GREPhysics.NET: GR0177 #78

GR0177 #78

Problem

GREPhysics.NET Official Solution

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Advanced Topics $\Rightarrow$ }Particle Physics

A few lines from the Particle Physics chapter of my upcoming book (title tentative, to be publicized on this website eventually), Reviewing Forgotten Physics for the GRE, Prelim's and Qualifier's:

Baryons and mesons are hadrons. (Hadrons are particles that interact with the strong nuclear force.) Baryons have a baryon number of $B=\pm 1$ , while mesons have $B=0$ . Nucleons (which are baryons) have $B=1$ , while antinucleons (antineutrons and antiprotons) have $B=-1$ .

Leptons (electrons, neutrinos, muons, and tau's) interact with the weak nuclear force. There are three kinds of lepton numbers. There is the electron lepton number $L_e$ and there is a muon lepton number, as well as a neutrino lepton number.

Electrons $\beta^{-}$ and the electron neutrino $\nu_e$ have $L_e=1$ , while positrons (antielectrons) $\beta^{+}$ and the electron antineutrino $\bar{\nu}_{e}$ have $L_e=-1$ . All other particles have $L_e=0$

Conservation of baryon number or lepton number is just summing up either the baryon number or the lepton number on both sides of the reaction.

Conservation of the numbers above explain for why a reaction like $\mu \rightarrow \beta^{-} + \nu_\mu + \bar{\nu}_e$ must occur in lieu of $\mu \rightarrow \beta^{-} + \bar{\nu}_e$ or $\mu \rightarrow \beta^{-} + \nu_e$ or $\mu \rightarrow \beta^{-} + \bar{\nu}_{\mu}$ or $\mu \rightarrow \beta^{-} + \nu_{\mu}$ .

For the first reaction, the electron Lepton numbers are $0 \rightarrow 1+0-1$ , and thus $L_e$ is conserved. (Muon Lepton numbers are $1 \rightarrow 0+1+0$ )

For the second reaction, the muon lepton numbers are $1 \rightarrow 0 + 0$ , but that does not add up, and so $L_\mu$ is not conserved.

For the third reaction, the electron Lepton numbers are $0 \rightarrow 1+1$ , but that does not add up, and so $L_e$ is not conserved.

As an exercise, one can calculate the fourth and fifth reaction.

(If you're beginning to feel like this is more alchemy than physics, then note that the Standard Model of particle physics is just a transitory theory, like alchemy was to chemistry. It is to be replaced soon with a less ad-hoc theory, perhaps by you.)

With respect to this question on the GRE exam, one sees that lepton number would not be conserved for any of the possible cases of the latter reactions.

Alternate Solutions

dham
2010-10-07 00:40:24

The important thing to remember is that muons, neutrinos and electrons are leptons.

The first reaction is valid because you start with 1 lepton and end with (1+1-1)=1 lepton, since an antineutrino counts as -1 leptons.

With the second reaction you start with one lepton (the muon) and end with either two or zero total leptons (either 1 neutrino or 1 antineutrino and one electron). This is doesn't work, since lepton number is not conserved. Choice E is then correct.

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dham
2010-10-07 00:40:24

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hanin
2009-10-05 19:17:50

Is there any way to employ MOE for this problem? I could immediately throw out A and D but not sure about the others..

*hey, where can I get that book you're writing? :)

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QM320
2008-11-05 21:32:35

What is the correct answer for this one? Is it (E)?

JMOAN
2008-11-06 21:02:58

Yes, the answer is (E).

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David
2006-12-01 22:39:14

Answer A (conservation of charge) calls for a listing of charges on the particles mentioned.

Your explanation of Baryons could be tied back to the test by stating that the particles involved in the question are not Baryons.

Apostrophes are for denoting possession and omitted letters in contractions, and are not related to pluralization.

sharpstones
2007-04-12 07:10:24

your mom

ayabep
2007-08-30 16:01:14

I needdddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd Reviewing Forgotten Physics for the GRE, Prelim's and Qualifier's

^.^

ayabep
2007-08-30 16:02:40

\(^.^)/

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