1.

A photon and an electron have energy 200 eV each. Which one of these has greater de-Broglie wavelength ?

Answer»

SOLUTION :`lambda=h/p` for PHOTON P `=E/C and 1=(hc)/E` for electron `P=sqrt(2M E)`
`lambda_("photon")=6.2xx 10^(-9)m, lambda_("electron") =0.86 xx10^(-10)m`


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