Raws li nthwv dej-particle duality, De Broglie wavelength yog wavelength manifestednyob rau hauv tag nrho cov khoom hauv quantum mechanics uas txiav txim siab qhov yuav tsum tau ceev ntawm kev nrhiav cov khoom ntawm ib qho chaw. ntawm qhov chaw configuration.
Dab tsi yog qhov txawv ntawm wavelength thiab de Broglie wavelength?
Qhov sib txawv ntawm De Broglie wavelength thiab wavelength yog De Broglie wavelength piav qhia txog nthwv dej ntawm cov khoom loj, whereas wavelength piav txog lub zog ntawm nthwv dej. … Yog li ntawd, peb tuaj yeem ntsuas nws raws li qhov kev ncua deb ntawm cov ntsiab lus sib law liag ntawm tib theem ntawm nthwv dej.
Dab tsi yog de Broglie wavelength?
Lub de Broglie wavelength ntawm ib qho particle qhia qhov ntev ntawm cov khoom zoo li nthwv dej yog qhov tseem ceeb rau cov khoom ntawd. De Broglie wavelength feem ntau yog sawv cev los ntawm lub cim λ lossis λdB. Rau ib tug particle nrog momentum p, lub de Broglie wavelength yog txhais raws li: λdB=h/p
Koj pom qhov wavelength ntawm ib de Broglie li cas?
The deBroglie wavelength yog txhais raws li nram no: lambda=h/mv, qhov twg greek tsab ntawv lambda sawv cev rau lub wavelength, h yog Planck tus contant, m yog particle qhov loj thiab v yog nws qhov nrawm.
Dab tsi muaj ib yam ntawm Broglie wavelength?
Yog hais tias ib tug proton thiab electron muaj tib lub de broglie wavelength lawv lub zog yuav sib npaug. Li no lo lus teb raug yog kev xaiv (A) Momentum of electron=momentum of proton.
