

The color of the photon is related to its frequency $f$, which can be related to the energy of the photon by the expression $E = hf$, where $h$ is Planck's constant. Thus the different colors of the emitted photons describes their different energies. The next step is to determine why specific elements emit certain colors. This has to do with the different ...

1

There are two mistakes. As AccidentalFourierTransform pointed out, the coefficient $7.181\times 10^{-16}$, when converted from MeV to eV, should give $7.181\times 10^{-46}$. Mega means a million, and it to the fifth power gives $10^{30}$, not just $10^{15}$. In this way, the OP has to add a $10^{-15}$ factor to his result. That makes his result $10^{-3}$ ...

1

I consider the scattering process $A+B \to 1 + 2$. The differential cross-section is always given by \begin{split}\label{eq1} d\sigma &= \frac{1}{(2E_A)(2E_B)|v_A - v_B|} \frac{d^3p_1}{(2\pi)^3} \frac{1}{2E_1} \frac{d^3p_2}{(2\pi)^3} \frac{1}{2E_2} \left| {\cal M} \right|^2(2\pi)^4 \delta^4( p_A + p_B - p_1 - p_2) \end{split} ...

Only top voted, non community-wiki answers of a minimum length are eligible