GR 8677927796770177 | # Login | Register

GR8677 #16
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Advanced Topics$\Rightarrow$}Particle Physics

The properties of a muon most closely resembles that of an electron. In the usual $\beta$-decay, an electron is emitted along with either a neutrino or an antineutrino. In a wildcard $\beta$-decay, a muon is emitted instead of an electron.

Alternate Solutions
 There are no Alternate Solutions for this problem. Be the first to post one!
Comments
 ernest212019-08-10 03:09:29 Your your equation is well solved. upcoming fighting gamesReply to this comment joshuaprice1532019-08-08 07:17:05 I am not positive the place you\'re getting your info, but good topic. I needs to spend a while finding out more or working out more. Thanks for excellent information I was looking for this information for my mission. tree pruningReply to this comment chemicalsoul2009-11-01 08:18:31 haha thats a really good use of the word Wildcard ! I like it.Reply to this comment justmax2009-10-19 19:32:29 Muons are leptons: A) electrons are leptons B) & C) graviton and photon are force mediating particles (bosons) D) Pions are quarks (a quark and an antiquark) E) Protons are a trio of quarks. A) fits best.Reply to this comment sonnb2007-05-20 02:08:40 A muon has charge -1 and spin 1/2. An electron has charge -1 and spin 1/2. A graviton has charge 0 and spin 2. A photon has charge 0 and spin 1. A pion has either charge +1 or 0 and spin 0. A proton has charge +1 and spin 1/2. With these criteria, the electron most closely resembles the properties of the electron. From Vol 2 of H/R/K. Reply to this comment comorado2006-10-23 12:39:26 I would say that muon is a partical from Lepton group so are electron, tauon, neutrino. So the answer must be (A).Reply to this comment

Post A Comment!
 Username: Password:
Click here to register.
This comment is best classified as a: (mouseover)
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...