Metodick vod ur enie kon tanty s
Download
1 / 15

Metodický úvod (určenie konštanty α s ) - PowerPoint PPT Presentation


  • 79 Views
  • Uploaded on

Metodický úvod (určenie konštanty α s ). Detektor DELPHI. SCHÉMA DETEKTORA VO FYZIKE ČASŤÍC. Identifikácia častíc. Rozpady Z 0 (10 -24 s). Pri E CMS =91 GeV e + e -  Z 0 Z 0  e + e - (3,67%) Z 0  μ + μ - (3,67%) Z 0  τ + τ - (3,67%) Z 0   qq ( 2 jety ) (60%)

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Metodický úvod (určenie konštanty α s )' - virote


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Metodick vod ur enie kon tanty s

Metodický úvod

(určenie konštanty αs)





Rozpady z 0 10 24 s

Rozpady Z0 (10-24s)

Pri ECMS=91 GeV e+e- Z0

Z0 e+e- (3,67%)

Z0 μ+μ- (3,67%)

Z0τ+τ- (3,67%)

Z0qq( 2 jety ) (60%)

Z0qq g( 3 jety ) (24%)

Z04 a viac jetov(6%)


Rozpad na elektr n a pozitr n e e z 0 e e

DELPHI

Rozpad na elektrón a pozitrón: e+e-→ Z0 → e+e-


Rozpad na mi n a antimi n e e z 0

DELPHI

Rozpad na mión a antimión: e+ e- → Z0 → μ+ μ-


e+e-Z0t+t-m+m-

častica tau sa rozpadne skôr,

než ju stihneme zaregistrovať

Tau častica sa rozpadá na:

t– -> nt + e– +net+ -> ne + e+ +nτv 17% prípadov

t– -> nt + m– +nmt+ -> nμ + m+ +nτv 17% prípadov

t– -> nt + hadróny t+ -> nt + hadróny v 60% prípadov

τ+τ– e+e- + 4 neutrína

τ+τ– μ+μ- + 4 neutrína

τ+τ– μ+e- + 4 neutrína

DELPHI

Z0 → τ+τ–


Z 0 e e
Z0→ τ+τ– → e+e-


Z0 → τ+τ–

jeden tau sa rozpadol na elektrón a druhý na hadrón


Rozpad na kvark a antikvark e e z 0 qq

DELPHI

Rozpad na kvark a antikvark: e+ e- → Z0→qq

e+e-Z0q+q-



DELPHI

Rozpad na kvark a antikvark:e+ e-→ Z0 → qq ggg ( s vyžiarením 3 gluónov )


S 0 2 n 3 jets n 2 jets
αs= 0,2.N3-jets / N2-jets



ad